Thank you for your time in making this video very helpful Just so I understand W/V if I want 1L of 25% sodium chloride solution I would calculate it like this 1000 ml x 25gr / 100ml = X/ 1000ml x 1000ml Answer would be 250gr of sodium chloride for 1000 ml of water
Thank you Ken for watching and for the kind comment. Your approach and solution are correct. You will weigh out 250 grams of sodium chloride and add sufficient quantity of water to make a 1 liter solution. Keep up the good work.
A brass sample was analysed for copper. A 0.7063 g sample of brass was dissolved in nitric acid and diluted to 1 litre. 1 ml of this solution was further diluted to 100 ml. The absorbance of this final diluted solution measured by atomic absorption was 0.443. A 4 ppm Cu standard solution measured under the same conditions gave an absorbance of 0.422. Assuming linearity between absorbance and concentration, calculate the % Cu in the brass sample. (5 marks)
Yes you can VG. You may also share any specific example questions you have so a more precise answer can be provided for those questions. Thank you for watching and for the comment.
can you explain this using percentages in dry raw material such as Calcium Amino Acid Chelate 20% in a 100 gram bag? How do I calculate a dose? Say I want to measure out 2000 mg. Is that 1/4 teaspoon or how can I set up and solve this equations? Thank you for doing this! Much appreciation.
THANKS SO MUCH! BUT I STILL GET STUCKED UP TRYING TO SOLVE A PROBLEM LIKE THIS QUESTION:Commercial Hcl is labelled 37%(wt/wt),its specific density is 1.18g/ml 1)find the molarity of hcl 2)find the mass of solution containing 0.1mol of Hcl 3)find the volume of solution containing 0.1mol of Hcl I will be Very delighted and Grateful if you help me with this problem...Godbless.
Hello, Assume you have 1 L (1000 ml) of solution so you first use the density equation (D=M/V) to find the mass of the solution. 1.18=m/1000 ml so m=1180 g. 1)HCl is only 37% (0.37) of the solution so 1180 g x 0.37 = 436.6 g Hal To get the molarity (mol/L) =mass/molecular weight so 436.6 g HCl/36.45 (MW of HCl) = 12 M HCl/L 2)to get the mass of 0.1 mol of HCl set up a proportion so 12 M/1180 g = 0.1 M/x. x=9.8 g of solution contains 0.1 mol of HCl 3)to get the volume plug 9.8 g back into the density equation to get the volume so 1.18 = 9.8/v v=8.3 ml of solution contains 0.1 mol of HCl
Thank you Aina for watching and for the question. First, we determine the amount of sodium chloride that is needed to prepare 50 tablets as follow: 50 tablets x 2.25 grams/tablet which is equal to 112.5 grams. Next, we use the 0.9% (w/v) concentration together with the 112.5 grams to determine the volume of the 0.9% solution needed. So we have: (0.9 g/100 mL) = (112.5/ Y mL). Solve for Y. So Y = ((100 mL x 112.5 g)/0.9 g) = 12,500 mL. Hope this helps.
How to prepare field (x dose) concentrations reagents by the formulae: C1V1 = C2V2??? chemicals.....For lab work. i am calculating chemical for trichograma chilonis lab experiment
Hello Sir! I do have a homework, I need an excel formula for calculating percentage composition of a compound, wherein you will only input the compound e.g. H2O. then the percentage will come out. Thanks.
The mineral water is fluoridated to 1.8ppm by adding NaF.calculate the amount of NaF you will add per gallon of treated water if the water already contain 0.4ppm flourides?
Hello sir can you solve? When 125g of a solution was evaported to dryness, 42.3g solute was recovered. What was the percentage concentration of the solution by mass?
But when , more than one constituent solute in solvent then find quntitiy of preparation solutions . ex NaOH. =10% HCl. =5% Preparation solution 2400L How calculate .....
Thank you Jacky for watching and for the comment. The 12,500 mL is obtained by setting up a proportion using the definition of 0.9% sodium chloride solution which is 0.9 g/100 mL and the ratio of amount of sodium chloride available (2.25g x 50 = 112.5 g) to volume of solution that can be made. Hence, you have (0.9 g/100 mL)=(112.5 g/X mL). When you solve for X you get 12,500 mL.
