Good evening sir 36:46 Q1 11.2g Q2 6.4g Q3 No. of atoms=GM/MM(N0) = 0.6/12*6.022*10²³=3.011*10²² 55:36 No. of moles =8.8/44=0.2moles No. of molecules=0.2*6.022*10²³=1.2044*10²³ CO2 molecules No. of oxygen atoms present in this=2*1.2044*10²³=2.4088*10²³ oxygen atoms Thnku so much sir 🙏🏻 Understood everything
No of mole =0.2and no of molecules=1.2044 × 10 *23and no of oxygen=2.4088×10*23 thanku sir for teaching and solve all the questions related mole concept
as we know, No. oo Moles = given mass / molar mass so no of moles is 0.2 no. of molecules will be = 0.2*6.022*10^23 answer for no of molecules : 1.2044*10^23 as we know two oxygen atoms there are so multiplying no of moles *2 0.2*2=0.4 no of atom = 0.4*6.022*10^23 that is 2.4088*10^23 sir aap ne bht accha samjhaya jo concept mein class 9 ke end take nhi samjh paya aaj minton mein samjh gya.
9:29 kya bolte munna Kya bolte munni 🤣🤣🤣 My parents also laughed and said "nice teacher. You should study from him only."🤣🤣😁🤭 I first time said something good about my teacher.🤣🤣🤣 You are the best teacher sirrrrr jiiiii.😄🥰 Edit: Thanks for 40+ likes🙃😅
One of the best teacher in my education and those who like edumantra dil se like thoko sir ko lagna chaiyhye me jo padha raha hu vo bachho ko samajh aa raha he salam he sir 🙏🙏
I'm just so shocked that how can a teacher be like awesome. No words for you sir. U r the best teacher of science in India. Hats off respect to you sir🙂 . Keep it up, continue this. I'm sure one day you will great what you really deserve. Thank you sir 🙂
55:38, Q is= Calculate number of moles and molecules present in 8.8g of carbon dioxide, also calculate the number of oxygen atoms present in this? 1st equation, number of moles= given mass/ molar mass. So, 8.8g/44g = 2.2 Ans=2.2 moles 2 equation, Number of molecules= given mass/ molar mass × Avagadro number. So, 8.8g/44g× 6.022×10²³ =2.2×6.022× 10²³ = 13.2484×10²³ And 3 equation, Calculate all number of oxygen atoms present in this, So, = 2× 13.2484×10²³ ( I multiply it by 2 because there are 2 atoms present in ) = 26.4968×10²³ ANSWER...... And thanks sir for this precious knowledge 😊
Tq tq tq tq so much sir.... Can't tell how much I used to fear from this chapter just before 1 h... My This weekend plan was really really successful just because of you... For because of u now I feel confident about this chapter tq so much sir
I was stressing so much over this mole topic and now I am shocked.. it was so easy and I wasted so much energy on it.. thankyou so much sir.. Ur a gr8 tchr
Sir you told about the magic in mole concept and when i understood the magic i realised you are a magician of one kind. Thanku sir. Pls keep your style.
