pV=nRT question

http://www.a-levelchemistry.co.uk/AQA%20AS%20Chemistry/Unit%203%20practicals/Unit%203%20Practical%202004.pdf

page 12.

1) Suitable volume, i said 100cm^3 as that is the maximum that u can get in a syringe.

2) I dont get why they gave the volume as 24 dm^3. Anyways i assumed the volume to be 100cm^3, the temperature to be 298 K and the pressure to be
1 X 10^5 Pa

Working :

pV=nRT
therefore

n= \frac{pV}{RT}

Unparseable latex formula:

n= \frac{1\times 10^5 \times 100 \times 10^-^6}{8.31 \times 298} =4.038 \times 10^-^3 moles

Therefore n(Sr) =

Unparseable latex formula:

4.038 \times 10^-^3 moles

Mr(Sr) = 87.6

Hence the mass of Sr = n X Mr = 4.038 X 10^-3 moles X 87.6= 0.354 grams.

The problem is that the markscheme :
http://www.a-levelchemistry.co.uk/AQA%20AS%20Chemistry/Unit%203%20practicals/Unit%203%20Practical%202004%20ms.pdf

page 8 says 0.37g Sr gives 100 cm3 of H2 ?!

Thanks!

wizz_kid

http://www.a-levelchemistry.co.uk/AQA%20AS%20Chemistry/Unit%203%20practicals/Unit%203%20Practical%202004.pdf

page 12.

1) Suitable volume, i said 100cm^3 as that is the maximum that u can get in a syringe.

2) I dont get why they gave the volume as 24 dm^3. Anyways i assumed the volume to be 100cm^3, the temperature to be 298 K and the pressure to be
1 X 10^5 Pa

Working :

pV=nRT
therefore

n= \frac{pV}{RT}

Unparseable latex formula:

n= \frac{1\times 10^5 \times 100 \times 10^-^6}{8.31 \times 298} =4.038 \times 10^-^3 moles

Therefore n(Sr) =

Unparseable latex formula:

4.038 \times 10^-^3 moles

Mr(Sr) = 87.6

Hence the mass of Sr = n X Mr = 4.038 X 10^-3 moles X 87.6= 0.354 grams.

The problem is that the markscheme :
http://www.a-levelchemistry.co.uk/AQA%20AS%20Chemistry/Unit%203%20practicals/Unit%203%20Practical%202004%20ms.pdf

page 8 says 0.37g Sr gives 100 cm3 of H2 ?!

Thanks!

The periodic table gives the RAM of strontium as 87.6

To collect 100cm³ of gas... 100/24000 moles = 100/24000 x 87.6 = 0.365g strontium = 0.37g (2 dp)

Reply 2

EierVonSatan

21

The problem here is the exact numbers you use to derive the 1 mole of gas = 24 dm³

P = 101325, R = 8.314, T = 293.15 (20^oC = room temperature), if you were to use these values you should obtain the same answer :smile:

Reply 3

wizz_kid

OP

16

EierVonSatan

The problem here is the exact numbers you use to derive the 1 mole of gas = 24 dm³

P = 101325, R = 8.314, T = 293.15 (20^oC = room temperature), if you were to use these values you should obtain the same answer :smile:

oh right. However i still dont understand why they said that 1mole of gas = 24 dm^3 as a volume when they expected us to use 100cm3 as the volume?

Reply 4

EierVonSatan

21

wizz_kid

oh right. However i still dont understand why they said that 1mole of gas = 24 dm^3 as a volume when they expected us to use 100cm3 as the volume?

It's just a standard conversion so you can do the calculation as charco described :yep:

Reply 5

wizz_kid

OP

16

EierVonSatan

It's just a standard conversion so you can do the calculation as charco described :yep:

srry dont realy get that calculation and stil dont understand why 24 dm^3 was given?!

Reply 6

EierVonSatan

21

wizz_kid

srry dont realy get that calculation and stil dont understand why 24 dm^3 was given?!

it's a standard conversion in the same way 1000g = 1 kg

one mole of gas at room temperature occupies 24 litres

so 100/24,000 = 0.00417 moles of gas

Reply 7

youseriousfoo?

^ hey i thought he assumed 100,000 Pa how can we use 101325 in the equation, is that standard? sorry dont get it, thanks.

Reply 8

EierVonSatan

21

You don't need to use the exact numbers, it's just the relation ''1 mole of gas = 24 litres'' comes form inputting those numbers into pV = nRT

1 moles of gas n = 1 => V = RT/p = (8.314)(293.15)/101.325 = 24.05 dm³ = 24 dm³ to two significant figures.