Thank you Sourav for watching and for the comment. In question four, the concentration is given on a weight-in-weight basis. Hence, we are interested in the amount of hydrocortisone in grams needed to prepare the suppositories and that is why the answer is in grams. Hope this helps.
plz i have a question in a exercice i hope that you respond because i tried so hard to get the solution the question is how many mililiters are needed to prepare a 1L of a solution of 1 M from 37 % w /w
I think you miscalculated the example somewhere at 9:00. It should be 1,250mL not 12,500 since, xmL = (112.5)(100)/9g = 1,250 in the calculator. Btw thanks for sharing your knowledge.
Thank you Jelly Fish for watching and for the comment. After doing a quick check the solution is correct in the video for the question you mentioned. Also, when you multiply 112.5 by 2.25g and divide by 9g you get 28.125 and not 1,250. Hope this helps. Let me know if we still missed something.
@@David-qh5eo Thanks for the response. However, from the question we have 0.9% normal saline. Hence, x=(112.5*100)/0.9 which is equal to 12500 mL. It looks like you are dividing by 9g instead of 0.9g.
The 500 g is obtained using the desired 2% w/v ketoconazole concentration and the 25 liter antifungal shampoo that we want to prepare. So we set up a proportion using the definition of 2% w/v and the 25 liters. But 25 liters is 25000 milliliters since 1000 milliliters makes 1 liter. Hence, we have (2g/100 mL)=(Xg/25,000 mL). Solve for X. So, X= (2 g x 25,000 mL)/100 mL which is equal to 500 g. Hope this helps. Thank you Margarett for watching and for the comment.
@@PharmaceuticalCalculationsEasy yes I figure it out ,thanks,but the second maths give me a headache because I don't know how u get the answer, n I watch it over n over n still dont
Hi sir I do have a question for u I could not answer it What weight of potassium permanganante is required to produce 300 ml of solution such that 5ml of this solution diluted to 250ml gives a 0.01%w/v solution? could please answer it for me I would be so thankful .
Hi, see the answer at th-cam.com/video/7S6xjsxKVHE/w-d-xo.html . Tell your friends and colleagues about the channel. You can also find more solutions at www.rxcalculations.com
Sir, If I will make a gallon of cough preparation that will 1 grain of active ingredient per teaspoonful. How many grams of active ingredient will I need?
Jas young The following information is pertinent: 1 teaspoonful = 5 mL; 1 gallon = 3.7854 L = 378534 mL and 0.0154 grains = 1 mg.First set up a proportion ((1 grain/1 tsp)=(y/1 gallon)) ---equation 1Now substitute relevant information from above into equation 1 as follows: ((1 grain/5 mL)=(y/3785.4 mL)).Solve for y to give: y = ((1 grain * 3785.4 mL)/5 mL) = 757.08 grains.Convert grains to grams as follows: 0.0154 grains = 1 mg; Hence 757.08 grains = zSolve for z to give:z = ((1 mg * 757.08 grains)/0.0154 grains = 49161 mg = 49.2 g
Pharmaceutical Calculations Again, thank you sir. Now, I know why I can't solve the problem because in my module, all the conversion wasn't right. The way I set the proportion was right but I always end up wrong because the conversion wasn't accurate. Sir, will you post some conversion especially the apothecary. Please!
Thank you Christian for watching and for the comment. Please turn of the closed captions (CC) button on the TH-cam video player to remove the subtitles. Hope this helps.
Please help me for making the dose I want to make 3 doses of bifenthrine pesticide the LD50 of bifenthrine is 70 i want to make three doses according to 1/5,1/10 and 1/20 of LD50 while the pesticide which i purchase from the market contain 10EC bifenthrine
Jas young Set up a proportion(60 grains/2 L) = (y/1 tbsp.) ------ equation 1But 1 L = 1000 ml which implies 2L = 2000 mLAlso, 1 tbsp. = 15 mLHence, substituting the above into equation 1 gives(60 grains/2000 mL) = (y/15 mL)Solving for y gives:y = ( (60 grains * 15 mL)/2000 mL) = 0.45 grainsNow convert grains to milligrams (mg) as follows:0.0154 grains = 1 mg Hence 0.45 grains = zSolve for z to give:z = ((1 mg * 0.45 grains)/0.0154 grains = 29.2 mg (answer).Hope this helps.