Sir the way you explain is very clear. As sometimes during class we cannot understand and we think that we will ask afterwards that doubt . But, we won't and because of that small doubt we don't understand whole concept. U explain even those small doubts too. Which helps us a lot. U r the best teacher .☺
I think you are failure in your class because students who have to study asks the doubts And I also ask my every doubt But sanjiv sir is best so I see the videos of sanjiv sir
00:00 Start 0:57 Avogadro number 8:50 Atomic mass 12:54 Molecular Mass 16:00 Molar Mass 26:17 textbook explanation 29:46 Question 1 31:54 Question 2 33:14 other questions 36:47 Formulae on the concept and questions 46:21 Textbook Exercise 52:02 Extra questions 55:14 Homework Questions 56:01 Outro
Video Timeline: 00:00 Start 0:57 Avogadro number 8:50 Atomic mass 12:54 Molecular Mass 16:00 Molar Mass 26:17 textbook explanation 29:46 Question 1 31:54 Question 2 33:14 other questions 36:47 Formulae on the concept and questions 46:21 Textbook Exercise 52:02 Extra questions 55:14 Homework Questions 56:01 Outro
Aaj tak mai 10 video chemisty k mole concept par dekha hu par aaj m pura mole concept samajha hu aur mera marks is bar science m 95 aaya tha o bhi aap ka kirpa se
Sir, I've got 50/50 in my science test , credit goes to u🙃, sir ur best teacher and Ur teaching style is amazing, 💕 keep supporting in our studies by making our study easier 🤗
Sir tomorrow is my offline exam and I have lot of doubt in mole concept and just because of your amazing explanation I understand it easily..thanku sir...🙏🙏
16:30 Soooooo helpful Last day before exam and opened I opened my book for the first time Understood nothing from the textbook Sir explained it sooo well that I couldn't believe I got it and cried tears of happiness!!:) 😍
OMG! Thank you so so so much sir! This explanation was the best. I saw so many other explanations, but just couldn't understand this concept. But now it so clear.👌🙏🙏
26:46 yes sir ,, I understood this .. Thanks a lot sir 🙏🙏 Your way of teaching is best for me. I am finding the best channel for my education since 2 months butt Today, I finded .... I'll always study from your channel ... Thank you so much...😊
26:47 Yes sir sab samj aa gya sir , aap our bkp dono he kamal ke teachers hai , backbencher ko toper banane ka dam rakhte hai aap dono ! Mere mujhe itna acche se samj nai aaya jitna yaha aaya!! Thank you very much sir☺️
What a way of teaching. Education will not be a burden for students. When every teacher's thinking will be like you. Highly inspired. Your method of teaching by laughing and saying jokes touches a student's heart and reduce the stress of that student during exams. Thank you sir, never give up. Keep going, you are doing a very good work, god is always with you.
Homework Question (Answer) -: i) No. Of Moles = Given Mass/Molar mass Given mass = 8.8 g of CO2 Molar mass of CO2= 12 + 16*2 Molar mass of CO2= 44g 8.8/44 = 0.2 Moles ii) No.of atoms = Given mass/Molar mass * N↓A Given mass/molar mass= 0.2 moles So, 0.2* N↓A = No.of atoms 0.2* 6.022*10^23= 1.2044*10^23 Oxygen Atoms present= (1.2044*2) 10^23 = 2.4088*10^23 O atoms. Solved.
Solution of last question No of moles present in 8.8g of CO2= g. M / M. M= 8.8/44g =0.2 moles No. Of molecules present in 8.8 g of CO2= no. Of moles * avagadro no = 0.2* 6.022*10^23= 1.2044*10^22 No of O atom present in 8.8 g of CO2= 2*1.2044*10^22 = 2.4088*10^22
samajh mein aa gaya sir
😡jhut mat bolh
Sir I am your biggest fan ❤️❤️😍😍
How can a teacher be this good????
not teacher legend
Exactly
@@saifamer1773 LEGENDS ARE SOMETHING WHICH ARE MOST PROBABLY NOT REAL LIKE TALES BUT SIR IS REAL
@@detectivejojo7025you are right
Why ? Teacher can't be good?
Mole concept understanding from
Self-2%
School-8%
Edumantra-90%
Good bro
Correct yr 😂😀
Self 99%
School - 1%
Yes
36:40 que. calculate the mass of 0.4 moles of oxygen atom = 6.4g
que. calculate the number of atoms present in 0.6g of carbon atom=3.011*10^22
Sir you EXPLAINATION is amazing. My Doubts are clear . Thank you sir
👍👍👍
26:26 understood the overall concept which sir tried to explain 🙏
Good evening sir
36:46
Q1 11.2g
Q2 6.4g
Q3 No. of atoms=GM/MM(N0) = 0.6/12*6.022*10²³=3.011*10²²
55:36
No. of moles =8.8/44=0.2moles
No. of
molecules=0.2*6.022*10²³=1.2044*10²³ CO2 molecules
No. of oxygen atoms present in this=2*1.2044*10²³=2.4088*10²³ oxygen atoms
Thnku so much sir 🙏🏻
Understood everything
How you got the answer of ques 3
Pls explain
@@suiii958 7.2g one?
