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Post your Chem doubts here!
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Solving Chemistry Doubts Now :)
- Thread starter SB29597
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plz send solved mcQ's papers of chemistry
- Messages
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The mark schemes are available, that may be easier for you.
Why don't you try solving them and check with the marking schemes and get back to me with whichever you don't get?
Why don't you try solving them and check with the marking schemes and get back to me with whichever you don't get?
- Messages
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- Reaction score
- 192
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- 53
Hi! So here goes.
To understand this, you first need to know that 1 mole of Ag will contain 6.02*10^23 atoms (Avogadro's constant). Part (b) examines that.
In part C (1) they expect you to use the data booklet to find the atomic radius of Argon- 0.192 nm. Convert this to dm as most Chem calculations are in dm. That will be 1.92*10^-9 dm. now you just substitute this in the place of 'r' in the formula- (4/3)*3.14*(1.92*10^-9)= 2.96*10^-26 dm3 which is your first answer.
For C(2) we had seen in (b) that one mole is 6.02*10^23 atoms right? So to calculate the volume of i mole, we simply multiply it by the constant. So, (2.96*10^-26)*(6.02*10^23)= 1.78*10^-2 dm3, your ans to the second part.
Part c(3) 24dm3- we use 24dm3 as volume of an ideal gas and even for mole calculations.
Part C(4) ((1.78*10^-2)/ 24))*100= 0.074 percent- ( actual volume/ ideal volume)* 100 as we req percentage.
Part C(5) In part (a), one of the assumptions we are supposed to make is that the volume of the molecules of the gas is negligible compared to the volume of the gas. As the molecule of argon occupies only 0.074 percent, this is justified.
To understand this, you first need to know that 1 mole of Ag will contain 6.02*10^23 atoms (Avogadro's constant). Part (b) examines that.
In part C (1) they expect you to use the data booklet to find the atomic radius of Argon- 0.192 nm. Convert this to dm as most Chem calculations are in dm. That will be 1.92*10^-9 dm. now you just substitute this in the place of 'r' in the formula- (4/3)*3.14*(1.92*10^-9)= 2.96*10^-26 dm3 which is your first answer.
For C(2) we had seen in (b) that one mole is 6.02*10^23 atoms right? So to calculate the volume of i mole, we simply multiply it by the constant. So, (2.96*10^-26)*(6.02*10^23)= 1.78*10^-2 dm3, your ans to the second part.
Part c(3) 24dm3- we use 24dm3 as volume of an ideal gas and even for mole calculations.
Part C(4) ((1.78*10^-2)/ 24))*100= 0.074 percent- ( actual volume/ ideal volume)* 100 as we req percentage.
Part C(5) In part (a), one of the assumptions we are supposed to make is that the volume of the molecules of the gas is negligible compared to the volume of the gas. As the molecule of argon occupies only 0.074 percent, this is justified.