Thank you for watching Harishkumar and for the comment. Is there a specific question that wasn't clear? We will be happy to provide additional explanation.
Sir kindly tell how can we calculate concentration of lol liquid compound in mg Eg. How to calculate concentration of linalool in mg/ml i.e. how many miligrams are present in 1 ml (as linalool is in liquid form )
@@PharmaceuticalCalculationsEasy density is 0.87 g/ml so kindly tell how many miligrams are in 1 ml of linalool Tell how to calculate it Give ur answer in comment, solve it so that I can calculate concentration of any liquid compound in mg
Pharmaceutical Calculations Hello sir, I bought your book "Pharmaceutical Calculations 1001 questions with answers." So far I have 2 questions; Chapter 4, Ways of Expressing Concentration, questions 63 and 72. Am not sure how you arrived at that answer. Can you please help explain how you solve it? Thank you.
Am sorry I got the answer 12500 voz I was thinking 1125 is less than 12500 I get by dividing with .9 the value will come less but it goes up as it decreases down than when I calculated I got it correct thank sir
Hi buddy, Could you help me preparing 1000 ppm/L of ammonium from NH4Cl the M.W. is 53.49 Also preparing 1000 ppm/L of NO3 from KNO3, the M.W. is 101.10, please? thanks
Hello Niema, the questioned asked for concentration in percent volume-in-volume. We have to "convert" the 225 g mass quantity given into its volume equivalent. Hence, we divided the 225 by 0.8 (its density) to obtain the 281.25 mL. Dividing the 281.25 by the total quantity of 1500 ml and scaling it to 100 mL gives the %v/v. Hope this helps. Let me know if you have additional questions.
((65 g/165 g)=(y/100 g)) => y = ((65 g/165 g)*100 g) = 39.4% w/w Also, because the density of water is 1 g/ml it implies that 100 g of water has a volume of 100 ml. So, ((65 g/100 ml)=(z/100 ml)) => z = ((65 g/100 ml)*100 ml) = 65% w/v
@@PharmaceuticalCalculationsEasy Oh , thank ......but my doctor give me the question without any another information ... However , please give me your account in telegram or instagram
Thank you MJolnir for watching and for the comment. You are completely right about the audio in the video. Audio quality of videos have improved significantly since those early videos. Here is one of the latest videos for your listening pleasure th-cam.com/video/ErtipYKWz0M/w-d-xo.html. Thank you for supporting the channel.
The 500 g is obtained using the desired 2% w/v ketoconazole concentration and the 25 liters of antifungal shampoo that we want to prepare. So we set up a proportion using the definition of 2% w/v and the 25 liters. But 25 liters is 25,000 milliliters since 1000 milliliters makes 1 liter. Hence, we have (2g/100 mL)=(Xg/25,000 mL). Solve for X. So, X= (2 g x 25,000 mL)/100 mL which is equal to 500 g. Hope this helps. Thank you Margarett for watching and for the comment.
0.9% (w/v) solution of NaCl is equal to 0.9 grams of NaCl in 100 mL solution. Also, the amount of NaCl in the 50 tablets is 2.25 g x 50 which is 112.5 g. So to determine the volume of 0.9% NaCl solution you can make from the 112.5 g, we set up a proportion and solve for the unknown which gives the 12500 mL (see 9:42 in the video). Hope this helps.
Hi sir I do have a question for u I could not answer it What weight of potassium permanganante is required to produce 300 ml of solution such that 5ml of this solution diluted to 250ml gives a 0.01%w/v solution? could please answer it for me I would be so thankful .
Hi, see the answer at th-cam.com/video/7S6xjsxKVHE/w-d-xo.html . Tell your friends and colleagues about the channel. You can also find more solutions at www.rxcalculations.com
Thank you for showing me step by step in your teaching how to do percentage concentration calculations. I understand it better and now I can do them.
You are welcome Janay. Thank you for watching and for the comment.
This type of Learning is so easy.. sir you've explained it very well Now I'm able to solve all percent problems ..thnkyou sir so much
Thank you Pradnya for watching and for the comment. Welcome to the Pharmaceutical Calculations community.