@@v3nshika yes
@@suiii958
I made some mistake... Dn the correction
@@v3nshika okay baby, no problem I suggest you how to solve the problem 😊
Outstanding, Speechless, Mind-blowing!!!!!
Bhai tumne crazy tips ki video dekh kar ye logo banaya hai na😂😂......maine bhi uski videos dekh ke bohot logo banaye hai😂😂😂
ABD❤❤❤❤
Tham jaa bhai....
Fabulous ,, awesome , that's Gajab 😂👍👍
DJ DEKH KAR AA RAHE HO ?
sir you are great . may god give you 1000 million subscribers . best teacher
1000 million mai kitne zero hote ha bol na phle😂
@@effectioneffector4162 you are right 😂😂
Sahi kaha vahi
Ans of 1st 3 questions
1)11.2gm
2)6.4gm
3) 3.011*10²²
thoda zaada nahi hogaya???
No of mole =0.2and no of molecules=1.2044 × 10 *23and no of oxygen=2.4088×10*23 thanku sir for teaching and solve all the questions related mole concept
Thanks 😎😎😎😊😎😎😎
Mera Bhi Sahi Hai 🙂
Lol me matching my answer with urs🤡
as we know,
No. oo Moles = given mass / molar mass
so no of moles is 0.2
no. of molecules will be = 0.2*6.022*10^23
answer for no of molecules : 1.2044*10^23
as we know two oxygen atoms there are so multiplying no of moles *2
0.2*2=0.4
no of atom = 0.4*6.022*10^23
that is 2.4088*10^23
sir aap ne bht accha samjhaya jo concept mein class 9 ke end take nhi samjh paya aaj minton mein samjh gya.
I like how he himself questions what he says...
He knows every question a student is most bound to ask after watching his video.
9:29 kya bolte munna
Kya bolte munni 🤣🤣🤣
My parents also laughed and said "nice teacher. You should study from him only."🤣🤣😁🤭
I first time said something good about my teacher.🤣🤣🤣 You are the best teacher sirrrrr jiiiii.😄🥰
Edit: Thanks for 40+ likes🙃😅
Sahi hai
😂😂👍🏻👍🏻
@@anshvashist5739 😅😅
@SHOOTER GAMING 😌
55:20 number of moles=0.2moles, number of molecules=12.044× 10raise to power22
I AM SPEECHLESS FOR THE EXTREME NEXT LEVEL EDUCATION THIS MAN IS PROVIDING 😭😭..!!
I was so devastated with the mole concept .. but HATS OFF 2 HIM 🤌🪄
No. Of moles 0.2
No. Of molecules of co2 is. 1.2044*10²³
no of oxygen atoms is. 2.4088*10²³
My answer is also same
My ans is also same great
Same answer... I just came here to confirm 😂
Well done bro
One of the best teacher in my education and those who like edumantra dil se like thoko sir ko lagna chaiyhye me jo padha raha hu vo bachho ko samajh aa raha he salam he sir 🙏🙏
HIS T - SHIRT ALWAYS HAVE A OUTSTANDING MORAL...!!😍
Really !
@@kirtisakhuja5666 ya😂😂😂😂😂😂 bro or sis😂😂😂😂
@@bikrampradhan618 simp
@@t-series2638 omg mind your language you ill manered peanut
Really bro
I'm just so shocked that how can a teacher be like awesome. No words for you sir. U r the best teacher of science in India. Hats off respect to you sir🙂 . Keep it up, continue this. I'm sure one day you will great what you really deserve.
