Mass percentage ,Molarity,Mole fraction and Molality Class XI Chemistry |General Basics of Chemistry
ฝัง
- เผยแพร่เมื่อ 6 ต.ค. 2024
- In this lecture we are going to cover the following topics :-
1.Molarity
2.Molality
3.Mass percentage
4.Mole fraction
Mole concept |Class XI Chemistry|
• Mole concept |Some bas...
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Worksheet 1.1 (Homework)
1. What is the mass percent of carbon in carbon dioxide?
(i) 0.034%
(ii) 27.27%
(iii) 3.4%
(iv) 28.7%
Mass percent of carbon = Mass of Solute/ Mass of solution×100
12/44×100= 300/11= 27.27%Ans.
Ans) 27.27% Ans.
2. If the concentration of glucose (C6H12O6) in blood is 0.9 g L-1 what will be the molarity of glucose in
blood?
(i) 5 M
(ii) 50 M
(iii) 0.005 M
(iv) 0.5 M
Mass of glucose= C6H12O6= 0.9g
Moller mass of Glucose= 180g
12×6+1×12+16×6= 180 g
Molarity= Number of moles of solution/Volume of solution in litre
0.9/180= 0.005MAnd.
Ans) 0.005 Ans.
3. What will be the molality of the solution containing 18.25 g of HCl gas in 500 g of water?
(i) 0.1 m
(ii) 1 M
(iii) 0.5 m
(iv) 1 m
Ans) Mass of HCL= 18.25g
Moller mass of Hcl= 36.5g
1×1+35.5= 36.5g
n= m/M=18.25/36.5= 5/10= 0.5 Moles.
Mass of water in kg= 500/1000= 0.5kg
Morality = number of moles / Mass of Solvent in Kilogram
= 0.5/0.5= 1m Ans
Ans) 1m
4. If 500 mL of a 5M solution is diluted to 1500 mL, what will be the molarity of the solution obtained?
(i) 1.5 M
(ii) 1.66 M
(iii) 0.017 M
(iv) 1.59 M
Ans) Molarity is= 5M
Volume of solutions are = 500 ml ans 1500ml
By using formula of molarity= M¹V¹=M²V²
5×500=M²(1500)
2500= M²×1500
M²= 2500/1500= 5/3= 1.66 M² Ans
Ans) 1.66.
5. What will be the molarity of a solution, which contains 5.85 g of NaCl(s) per 500 mL?
(i) 4 mol L-1
(ii) 20 mol L-1
(iii) 0.2 mol L-1
(iv) 2 mol L-1
Ans)Given mass of NaCl= 5.85g
Mollar mass of NaCl= 58.5g
23×1+35.5= 58.5
n= m/M= 5.85/58.5=1/10= 0.1 Moles
Volume= 500 ml= 500/1000= 0.5 L
Molarity= Number of moles of solution/Volume of solution in litre
0.1/0.5=1/5= 0.2 ML Ans.
Ans) 0.2 ML.
6.A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the mass per cent of
the solute.
Ans) Mass percent of solute= Mass of solute/Mass of Solution×100
2/20×100= 10%.
Q7) Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form
250 mL of the solution.
Ans) Mas of NaOH= 4 g
Moller mass = 40 g
n= m/M= 4/40= 0.1 Mole
So, Volume = 250/1000
Molarity= Number of moles of solution/Volume of solution in litre
Molarity = 0.1/0.25= 2/5= 0.4 Mole per litre.
Q8) What is the concentration of sugar (C12H22O11) in molL
-1 if its 20 g are dissolved in enough water to
make a final volume up to 2L?
Ans) Mass of sugar = 20 g
Moller mass = 342
n=m/M= 20/342= 0.058 Moles
So, Molarity =Molarity= Number of moles of solution/Volume of solution in litre
0.058/2 = 0.029 Moles.
Q9) . Calculate the molality of a solution where 0.5 grams of toluene (C7H8) is dissolved in 225 grams
of Benzene(C6H6).Calculate the moles of given solute.
Ans) Banzene in kilograms 225= 0.225 kg
Moles of Toluene =n=m/M
Moles of Toluene=0.5/92= 1/184=0.005 Moles
Molarity= Number of moles of solution/Volume of solution in litre
Mollality= 0.005/0.225= 0.022 Ans.
10). What is the unit of molarity and molality?
Ans) unit of mollarity us Mole per litre(M/L) and unit of mollality is Mole per kilogram (M/kg).
11.If 4 g of NaOH dissolves in 36 g of H2O, calculate the mole fraction of each component in the solution.