Thank you for your time in making this video very helpful
Just so I understand W/V if I want 1L of 25% sodium chloride solution I would calculate it like this
1000 ml x 25gr / 100ml = X/ 1000ml x 1000ml
Answer would be 250gr of sodium chloride for 1000 ml of water
Thank you Ken for watching and for the kind comment. Your approach and solution are correct. You will weigh out 250 grams of sodium chloride and add sufficient quantity of water to make a 1 liter solution. Keep up the good work.
@@PharmaceuticalCalculationsEasy thank you very much for your help
This is very straight forward and helpful. Thank you.
What a sweet intro song considering how stressful studying can be ❤️
Thanks for your kind words and for watching.
Thank you sir, God bless!
Each & everyone can easy to understand ur teaching
Thank you for your kind words Deva slr and thanks for watching.
I did not understand why you're in the exact 8:44 to hit his sodium at 50 ??Thank you for this wonderful explanation
My Ghanaian uncle 😍
I KNOW RIGHT
Finally i got best chanel for my SPLE preparations
Thank you Atiq for watching and for the kind comment. Welcome to the Pharmaceutical Calculations community.
Oh thank you ❤️
Thank you sir for the clearl explanation of the topic., appreciate it
Glad you found the video helpful Hakimin. Thank you for watching and for the comment.
Hello sir you are best 🤠🤠
My name is milli
شكرا جزيلا
Thanks a great explanation❤️
Thank you for watching and for the comment.
Thank you sir! Very easy to understand!
Glad you found the video helpful. Thank you for the comment.
A brass sample was analysed for copper. A 0.7063 g sample of brass was dissolved in nitric acid and diluted to 1 litre. 1 ml of this solution was further diluted to 100 ml. The absorbance of this final diluted solution measured by atomic absorption was 0.443. A 4 ppm Cu standard solution measured under the same conditions gave an absorbance of 0.422. Assuming linearity between absorbance and concentration, calculate the % Cu in the brass sample. (5 marks)
Gb3
@@ofeiconstantine1974? Say what?
very nice concept
Very informative video, can i also use this formula of w/v for concentrates for dairy products?
Ur advise shall be a great help.
Yes you can VG. You may also share any specific example questions you have so a more precise answer can be provided for those questions. Thank you for watching and for the comment.
thank you so much sir. you explain it so well.
Thank you Fatoumata for watching and for the comment. Welcome to the pharmaceutical calculations community.
can you explain this using percentages in dry raw material such as Calcium Amino Acid Chelate 20% in a 100 gram bag? How do I calculate a dose? Say I want to measure out 2000 mg. Is that 1/4 teaspoon or how can I set up and solve this equations? Thank you for doing this! Much appreciation.
this is so helpful thank you very much
Glad you found the video helpful SE. Thank you for watching and for the comment.
THANKS SO MUCH! BUT I STILL GET STUCKED UP TRYING TO SOLVE A PROBLEM LIKE THIS
QUESTION:Commercial Hcl is labelled 37%(wt/wt),its specific density is 1.18g/ml
1)find the molarity of hcl
2)find the mass of solution containing 0.1mol of Hcl
3)find the volume of solution containing 0.1mol of Hcl
I will be Very delighted and Grateful if you help me with this problem...Godbless.
Hello,
Assume you have 1 L (1000 ml) of solution so you first use the density equation (D=M/V) to find the mass of the solution. 1.18=m/1000 ml so m=1180 g.
1)HCl is only 37% (0.37) of the solution so 1180 g x 0.37 = 436.6 g Hal
To get the molarity (mol/L) =mass/molecular weight so 436.6 g HCl/36.45 (MW of HCl) = 12 M HCl/L
2)to get the mass of 0.1 mol of HCl set up a proportion so 12 M/1180 g = 0.1 M/x. x=9.8 g of solution contains 0.1 mol of HCl
3)to get the volume plug 9.8 g back into the density equation to get the volume so 1.18 = 9.8/v v=8.3 ml of solution contains 0.1 mol of HCl
oh am indeed grateful...Godbless you...i'd use this method to solve other problems...thanks again😃
The song for your intro was interesting
Glad you enjoyed it Ivonne. Thank you for watching and for the comment.