Thank you sir 🙂
55:45 ans is
O.2 moles of CO²
12.044×10²² molecules of CO²
24.088×10²² molecules of oxygen
Am I right sir ??
Plss ans this sir.... Btw ur the best sir
U are correct bro
You are correct 👍👍👍
Thank u 🤗
No bro molecular mass is 1.2046 x1023
@@ironman-gt8fk both answers are correct
Calculate the mass of 0.8g of Nitrogen atom = 1mole = 14g of Nitrogen
0.8mole = x g
1/0.8 = 14/x
= 14 X 0.8
= 11.2 g of Nitrogen atom
❤️❤️❤️❤️❤️
Wah
55:26 answer no. of moles = 0.2g, no. of molecules = 12.044*10^22 , oxygen molecules = 24.088*10^22
sir it it correct plz tell if you see
yes,aniruddh you are right
No of molecules=No. Of moles × arogardo no.=0.2×6.022×10²²=1.2044×10²²
But how oxygen atom became that plz help~
I know that it has been multiplyed by 2 but why??
@Varsha Bhargava [UHA] thanx bro XD
@@johnking8689 u gotta make it correct bro
Its 0.2 x 6.022 x 10²³ = 12.044 x 10 ²²
55:38, Q is= Calculate number of moles and molecules present in 8.8g of carbon dioxide, also calculate the number of oxygen atoms present in this?
1st equation,
number of moles= given mass/ molar mass.
So, 8.8g/44g = 2.2
Ans=2.2 moles
2 equation,
Number of molecules= given mass/ molar mass × Avagadro number.
So, 8.8g/44g× 6.022×10²³
=2.2×6.022× 10²³
= 13.2484×10²³
And 3 equation,
Calculate all number of oxygen atoms present in this,
So,
= 2× 13.2484×10²³ ( I multiply it by 2 because there are 2 atoms present in )
= 26.4968×10²³
ANSWER......
And thanks sir for this precious knowledge 😊
Your teaching and T shirt are all rounder💯💯
A big thanks to Sanjiv Sir and his team
H.W Question answer
No. of moles = 0.2 moles
No. of molecules= 12.044 * 10^22
No. of O atoms= 24.088 * 10^22
Ki
Great
How
Tq tq tq tq so much sir.... Can't tell how much I used to fear from this chapter just before 1 h... My This weekend plan was really really successful just because of you... For because of u now I feel confident about this chapter tq so much sir
Salute hai sar aapko aapke jaisa koi nahin padhta hai world ke best teacher
I was stressing so much over this mole topic and now I am shocked.. it was so easy and I wasted so much energy on it.. thankyou so much sir..
Ur a gr8 tchr
Ha great ki spelling to pata nahi hai
Aap jaise log hi science ke paper me maths ke question likh kar aate hai
Haa
Sanjiv sir soo best teacher , 😀
And your English teacher is stressed about your English..😂😂
waste some energy on me to bay
Sir you told about the magic in mole concept and when i understood the magic i realised you are a magician of one kind. Thanku sir. Pls keep your style.
Sir the way you explain is very clear. As sometimes during class we cannot understand and we think that we will ask afterwards that doubt . But, we won't and because of that small doubt we don't understand whole concept. U explain even those small doubts too. Which helps us a lot.
U r the best teacher .☺
I think you are failure in your class because students who have to study asks the doubts
And I also ask my every doubt
But sanjiv sir is best so I see the videos of sanjiv sir
Very good explanation Sir
@@usharawat2468 some students are shy and cannot Ash doubts and the real failure is you who are making fun of others
@@gwsquad3457 abe to failure hai jo yaha aake reply padh raha
Hai
Outstanding sir I have no doubts, and tomorrow is my science pre - board
Sir smjh aagya but revision karte rahna hoga warna puri mehnat pr paani phir jaega..thnk u sir aap pahle insaan ho jissne meko ache se smjha diya😍😍😍😍
Sir answer of the question asked by you at the end (56:02) is
(i). 4.4 moles
(ii). 4.4 × 6.022 × 10^23...
