Mole Concept 06 || Application of Limiting Reagent || Percentage Purity || Class 11/JEE || RAFTAAR

[@princebaghel6697]

21:10-ans 2 mole of nahco³=1mole of na²co³
1mole of nahco3=1/2mole of na2co30.
0.2mole of nahco3=1/2×0.2 mole ofna2co3
Final ans =0.1mole of na2co3.

[@princebaghel6697]

The most lovely line by sir is "me harsh Tyagi Bahubali.

[@akshathudda1546]

akshat hudda
21:22 a) 0.1 moles 39:45 Na2CO3-90%
CaCO3- 10%
42:42 question alpha - 66.66%
58:10 - 106.4kg
1:12:08 - b)54.34

[@mr.pankajsingh2434]

We are same bro.

[@sDa_E_23]

👉🏽👉🏽SUPRAVA DUTTA👈🏽👈🏽
🅷🅾🅼🅴🆆🅾🆁🅺 🅰🅽🆂🆆🅴🆁🆂👇🏽👇🏽
21:23 answer is 0.1 mole
39:56 10% calcium carbonate, 90% sodium carbonate
42:43 alpha question- 66.67%
58:09 106.4 kg
1:03:18 25% of calcium carbonate
1:12:09 option b is correct 54.34%
1:16:50 1.12L carbon dioxide at STP
1:19:34 50% calcium carbonate

[@sumitkumar-bk8bp]

Gajab

[@himanshupandey4268]

Bhai yeh percentage purity ka topic pure jee advanced ka topic hai ya mains ka ???

[@kajukashyap5934]

1:12:09: option a is correct cuz x=1 and mass of MgCO3 is 0.84g (1.84 - x ) then ,mass % = (0.84 /1.84)*100

[@buddhadebbal]

(21:15) 0.1 moles
Alpha - 66.67%
(58:16) d. 106.4 kg
(1:12:13) x=1, % is 45.65
Thank you Sir for this amazing lecture

[@ManishSharma-hj8xj]

Manish Sharma
Homework Answers 👇👇
(21:15) = 0.1moles
alfa question = 66.67%
(58:10)= 106.4 kg
(1:11:40)= 45.66%
SIR I SOLVE ALL THE QUESTIONS BECAUSE OF YOU ❤❤
THANK YOU SIR

[@rahulrudraaa1818]

broo but 1:11:40 me x is 1 aa raha haa

[@Doraemon6.626]

Mera bhi x=1 aa raha h

[@ayushraut2005]

@@rahulrudraaa1818 yes

[@covidrangeilundablock1674]

B hoga

[@rishavkumar4916]

Yes

[@yashmangla4673]

21:15. HW ans 0.1 mole
Percentage purity of CaCO3 is 66.67%
1:12:15 %of MgCO3 is (a) 45.66%
Yash Mangla

[@adrikabanerjee3223]

Adrika Banerjee
21:23:a) 0.1 moles
39:45: Na2CO3-90%
Caco3-10%
42:42 question alpha-66.66%
58:10 106.4
1:12:08: b)54.34%

[@Lakshya-PW-]

1:12:08
45.66%

[@sarkariseva2875]

Sir ve 4 alkali metal ke carbonates ka decomposition kyu nahi hota hai kripya bateye sir

[@earningtips4408]

@@Lakshya-PW- matlab jaisa question ma bol 60g sodium phosphate ko 20g naoh sa milata ha , toh jab sir jo chemical formula likhtha for usaaa balance karta ha woh kaisa nikalta ha ? Malab ya wala H2So4+Naoh ----- Na2so4 +2h20 ?