Example batch size 400 letters
Hcl standard quantity is 121.5
How to desided the standard qty of dispensing in rawmaterile
thanks from 🇮🇶 iraq it was very helpful
th-cam.com/video/XSAen2o-TLE/w-d-xo.html
Thank u! You saved my life
Thank youuuuuuuuuu sooo muuuuchhh!!
Thank you @allysonguiriba58 for watching and for the comment. Welcome to the Pharmaceutical Calculations community.
Thank you sensei
Thank you, @Reechaan for watching and for the comment. Welcome to the Pharmaceutical Calculations community.
love you . thanks
Thank you Maria for watching and for the kind comment.
sir i have a question, in the second question of w/v how did you calculate to get the answer of 12500mL
Thank you Aina for watching and for the question. First, we determine the amount of sodium chloride that is needed to prepare 50 tablets as follow: 50 tablets x 2.25 grams/tablet which is equal to 112.5 grams.
Next, we use the 0.9% (w/v) concentration together with the 112.5 grams to determine the volume of the 0.9% solution needed. So we have: (0.9 g/100 mL) = (112.5/ Y mL). Solve for Y. So Y = ((100 mL x 112.5 g)/0.9 g) = 12,500 mL. Hope this helps.
I thought it would be like 112.5 × 0.9 then divided by 100 to make it 1.0125 ml
@@jomarieangelcorpuz7848 right ! me too!!
@@PharmaceuticalCalculationsEasy Thank you!
@@PharmaceuticalCalculationsEasy Hello Sir. Did you use the cross method for this one?
You are great
Thank you Mohammed for watching and for the comment.
How to prepare field (x dose) concentrations reagents by the formulae: C1V1 = C2V2??? chemicals.....For lab work. i am calculating chemical for trichograma chilonis lab experiment
It's really fun. Thanks
Hello Sir! I do have a homework, I need an excel formula for calculating percentage composition of a compound, wherein you will only input the compound e.g. H2O. then the percentage will come out. Thanks.
thank you
The mineral water is fluoridated to 1.8ppm by adding NaF.calculate the amount of NaF you will add per gallon of treated water if the water already contain 0.4ppm flourides?
But concentration is in % should we write by 100 while calculating solute?
If I understand your question correctly, then yes. Thank you for watching and for the comment.
thanks
Thank you so much sir!
Thank you Mika for watching and for the comment.
thank you thank you
Thank you @MUBARIKMUHIDIN for watching and for the comment.
Hello sir can you solve? When 125g of a solution was evaported to dryness, 42.3g solute was recovered. What was the percentage concentration of the solution by mass?
Thank you est. for watching and for the comment. See the solution to your question here: bit.ly/pcwbw .
@@PharmaceuticalCalculationsEasy omo thank you so much, I so thought my answer was wrong, turns out it wasn't! ♥️
Thank you very much
But when , more than one constituent solute in solvent then find quntitiy of preparation solutions . ex
NaOH. =10%
HCl. =5%
Preparation solution 2400L
How calculate .....
th-cam.com/video/XSAen2o-TLE/w-d-xo.html
thank you it's clear but how you got the answer 12,500 mL?
Thank you Jacky for watching and for the comment. The 12,500 mL is obtained by setting up a proportion using the definition of 0.9% sodium chloride solution which is 0.9 g/100 mL and the ratio of amount of sodium chloride available (2.25g x 50 = 112.5 g) to volume of solution that can be made. Hence, you have (0.9 g/100 mL)=(112.5 g/X mL). When you solve for X you get 12,500 mL.
Thank You
Thank you🙏
thank you for
Thank you @mohamedhussein9747 for watching and for the comment.
Thanks 👍
Thank you Favour for watching and for the comment.
I think In fourth questions solution the unit of volume will be in ml not in gram.
Thank you Sourav for watching and for the comment. In question four, the concentration is given on a weight-in-weight basis. Hence, we are interested in the amount of hydrocortisone in grams needed to prepare the suppositories and that is why the answer is in grams. Hope this helps.
Thank you very much but how can i calculate the percentage of other solute in the preparation
Thank you. this is an awesome piece of information
plz i have a question in a exercice i hope that you respond because i tried so hard to get the solution the question is how many mililiters are needed to prepare a 1L of a solution of 1 M from 37 % w /w
Thank you!!!!