Ans) Mass of NaOH= 4 g
Moller mass = 40 g
n= m/M= 4/40= 0.1 Mole
Mass of H²O = 36g Molecular Mass Of H²O = 18 g
1×2+16= 18g
Number of moles present in H²O = 36/18= 2 Moles
Mole fraction of NaoH= x¹= n¹/n¹+n²
So, Mole fraction of NaoH= Number of moles of NaoH/ number of moles of NaoH+ number of moles of water.
= 0.1/2+0.1= 0.1/2.1= 1/21= 0.047 Moles
Moles Fraction of water = number of moles of water/Moles of NaoH+ moles of H2o
= 2/0.1+2= 2/2.0.095 Moles
So, mole fraction of NaoH= 0.047 Moles and mole fraction of water is 0.095 Moles.
Q12)Calculate the mass percent of calcium, phosphorus and oxygen in calcium phosphate Ca3(po4)².
Ans) 1) Mass percent of calcium= Mass of solute/Mass of Solution×100
120/310×100= 1200/31 38.79%.
2) Mass percent of phosphorus= Mass of solute/Mass of Solution×100
= 62/310×100= 620/31=20%.
3) Mass percent of Oxygen= Mass of solute/Mass of Solution×100
= 128/310×100= 1280/31= 40.12% Ans.
Q13)Calculate the concentration of nitric acid in moles per liter in a sample which has a density, 1.41 g mL-1
and the mass per cent of nitric acid in it being 69%.
Ans) Mass of HNo³= 69%
So, Density= Mass/Volume
D= m/v
V= m/D= 100g/1.41gml= 70.92 ML Ans.
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Q) In H20 To find the mass percentage of Hydrogen and Water
Sol :-
Mass percentage = Mass / Total Mass
Hydrogen = 2×100/18
Hydrogen = 11.11 %
So oxygen percentage = 88.89 %
Q) How to dilute acid
Ans ) If you Dilution of acid We add acid drop in water and shake the glass we are not add water in acid otherwise Heat generate because it is very exothermic and blash your glass .
Molarity = 0.1/0.25
= 0.4 mol/liter
MOLARITY 0.04 mol/liter DEAR
In fact hydrogen and oxygen in water ar present in 1:8 ratio by mass
Ques - 2 What is the mass percentage of oxygen in H2O ?
Sol- mass percent = mass of salute / mass of solution × 100
Mass percentage of oxygen = 16 / 18 × 100
Mass percentage of oxygen = 88.88%
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Q2) Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form
250 mL of the solution.
Ans) Mas of NaOH= 4 g
Moller mass = 40 g
n= m/M= 4/40= 0.1 Mole
So, Volume = 250/1000
Molarity= Number of moles of solution/Volume of solution in litre
Molarity = 0.1/0.25= 2/5= 0.4 Mole per litre.
Thanks 👍
@Sam Saizan Me too 😊
Same answer
Yes we get answers 0.4
16:40 Molarity will be 0.029 mol /L
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Ques. What is mass % of H and O in H2O ?
Ans.Mass %= Mass of Solute /Mass of Solution×100%
(1) Mass % of H = 2/18 × 100% = 11.11%
(2) Mass % of O = 16/18 × 100% = 88.88%
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Q3) What is the concentration of sugar (C12H22O11) in molL
-1
if its 20 g are dissolved in enough water to
make a final volume up to 2L?
Ans) Mass of sugar = 20 g
Moller mass = 342
n=m/M= 20/342= 0.058 Moles
So, Molarity =Molarity= Number of moles of solution/Volume of solution in litre
0.058/2 = 0.029 Moles.
Thank you so much ❤️👍
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0.029 M/L
Well done 😊😊😊😊
H2=11.11 percent
O=88.88 percent
24:11 ans = 0.0241
11:38 answer -- acid should be added in water .
H2O :-. 11.11 g. And. O2:- 88.88g
Q4) . Calculate the molality of a solution where 0.5 grams of toluene (C7H8) is dissolved in 225 grams
of Benzene(C6H6).Calculate the moles of given solute.
Ans) Banzene in kilograms 225= 0.225 kg
Moles of Toluene =n=m/M
Moles of Toluene=0.5/92= 1/184=0.005 Moles
Molarity= Number of moles of solution/Volume of solution in litre
Mollality= 0.005/0.225= 0.022 Ans.
Aapne molarity bata diya molality batana hai dhyaan se padhiye question ko
Sahi me wrong m hai😅😅😅
Molality = No. Of mole of solute /mass of solvent (kg).