[@Lakshya-PW-]

@@earningtips4408 I'm not getting your point...
Thoda ache se likh ke btao apna doubt

[@earningtips4408]

@@Lakshya-PW- matlab jo chemical reactant formula aur product formula kaisa nikala yaaani example yah :- H2So4 + Naoh ------ Na2so4 + 2h20 ya toh salt ka tha baaki aisa dusro ka kaisa nikalta ha

[@Mr.N_204]

1:04:09 the answer to the question is 45.66%. I'll tried till max of half an hour and I have succeed 😉❤️👍

[@gurnoorsingh4443]

Gurnoor Singh
21:25 ans is (a)
Q(alpha)=> 66.66%
58:14 ans is d
1:11:31 ans is 54.34 Percent

[@MohitSingh-qk9jb]

Bhai sir ne value galat bataya tha x ka value 1 aayega, jisse mgco3 ka mass 0.84 aayega
Jisse percentage 45.6

[@beautifullife.2100]

( 1:12:00 )
Answer is Opt (a) i.e 45.66%
Value of x comes 1g
Name-Shubham

[@mohdhamza3799]

1:12:15 MgCO3 weight% is 45.65 value of x is 1 Mohd Hamza

[@zigyashakumari3795]

21:13 A 0.1 mole
2 mole NaHCO3= 1 mole Na2CO3
0.2mole NaHCO3 =1/2×.2 Mole Na2CO3
.1 mole Na2CO3 is formed
39:41 90% Na2CO3 and 10%CaCO3
42:41Percentage purity of CaCO3 =66.6%
59:13 106.4 kg
1:12:13 54.34% of MgCO3
Zigyasha Kumari:)

[@MRX-jp7be]

HW:- 0.1(A)
HW (alpha) :- 66.67%
HW (caco3+mgco3 wala) :- 45.6%
--------Madhav Kohli

[@Doraemon6.626]

21:13
Answer
A) 0.1 moles
Alpha Q ans.) 66.67%

[@pragatiupadhyay]

Pragati
21:15-0.1 mole 👍
At 39:12-Na2CO3=90%
CaCO3=10%
42:34:Purity %of CaCO3=66.67%
58:9-(d)
1:12=(a)

[@Shivkumar-123-yadav]

0.1 mole kaise aya hum se nhi bna..help please..

[@pragatiupadhyay]

Bhai pahle mole nikalo H2 ka phir stoichiometry coefficient ko compare karoge to ans aa jayega 👍

[@yuvrajrawat8746]

1:12:13 B for bombay --- YUVRAJ

[@aim_t_be_5568]

Question no. --› 21:16 min. Ka Ans. ---›› (a) 0.1 mol
Q. --› 2NaHCO3 ----------› Na2CO3 + CO2 + H2O

[@rajaryansingh3216]

21:23 0.1 mole of sodium carbonate

[@swathil3232]

Thank you sir 🙏
Swathi
21:27 (a) 0.1
39:50 Na2CO3 90% CaCO3 10%
58:10 106.4
Alpha question : 66.66%

[@Shivkumar-123-yadav]

0.1 kaise aya explain please!

[@swathil3232]

1 mol NaHCO3 consumed= 1/2 mol Na2CO3 comes
So when 0.2 moles of NaHCO3 consumed,
0.2 × 1/2 mol Na2CO3 comes
= 0.1 mol

[@Shivkumar-123-yadav]

Thanks swathi...

[@earningtips4408]

@@swathil3232 tum yaha bata sakti ho kya chemical reaction kaisa nikalna ha ? 😥 Yaani woh sir jo har que ma nikalata ha Sabsa pahla phir uska product nikalata woh kaisa nikalna ha ?

[@swathil3232]

I didn't understand

[@arvind2672]

Arvind kumar
21:23 a) 0.1 moles 39:45 Na2CO3-90%
CaCO3- 10%
42:42 question alpha - 66.66%
58:10 - 106.4kg
1:12:08 - b)54.34%
Bahubali op♥️♥️

[@anythingforfun4320]

Bhai answer A ayega 45.66%

[@arvind2672]

@@anythingforfun4320 1/1.84 ×100 = 54.34%

[@anythingforfun4320]

@@arvind2672 you may have done any mistake. You may see the answer on google

[@khushalpande6123]

Ultra, great, outstanding, OP class ............got confidence...........thanks a lot

[@earningtips4408]

Bhai tum yaha bata sakta ho ya chemical reaction kaisa nikalna ha ? 😥 Yaana woh sir jo har que ma nikalata ha Sabsa pahla phir uska product nikalata woh kaisa nikalna ha ?