You're welcome Meghan. Thank you for watching and for the comment.
Nice
Thanks for watching Vikrant and for the comment too.
How to prepare field (x dose) concentrations reagents by the formulae: C1V1 = C2V2 ?????
kindly dear Sir
I think you miscalculated the example somewhere at 9:00. It should be 1,250mL not 12,500 since, xmL = (112.5)(100)/9g = 1,250 in the calculator.
Btw thanks for sharing your knowledge.
Thank you Jelly Fish for watching and for the comment. After doing a quick check the solution is correct in the video for the question you mentioned. Also, when you multiply 112.5 by 2.25g and divide by 9g you get 28.125 and not 1,250. Hope this helps. Let me know if we still missed something.
@@PharmaceuticalCalculationsEasy sorry i mistyped my concern.
@@PharmaceuticalCalculationsEasy what i mean is xgrams = (112.5)(100)/9 = 1250 tabs. Sorry I mistyped my concern.
Btw thanks for replying
@@David-qh5eo Thanks for the response. However, from the question we have 0.9% normal saline. Hence, x=(112.5*100)/0.9 which is equal to 12500 mL. It looks like you are dividing by 9g instead of 0.9g.
@@PharmaceuticalCalculationsEasy auh ok thanks for clearing my confusion.
How many grams of NaHCO3 are needed to be added to 150 mL of 0.4 v/v solution
It appears some information is missing from the question. What is the concentration of the final product? How much are you making?
th-cam.com/video/XSAen2o-TLE/w-d-xo.html
Sir, thank you for the reply. unfortunately it didn't show all the message.
Thank you you are the bomb - thank you very much indeed
Sir, I have a question again. A drug is available as 5 gr/ fluid dram, and the order is for 20 ml. How many milligrams will that be?
th-cam.com/video/XSAen2o-TLE/w-d-xo.html
i cant understand the how u get 500 in the first maths
The 500 g is obtained using the desired 2% w/v ketoconazole concentration and the 25 liter antifungal shampoo that we want to prepare. So we set up a proportion using the definition of 2% w/v and the 25 liters. But 25 liters is 25000 milliliters since 1000 milliliters makes 1 liter. Hence, we have (2g/100 mL)=(Xg/25,000 mL). Solve for X. So, X= (2 g x 25,000 mL)/100 mL which is equal to 500 g. Hope this helps. Thank you Margarett for watching and for the comment.
@@PharmaceuticalCalculationsEasy yes I figure it out ,thanks,but the second maths give me a headache because I don't know how u get the answer, n I watch it over n over n still dont
let me please know that kotoconazole of puritity is 100% ?? If it is eastime 95% puritity , how can I calculate w/v for 2% concentration ?
how would u know the solute and preparation
Hi sir
I do have a question for u I could not answer it
What weight of potassium permanganante is required to produce 300 ml of solution such that 5ml of this solution diluted to 250ml gives a 0.01%w/v solution?
could please answer it for me I would be so thankful .
Hi, see the answer at th-cam.com/video/7S6xjsxKVHE/w-d-xo.html . Tell your friends and colleagues about the channel. You can also find more solutions at www.rxcalculations.com
Thank youuuu
thank u so much!
You're welcome! Glad you liked the video. Thank you for watching and for the comment.
What is the final concentration if 200 mL of a 1 in 40 solution is diluted to 550 mL?
Thank you for watching and for the comment. The answer is 1:110. See the solution here: jo.my/3xdkoe
Sir,
If I will make a gallon of cough preparation that will 1 grain of active ingredient per teaspoonful. How many grams of active ingredient will I need?
Jas young The following information is pertinent: 1 teaspoonful = 5 mL; 1 gallon = 3.7854 L = 378534 mL and 0.0154 grains = 1 mg.First set up a proportion ((1 grain/1 tsp)=(y/1 gallon)) ---equation 1Now substitute relevant information from above into equation 1 as follows: ((1 grain/5 mL)=(y/3785.4 mL)).Solve for y to give: y = ((1 grain * 3785.4 mL)/5 mL) = 757.08 grains.Convert grains to grams as follows: 0.0154 grains = 1 mg; Hence 757.08 grains = zSolve for z to give:z = ((1 mg * 757.08 grains)/0.0154 grains = 49161 mg = 49.2 g
Pharmaceutical Calculations Again, thank you sir. Now, I know why I can't solve the problem because in my module, all the conversion wasn't right. The way I set the proportion was right but I always end up wrong because the conversion wasn't accurate. Sir, will you post some conversion especially the apothecary. Please!