You explained awesome sir.. 👍
Everything is clear sir
I am not able to understand the mole concept but now It has cleared everything🤗
Numerical : Calculate the mass in milligrams of (I) 10*21 atoms of U-238 (ii) 10*20 molecules of oxygen gas.
26:40 sir everything is clear like a ash. (I am not saying other bcue I love chemistry and according to chemistry nothing is more clear than ash.)
Itni acchi English mat bolo kariye uper se Nikal jati hai
@@middleclassgamer7567 haha 😅🤣🤣🤣
Sir answer of H.W. question is
12.044*10^22 molecules of CO2
0.2 moles of CO2 molecules
24.088*10^22 atoms of OXYGEN
00:00 Start
0:57 Avogadro number
8:50 Atomic mass
12:54 Molecular Mass
16:00 Molar Mass
26:17 textbook explanation
29:46 Question 1
31:54 Question 2
33:14 other questions
36:47 Formulae on the concept and questions
46:21 Textbook Exercise
52:02 Extra questions
55:14 Homework Questions
56:01 Outro
26:33
I understood.😊
Itna baar koun likhega " I understood" ,, sidhe like karo yaaro... Jisse sir ko pata chl jaaye.
@ULTRA GAMING 《RAJ》 like toh kr de bhai phir
I'm speechless, I can't explain about Ur teaching methods.sir U are the great 🙌🙌🙌
26:20 aara sir apna aap ko badlne ki koi jarurat nahin hai apne aapko Jaise Ho vaise badhiya Ho😘😆👍👍👍
@Ritika jain thanks
Video Timeline:
00:00 Start
0:57 Avogadro number
8:50 Atomic mass
12:54 Molecular Mass
16:00 Molar Mass
26:17 textbook explanation
29:46 Question 1
31:54 Question 2
33:14 other questions
36:47 Formulae on the concept and questions
46:21 Textbook Exercise
52:02 Extra questions
55:14 Homework Questions
56:01 Outro
Answer of the question to calculate mass of 0.8 moles of nitrogen atom = 1/0.8 =14/x
= x = 14 × 0.8
= 11.2
we know we also have mind
Bhai bhai 😂
@@Disappointment_official tab comment mai kyu aaya khud krleta
Yes sir I understood. Before watching this video i was in doubt now Im able to understand everything
20:07 waah! Sir kya baat batai.👏👏
Clearly understood😃😃
Sir ko patalooo 😂 saadi karlrna
Aaj tak mai 10 video chemisty k mole concept par dekha hu par aaj m pura mole concept samajha hu aur mera marks is bar science m 95 aaya tha o bhi aap ka kirpa se
Awesome sir , i understood everything best teacher. Thanks a lot sir i am doubtless🙏🙏
Sir you are one of the best teachers I had ever seen in my life😊🤗😘😘
Sir you explained in such a great manner that i will never forget this concept
It means bro that the Avagadro Number is equal to 1 gram??? Isn't it
😂🤣 yee budy
@@AdilBains 😅😅
Bro are you active on yt, i want to ask something
Sir aap ne pura concept clear kar diya. Aur aap bahot easy way me samjhate ho. Thankyou sir.
Tommorow is my exam and this video helped me a lot.Thank u very much!!
I have also a exam of science by tomorrow
Speechless! Understood 1st half
What about other half..?