[@3psgaming343]

@@earningtips4408 bhai kuch reactions ko to Ratna Padta hai air kuch reactivity series ki help Se banana padta hai. Jis element ka high reactivity series hogi wo dusre element ko Uske compound me Se HTa k apne sath uska compound bna lega product side me..

[@vanshtyagi6805]

Homeeork question answer at 21:15
Option-(a) 0.1 moles

[@priyanshupandey5708]

58:08 answer :- option(b) 105.4kg

[@yuvrajrawat8746]

SIR THANKS FOR THIS SERIES MUJHE MOLE CONCEPT SE BOHIT DARR LAGTA THA LEKIN AB MOLE CONCEPT HALWAA HOGYA HAI THANKS SIR AND THE WHOLE PW TEAM LOVE YOU ❤️❤️❤️❤️ -Yuvraj

[@chiragjagwani1042]

True

[@sakshamjeswani8603]

1:11:40 saksham jeswani
Answer 45.66%

[@malik92727]

Sujal malik, 1:12:12 mass percent of mgco3 is 45.66%

[@SumitSharma-gc7op]

21:13 - Ans - 0.1 Mole
-Sumit Sharma

[@ShivamSrma]

Shivam Kumar Sharma
21:10 (A) 0.1 mol
42:45 66.6 %
1:11:56 (B) 54.34 %

[@sreekalarani9309]

21:14
Homework question
Answer : 0.1 mole

[@prajjwal2519]

Prajjwal
HOME WORK ANSWER IS (a) 0.1 mole
Alpha question answer is 66.66%
55:00 answer is d) 106.4 kilograms
1:06:00 ANSWER is a) 45.66%
Thanks....♥️

[@Godsaybehappy]

Harshit Dubey
21:21 ans is 0.1 mole
42:33 alpha question answer is 66.66%

[@epic844]

39:46 9gram

[@userxyz321]

Sir you are the best teacher in world 😍💖✨You gave me the strength I needed to take the next steps towards my dream
🔥🥰❤

[@AnuragSingh-gr2ob]

21:14 Que's Ans: 0.1 moles of sodium carbonate

[@chaitanyachinchole_1476]

11:10 sir ye question Mai Maine pehle total resedue nikala fir resedue mass diya hua tha to usee moles nikala fir MgCO3 ke moles nikle fir mass Nikla fir mass percentage .....ANS IS OPTION A 45.65 perfect. CHAITANYA

[@tiwariji.789]

21:12 0.1 mol
42:34 (alpha question) 66.667%
57:49 106.4 Kg
1:11:05 x= 1 hoga
and hence % MgCO3 by mass = 45.65217%
NAME => AAKASH TIWARI
FROM PATNA (BIHAR)

[@harikeshkushwaha3239]

21:25 mole of sodium corbonate a) 0.1

[@Lakshya-PW-]

42:30
Alpha question
66.67% CaCO3 purity percentage

[@AryanSingh-pl2yl]

I also got 66.67percent

[@ghostmedia7768]

DARRANE WALLAH
21:25 ans is (a)
Q(alpha)=> 66.66%
58:14 ans is d
1:11:31 ans is 54.34 Percent

[@kajukashyap5934]

1:11:31 : option a is correct cuz mass % of MgCO3 = (0.84 /1.84)*100

[@helloguysyt5669]

21:16 ans is (a) 0.1❤❤💕💕💕💯💯🔥🔥love you sir ❤❤

[@mitunkumar2181]