Thank you sir for making this video o❤️❤️
Thank you sirji🥰🥰
Acid is added in water because in the dilution process a large amount of heat is produced which is absorbed by water.
11:27 While dilution acid should be added to water
15:54 - 0.4
16:42 - 0.029
Hi
Thnx sir
15:58 Answer 0.4 mol/L
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Q1) What is mass percent of Hydrogen in water(H²O)?
Ans) Mass percent=Mass of solute /Mass of solution×10.
Mass percent of Hydrogen=2/18×100=11.11% Ans.
Q2) What is mass percent of Oxygen in water(H²O)?
Ans) Mass percent=Mass of solute /Mass of solution×10.
Mass percent of Oxygen=16/18×100=88.88% Ans.
When we dilute acid then acid is added to water.
Q2) Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form
250 mL of the solution.
Ans) Mas of NaOH= 4 g
Moller mass = 40 g
n= m/M= 4/40= 0.1 Mole
So, Volume = 250/1000
Molarity= Number of moles of solution/Volume of solution in litre
Molarity = 0.1/0.25= 2/5= 0.4 Mole per litre.
Q3) What is the concentration of sugar (C12H22O11) in molL
-1
if its 20 g are dissolved in enough water to
make a final volume up to 2L?
Ans) Mass of sugar = 20 g
Moller mass = 342
n=m/M= 20/342= 0.058 Moles
So, Molarity =Molarity= Number of moles of solution/Volume of solution in litre
0.058/2 = 0.029 Moles.
Q4) . Calculate the molality of a solution where 0.5 grams of toluene (C7H8) is dissolved in 225 grams
of Benzene(C6H6).Calculate the moles of given solute.
Ans) Banzene in kilograms 225= 0.225 kg
Moles of Toluene =n=m/M
Moles of Toluene=0.5/92= 1/184=0.005 Moles
Molarity= Number of moles of solution/Volume of solution in litre
Mollality= 0.005/0.225= 0.022 Ans.
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Molarity of NaoH====0.4 mole/litter
Thankyou so much sir...
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sir mera doubt ye tha ki mass of solution = solute + solvent to apne q no 2 me kiya par 1 me nahi kiya
17:12 0.029 mol/l
0.00295*
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7:11 mass percent of hydrogen = MASS OF SOLUTE \ MASS OF SOLUTION
2\18* 100 = 11.1%
MASS PERCENT OF OXYGEN =MASS OF SOLUTE \MASS OF SOLUTION
16\18*100 =88.8%
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हरे कृष्णा सर ❤️
Mass Percentage of H = 11.11u
Mass Percentage of O = 88.88u
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16:23 10 ans
Respect your answer but my answer is "0.4M""...
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Q1) What is mass percent of Hydrogen in water(H²O)?
Ans) Mass percent=Mass of solute /Mass of solution×10.
Mass percent of Hydrogen=2/18×100=11.11% Ans.
F
sir you did support us very well in 2023-24 batch, i scored 96 prcntage because of u only.. all thanks to u..but please help us with sci in 11th 12th..
When we dilute acid then acid is added to water.
Class 11 ... 2023-24 bale attendance lagayen ------>
But I m in class 10
Main
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Acid into water sir
Mass percentage of hydrogen in H2O is 11.1% .
Mass percentage of oxygen in H2O is 88.8%.
Sir please class 11 physics also
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11 :33 we add acid 2 water by constant stirring if we add water to acid it could be highly exothermic and it can splash the container
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Ans - 3 In the dilution we added acid into water
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Sir chapter 1 se topic Wise start Karo starting se for jee / neet preparation
Molality -
0.0241 mol/Kg or m
11:27 When we dilute acid we add acid to water
Very good
thank you sir
for these lession
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Question number 4 ka molar mass 342 and moles is 0.058 and volume 50 and molarity is 0.001
Sir some basic concept of chemistry chapter start krwado . Please Sir 🙂
Molarity[M] of C12H22O11 = 2.967 moles/liter !!?______right answer?
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17:10 Answer..:- 0.0295 M / L..
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jaha tak mujhe yaad hai bhaiya acid ko water me daalte hai, drop by drop
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0.4 mols/litter
Excellent approach 👍
Who is watching this just before your exam day🤪
Hydrogen mass percentage=11.76
Oxygen mass percentage= 88.89
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How are 0.5 mol Na2co3 and 0.5 M na2co3 diffrent
Mass percent of H=11.21% O=88.88%
Acid into water
HCl is added in water drop by drop