Better to remove subtitle can't see the formula
Thank you Christian for watching and for the comment. Please turn of the closed captions (CC) button on the TH-cam video player to remove the subtitles. Hope this helps.
thanks for your video. please how do I calculate how mang grams of 10mM Nacl needed to make 15ml solution
th-cam.com/video/XSAen2o-TLE/w-d-xo.html
please convert the 103mg%w/v into miligram per 100 mililitre
Please help me for making the dose
I want to make 3 doses of bifenthrine pesticide the LD50 of bifenthrine is 70 i want to make three doses according to 1/5,1/10 and 1/20 of LD50 while the pesticide which i purchase from the market contain 10EC bifenthrine
2 L of solution with 60 grains of active ingredient. If a dose is tbsp., how many milligrams are in a dose? please help
Jas young Set up a proportion(60 grains/2 L) = (y/1 tbsp.) ------ equation 1But 1 L = 1000 ml which implies 2L = 2000 mLAlso, 1 tbsp. = 15 mLHence, substituting the above into equation 1 gives(60 grains/2000 mL) = (y/15 mL)Solving for y gives:y = ( (60 grains * 15 mL)/2000 mL) = 0.45 grainsNow convert grains to milligrams (mg) as follows:0.0154 grains = 1 mg Hence 0.45 grains = zSolve for z to give:z = ((1 mg * 0.45 grains)/0.0154 grains = 29.2 mg (answer).Hope this helps.
Pharmaceutical Calculations Thank you. I was confused and I don't know the conversion of grains to mg. How do you get 0.0154 grains?
No connection between steps.... I can't understand...
Thank you for watching Harishkumar and for the comment. Is there a specific question that wasn't clear? We will be happy to provide additional explanation.
Sir kindly tell how can we calculate concentration of lol liquid compound in mg
Eg. How to calculate concentration of linalool in mg/ml
i.e. how many miligrams are present in 1 ml (as linalool is in liquid form )
Thank you S Dar for watching and for the comment. You will need to know the density of the liquid. Hope this helps.
@@PharmaceuticalCalculationsEasy density is 0.87 g/ml so kindly tell how many miligrams are in 1 ml of linalool
Tell how to calculate it
Give ur answer in comment, solve it so that I can calculate concentration of any liquid compound in mg
@@Darjasa density = m/v
@@Darjasa substitute then convert to mL
how much is 80% of 40 is greater then 4/5 of 25.
How can solve sir
Thank you Prince for the comment. Please clarify the question as it is a little unclear.
Hhcl % + 2%ppt sulpur + ketacanozole to scale be * 2 week how to make that mixture
how to calculate H20?
I don’t understand on how you get the answer you simply put the answer by not explaining on what to multiply or divide
Thank you Affewang for watching and for the comment. Please specify which question you are referring to so your question can be better addressed.
how do you get x=12,500ml for the second equation? ive got the correct order but not sure on the calculation to get 12,500 :(
You solve for x by cross multiplying and dividing as follows: x= (100 mL x 112.5 g)/0.9 g = 12,500 mL. I hope this helps
Pharmaceutical Calculations Hello sir, I bought your book "Pharmaceutical Calculations 1001 questions with answers." So far I have 2 questions; Chapter 4, Ways of Expressing Concentration, questions 63 and 72. Am not sure how you arrived at that answer. Can you please help explain how you solve it? Thank you.
Am sorry I got the answer 12500 voz I was thinking 1125 is less than 12500 I get by dividing with .9 the value will come less but it goes up as it decreases down than when I calculated I got it correct thank sir
Thank you
You are welcome Natalia. Thank you for watching and for the comment.