सर आप बहुत अच्छे पढ़ाते हो ये बात हमें ही नही बल्कि पूरे हमारे TH-cam family को पता है। But सर आप थोड़ा zero level से पढ़ाओगे तो अच्छा रहेगा क्योंकि इस प्लेट फॉर्म पे कुछ ऐसे हमारे yotube family है जो fast बार 11 th पढ़ रहे है। कुछ ऐसे भी family है जो jee का prepare कर रहे हैं । आप थोड़ा zero level से पढ़ाओगे तो शायद 11 th वालों के लिए अच्छा रहेगा बिल्कुल निखिल सर की तरह pleace सर
❤️❤️❤️❤️❤️ Thank you sir ❤️❤️❤️❤️

[@Factsfactory67]

Nikhil sir hero h aapne

[@aim_t_be_5568]

Ans. Q. 35:05--› Na2CO3 ----›90%
CaCO3 ----›10%

[@saritayadav1266]

40:23 answer is 10percent CaCO3 and 90percent Na2CO3

[@gammingquarter6127]

1:12:14 Answer is option (b) 54.34% , solved by Arzoo

[@abhinavpandey7221]

42:38 ans alpha =66.6%
58.22 ans is (d)=106.4
1:12:14 ans is (a) =45.66

[@krishnamlad8033]

Krishnam Lad
21:21 a)0.1
40:50 % purity ofNa2CO3= 90% CaCO3 10%
Alpha question %CaCO3=66.67%
1:10:43 a 45.66%

[@thejuskk6931]

Hw ans. 0.1 moles na2co3
Alpha q ans 66.66%
Next hw ans 45.66%
Thejus kk

[@yadavji5739v]

Sir I am form punjab I am studying in +1standard studies from PW platform you clears my all concepts I started my jee preparation form this class and I hope I would succeful in my goal ,you very great sir

[@chauhanraviiii]

sir my all answers are correct except the caco3 and mgco3 combined decomposition question i attempted that by pausing your video . Thanks all PW team for making this golden oppurtunity for us .

[@sportandstudychannel7483]

Time58:11 answer (d)is correct

[@muskanagrawal9406]

Muskan Agrawal
(1:12:05) Ans-45.66%

[@HimanshuKumar-zv9jp]

39:45 answer is 10% of caco3 and 90% of na2co3

[@riddhimaheshwari6657]

Thank you sir. 1:12:14 Answer= 54.34% -Riddhi Maheshwari

[@techtokazaan1907]

Wrong because question asked percentage of mgco3 not caco3

[@arbindkumar-hb3vb]

Time 1:10 question ans 54.34% option no (b)
Name Manoj kumar

[@dontreply395]

1:11:06 the value of x is 1....0.84 is the mass of MgCO3

[@harshshah8751]

harsh , sir on 1:12:11 value of x is coming out as 1. so % mgco3 is 45.66 %

[@ayushraut2005]

42:44
%CaCO3 = 66.67%
1:11:58
%MgCO3=45.65%

[@ayushraut2005]

1:11:58- x=1

[@piyushjha5425]

1:11:31 ans B

[@priyanshupandey5708]

21:14 answer:- (a)0.1mol

[@priyanshupandey5708]

1:12:09 answer:= option(b) 54.34%

[@dr.tn12]

SIR UR BEST 👌👌👌👌CHEMISTRY TEACHER I'VE STUDIED FROM I'M NEET ASPIRANT
WHEN I WATCHED UR 🔔🔔🔔CHEMICAL BONDING LECTURE I FALL IN LOVE 😍😍😍WITH CHEMISTRY📒📒📒💓💓💓
BY THE WAY ANSWERS R
21:00 pr 0.1 m
Alpha 66.66%
39:00 pr CAC03 10% and na2co3 is 90%
58:00 pr 106.4
1:12 pr 54
NAME TAHA 😊☺😚🙂😉🙃🤗

[@664arhaannawab8]

Same here

[@koshichauhan8504]

Lokesh
1:12:02
Sir x = 1
So % of MgCO3 is 45.66%

[@namanmagotra4033]

@42:45, percentage purity of CaCO3 =66.66%

[@Cut_tale_himanshu]

21:24 a) 0.1
39:50 CaCO3 10%. Na2CO3 90%
58:04 b) 105.4 kg
1:03:09 %CaCO3 = 25%

[@darkpulse4244]