Sir you are better than my school teacher . your are amazing teacher thank u 😊
34:46 that "are kar lenge yaar" was really motivating 😄
26:32 Aagya sir smjh me
Sir I understood everything thank you very much sir ❤️❤️
Sir, I've got 50/50 in my science test , credit goes to u🙃, sir ur best teacher and Ur teaching style is amazing, 💕 keep supporting in our studies by making our study easier 🤗
Online me mere bhi aye the
@Shakti Techno SAVAGE OP RESPECT +
Mai bhi dakata hu sir ke video muzai to 25/2 and half hi milai
Also got 80/80 and 100/100 on my test
@@omkarmohite9955 kyunki tu khud se nhi padhta h
Sir ,smjh me achhe se aa gya
Sir tomorrow is my offline exam and I have lot of doubt in mole concept and just because of your amazing explanation I understand it easily..thanku sir...🙏🙏
I am watching on examday morning 4 am😅😂😂😂
this is lecture - legend
@@Disappointment_official ＳＡＨＩ
@@adarshanand3989 i know
@@adarshanand3989 there is one more legend then me
33: 50
1-11.2gm
2-6.4gm
3-.3011×10²³
U r right 👍👍👍👍👍👍
Yes u are really right but muje 3rd samaj nahi nahi aaya
Can anyone please explain how the last answer came? I’m a lil’ confused
Ans. 3rd is 3011×10ki power 19
@@harshalmudgal9558 ha aur agar 3.011 lenge to power 22 ayegi
*55:7* No. Of moles=0.2g
No. Of atoms in O2=1.2044×10^23
Ryt
I got the concept sir 😊 jo mujhe 11th ki class mein nhi itna easily smjha wo yha smjh gya
26:30 good teaching really got the concept
One of the easiest video of mole concept which i found on utube ...😊♥️😘
Realy superb teaching😍..appreciate it👏
Yaha v youtube me you short hai,,,, pata nahi aur kitna short cut hogi ye duniya😶😑😕
16:30
Soooooo helpful
Last day before exam and opened I opened my book for the first time
Understood nothing from the textbook
Sir explained it sooo well that I couldn't believe I got it and cried tears of happiness!!:)
😍
OMG! Thank you so so so much sir! This explanation was the best. I saw so many other explanations, but just couldn't understand this concept. But now it so clear.👌🙏🙏
U r gajab sir phle Mai samjh rha tha ki ye bahut hard topic hai but after this video I clear all dought
amazing session !
perfect for pre exam revisions
thank you sir!
Good teacher quality looks on his t shirt ..... motivation 💪
26:46 yes sir ,, I understood this ..
Thanks a lot sir 🙏🙏
Your way of teaching is best for me.
I am finding the best channel for my education since 2 months butt Today, I finded .... I'll always study from your channel ... Thank you so much...😊
U got a lots of blessing🙏🙏🙏🙏 😇😇😇😘😘😘😘
finally sir you make me understand this so called mole concept .......you are great sir ......hats off to you🙏🙏😍😍
Are u in kv school ???
@@chandrashekarrao2906 yes
Wheare do you leave
Sir u r a great teacher the teaching way of urs is superb
26:48 HAA SIR, BOHT ACCHESE SAMJHA 🙌🔥
Yes
This is the best platform of education in free of cost i recommend to all of you ❤❤❤❤❤❤❤
36:28 sir answer 3.011* 10^22
It's worng
Answer is 3011 × 10^19
Sir plz cover the whole ncert book
And your explanation is awesome
Yes
26:28 everything understood crystel clear sir
we all have watched mind your studies
Sir I love your teaching classes.....😀😀😀😀
Sir I am from Afghanistan , I really appreciate with your video ! Thank you sir I am doubtless 😘😘😘😘
Thanks bhai, Love to Afghanistan and You from India
Tumhare number dedo me afganistan ke bare me kuch Janna chata hu cricket ke bare me bhi
🙄🙄🙄🙄 jhutha
@@Mansi7705 sahi pakde hain. Op bolte.
@ek inssan haan yaar pata nhi game me bhi aate hain aur kehte hain me tumhare india se nhi hu phir bhi tumhari bhasha aati hai.
Sir everything is the best 1 thing that decrease our concentration is too many ads..