Hi buddy,
Could you help me preparing 1000 ppm/L of ammonium from NH4Cl the M.W. is 53.49
Also preparing 1000 ppm/L of NO3 from KNO3, the M.W. is 101.10, please?
thanks
th-cam.com/video/XSAen2o-TLE/w-d-xo.html
Plzz tell me agar hamra pas solvent ha aur solute be hama concentarion of solution find karna ha what formula apply plzz ayone whose help me
Hello, how did you get 281.25? I'm a little lost on how you got that answer.
Hello Niema, the questioned asked for concentration in percent volume-in-volume. We have to "convert" the 225 g mass quantity given into its volume equivalent. Hence, we divided the 225 by 0.8 (its density) to obtain the 281.25 mL. Dividing the 281.25 by the total quantity of 1500 ml and scaling it to 100 mL gives the %v/v. Hope this helps. Let me know if you have additional questions.
ok, I understand now but the answer came up to 0.1875 but I'm assuming you move the decimal point 2 places over. Thank You.
Yes, that is correct. The decimal point moves two places to the right since you multiply 0.1875 by 100 mL.
ok now I understand better
Shift each side but I didn't understand the 12500 ml of the second question that with which unknown he made it
65 g of sugar in 100g of water what is the concentration's percentage?
Please do help me
((65 g/165 g)=(y/100 g))
=> y = ((65 g/165 g)*100 g) = 39.4% w/w
Also, because the density of water is 1 g/ml it implies that 100 g of water has a volume of 100 ml.
So, ((65 g/100 ml)=(z/100 ml))
=> z = ((65 g/100 ml)*100 ml) = 65% w/v
If its okay how about 25g of alcohol in 150g of water?
((25 g/175 g)=(y/100 g))
=> y = ((25 g/175 g)*100 g) = 14.3% w/w
See updated solution to your earlier question.
How calculate the strength of NaOH in g/l ????
Please help me now
In its current form there is not enough information to proceed. So provide more information about the NaOH (e.g., percentage strength, volume).
@@PharmaceuticalCalculationsEasy
Oh , thank ......but my doctor give me the question without any another information ...
However , please give me your account in telegram or instagram
u sounds like Barry Kripke
5ppm en poids comment converti au ppm en volume
whats up with that Mic, about broke my eardrums. Good stuff otherwise
Thank you MJolnir for watching and for the comment. You are completely right about the audio in the video. Audio quality of videos have improved significantly since those early videos. Here is one of the latest videos for your listening pleasure th-cam.com/video/ErtipYKWz0M/w-d-xo.html. Thank you for supporting the channel.
cant understand how u get 500 in the first part
The 500 g is obtained using the desired 2% w/v ketoconazole concentration and the 25 liters of antifungal shampoo that we want to prepare. So we set up a proportion using the definition of 2% w/v and the 25 liters. But 25 liters is 25,000 milliliters since 1000 milliliters makes 1 liter. Hence, we have (2g/100 mL)=(Xg/25,000 mL). Solve for X. So, X= (2 g x 25,000 mL)/100 mL which is equal to 500 g. Hope this helps. Thank you Margarett for watching and for the comment.
you do
🙄
Thank you for watching and for the feedback.
How you got 12,500
0.9% (w/v) solution of NaCl is equal to 0.9 grams of NaCl in 100 mL solution. Also, the amount of NaCl in the 50 tablets is 2.25 g x 50 which is 112.5 g. So to determine the volume of 0.9% NaCl solution you can make from the 112.5 g, we set up a proportion and solve for the unknown which gives the 12500 mL (see 9:42 in the video). Hope this helps.
What's the unknown
Pls how can I contact you directly
Thank you Clair for question. Please send all communication to info@rxcalculations.com
Ok ✅
Pls check your email . Thank you
i like the accent
in hindi
timepass
اشرح ب العربي
ENGLISH please. "Consequence"??????
Very confusing,you r not using conc formula, these short cut are confusing
All of this is pharmacist math... and math tutors can't do this unless is Chem or nursing tutor 😞💔
Hi sir
I do have a question for u I could not answer it
What weight of potassium permanganante is required to produce 300 ml of solution such that 5ml of this solution diluted to 250ml gives a 0.01%w/v solution?
could please answer it for me I would be so thankful .
Hi, see the answer at th-cam.com/video/7S6xjsxKVHE/w-d-xo.html . Tell your friends and colleagues about the channel. You can also find more solutions at www.rxcalculations.com