How you get the time for revision

[@Cut_tale_himanshu]

@@darkpulse4244 Bhai Mai school nahi jata

[@abskshaw6221]

hw ans 0.1 moles.of Na2CO3

[@anshakki7838]

1:12:12 answer is b 54.34
Ansh mahajan

[@Sslm07]

Alpha answers
Impurity% 33.33
And calcium carbonate % 66.67

[@athangop]

21:00 answer 01. mole of Na2CO3
Aap best ho sir I can crack JEE with your guidance easily

[@arpitsinghrajpoot4848]

1:12:14: %of MgCO3 45.65% Arpit Singh Rajpoot

[@saritayadav1266]

1:10:59 answer is 54.3 percent
Sir my name is "TANISH YADAV"

[@ranjanamaurya1051]

Time 21:18 answer is 0.1 mole of Na2CO3.

[@krishsharma7088]

1 12 00
Ans (a) and sir x = 1 ayega apne glti se 0.84 bol diya tha.....great lecture by the way

[@sakshigupta1764]

(21:26)answer=0.1moles of na2co3

[@krishnaagarwal7533]

21:15 answer is a) 0.1 mole

[@TechPK605]

1:12:04 B) 53.34% percent

[@pramilapandey409]

Sir ji 58 min ke questions me ans hai Options (D) 106.4 kg means iit Delhi 🙏🙏🙏🙏🙏🙏

[@badshahkhan7555]

(1:11;40)ka 54.34 percent aayega. With 100percent surity.

[@rdsharma8703]

40:19 - ans 90% of sodium carbonate and 10% of calcium carbonate

[@princebaghel6697]

Sir jab se apki video dekhi h chemistry padne me maja aa raha h.

[@motivationalgalaxy]

Home work answer
M=1
W(CaCO3)= 100g
% of CaCO3=66.3%

[@alakhsir9304]

Pankaj
1:11:56 if x = 0.84 option b is correct ✔ 54.34%
If x =1 then option a is correct ✔ 45.66%

[@SSYADAV-xi1qj]

Sumit Yadav sir aaj aap ne itne sare questions kra diye ki confidence ata jab rha bhout intersting way me aap padhaya sab kuch samj me aya thankyou sir 🙏

[@gta5fun124]

21:17 Answer {option A}

[@sidrawajeeh7494]

x=1 g
Mass of MgCO3= (1.84-x)g
= 0.84 g
Mass% of MgCO3= 0.84×100/1.84= 45.65%
Sidra Wajeeh

[@vanshtyagi6805]

Answer of question at 58:11 is option (d)

[@Xyz54699]

21:20 hw question answer
(a) 0.1

[@nirajgupta5736]

Sir i completely understand the lecture Dheeraj kumar Gupta

[@sshorts81]

58:10-106.4kg option(D)

[@AnuragSingh-gr2ob]

1:12:15 Ans is 45.66%

[@AnuRadha-os5us]

The answer at 21:18 is option A 0.1

[@abdullahshaikh8166]

1:11:56
Answer =45.6%
Name is Abdullah

[@adityasinghkurmi]

Hello sir. Question alpha answer is 33.34% percentage of purity of caco3 in given sample

[@mdkaif-bb4vo]

Rong

[@tjbhai3216]

21:13 Ans.) a)0.1
58:10 Ans.) d)106.4 kg

[@princebaghel6697]

58:16 ans 106.4kg.

[@Prabhatsingh-mr1bv]

1:11:40 answer is option A 45.66%
👉👉 PRABHAT SINGH👈👈

[@sreekalarani9309]

58:17
Weight of lime : option (b) 105.4 kg

[@ranjanamaurya1051]

Sister perhaps your answer is wrong.

[@Kidszilla47]

1:11:06 x=1 ans 45.66 percent

[@pramilapandey409]

Sir ji we are in full confidence ... 🙏🙏🙏🙏🙏🙏
I am Shri Krishna Pandey 🙏🙏