HE ALSO WANTS MONEY BRO/SIS ...... THIS IS LIKE HIS PROFESSION
SIR KO SIRAF YT ADS SE PAISE DETA HAI
Video download karlo then data off kar ke video dekho
yes sir please reduce adds please sir
Sir is giving the education free it means a lot to us and if we can pay him by ads it will be benefited to sir also
@@GRANTHBHARDWAJ31 ur absolutely right
Sir u are a great teacher I can't explain in word ....n I will pray to God that u always success in your life ❤️thanks alot from your student💝
Shi hai
Right bhai i also pray to God 😇🙏👼
Sir aap hi sabse achha padhate ho in short time varna baaki online educators large time me small concept padhate hain
26:47
Yes sir sab samj aa gya sir , aap our bkp dono he kamal ke teachers hai , backbencher ko toper banane ka dam rakhte hai aap dono !
Mere mujhe itna acche se samj nai aaya jitna yaha aaya!!
Thank you very much sir☺️
Yes bro 😎........
BKP+SANJIV SIR= MAKES ANY OF US TOPPER
Sanjiv sir j cbse class vedios ❤️❤️ best
26:28 sb smjh aagya sir yha tk
Sir Answer of Your question is -
3011*10^19 = x
Best teacher in the world 🌍🌍🌍💪🏻💪🏻💪🏻 of edumantra
What a way of teaching. Education will not be a burden for students. When every teacher's thinking will be like you. Highly inspired. Your method of teaching by laughing and saying jokes touches a student's heart and reduce the stress of that student during exams. Thank you sir, never give up. Keep going, you are doing a very good work, god is always with you.
Oversmart kahi ke tujhe yha bhasan deneki jaroorat nhi
Oversmart apne aap ko jyada talented samajhti he kya
@@crisworld8160 your exam are coming KYA??¿
@@shubhangishubhangi5654 no siso after vacation
@@crisworld8160 mtlb final exam???
19:50 wallahh...!!!😂😂
i loved you explaination sir... :)
35:46 answer is 3.011*10^23
3.011*10^22 Hoga na
a
samajh mein aa gaya sir
H.w
No.of moles present in 8.8g of CO2 is 0.2
And no. Of molecules are 12.044×10²²
No. Oxygen atoms are 6.022×10²²
Maybe...
No. Of Co2 molecules=12.044×10^21
No. Of oxigen atoms =24.088×10^21
10^23
if I say true when I was in 9 I did't understand that but now after 2 year in class 11 I understand 😁 from this same video .
Sir You teach very well . Like Crystal Clear
👍🏻👍🏻👍🏻👍🏻❤️❤️❤️❤️
ok
When you are in 11th but still dont know about Mole concept 🙂
Dropper bro 😐
Hum bhi h ..
Mai bhi
Kya mai aapka help kr skta hu
Lol thats exactly why i am here
Homework Question (Answer) -:
i) No. Of Moles = Given Mass/Molar mass
Given mass = 8.8 g of CO2
Molar mass of CO2= 12 + 16*2
Molar mass of CO2= 44g
8.8/44 = 0.2 Moles
ii) No.of atoms = Given mass/Molar mass * N↓A
Given mass/molar mass= 0.2 moles
So, 0.2* N↓A = No.of atoms
0.2* 6.022*10^23= 1.2044*10^23
Oxygen Atoms present= (1.2044*2) 10^23
= 2.4088*10^23 O atoms.
Solved.
GR8 BRO
That's correct 💯
Before this I am struggling with this concept now my all doubt get cleared ☺️
Solution of last question
No of moles present in 8.8g of CO2= g. M / M. M= 8.8/44g =0.2 moles
No. Of molecules present in 8.8 g of CO2= no. Of moles * avagadro no
= 0.2* 6.022*10^23= 1.2044*10^22
No of O atom present in 8.8 g of CO2= 2*1.2044*10^22 = 2.4088*10^22
Outstanding performance sir😍😍😘