A lvl Chemistry molecular mass

Bigcable22 SG

A sample of 0.16g of liquid X is vapourised in a graduated syringe at 100degree celsius and 1 atm. The intital volume of air in the syringe recorded is 10cm^3 and the final recorded volume is 56cm^3. What is the relative molecular mass of X?

UltimaOnline SG

You know pressure in atm.

You know temperature in Kelvins.

You know volume of gas X (final vol - initial vol) in dm3.

Substitute into PV = nRT, and make no. of moles the subject.

You can also find no. of moles of gas X (in terms of molar mass).

Equate no. of moles of gas from PV=nRT, with no. of moles in terms of molar mass.

Solve for molar mass.

UltimaOnline SG

Note that you have two different values for gas constant R, depending on your units for pressure and volume.

(Standard) 'A' Level Qn.

(Challenging) 'O' Level Qn.

For gases, there exists a formula that describes the relationship of the variables of Pressure, Volume, no. of moles, Temperature :

P V = n R T (where R is a constant mathematical value).

If these units (pressure in Pa; volume in m³) are used, the value of the gas constant R is 8.314. If these units (pressure in atm; volume in dm³) are used, the value of the gas constant R is 0.08206. Temperature is always measured in Kelvins (where 298K = 25°C; 273K = 0°C).

a) A bubble of gas (initial diameter 1.585 cm; mass 2.0973 x 10^-2 g) that contains twice as much oxygen as it does carbon dioxide, emerges from a photosynthesizing aquatic plant (there’s a plesiosaur reptile aka “Loch Ness monster” right next to the plant, btw) at the bottom of a lake, where the temperature is really cold and the pressure is 6.4 times greater than at the lake’s surface; and the bubble rises until it reaches the surface of the lake (which happens to be at sea level, and at that moment is equivalent to room temperature), and bursts “pop!”.

(i) Calculate the diameter (in cm) of the bubble just before it bursts.

(ii) Calculate the body temperature (in °C) of the plesiosaur at the time the bubble emerges.

b) At sea level, standard atmospheric pressure (ie. 1.01325 x 10⁵ Pa) causes mercury in a dish to rise 760 mm up a glass column. A mixture of two alkanes (with molar masses 16.0 g and 30.0 g respectively) is stored in a container at 294 mmHg. The gases undergo complete combustion to produce CO₂ that has a pressure of 356 mmHg when measured at the same temperature and volume as the original mixture. Calculate the percentage composition of the mixture.

Solution :

ai) 3.0cm3

aii) 6 deg C

b) 21.1% and 78.9%

Bigcable22 SG

as for my question first hor i usedp = 101000pa

V=(4.6 X 10^ -5) m^3

m=0.16g

R=8.314 (so do i use the one for stp and rtp?)

T=373K

my answer i get like 106.7 but the answer is 106.4

???

eagle SG

Interesting

didn't know A lvl chem also hv PV = nRT formula, just like A lvl physics, Ideal Gas

Bigcable22 SG

yeah gaseous state but physics learn more, in chem no need to learn pv=NkT and 1/2m<c2> stuffs

UltimaOnline SG

as for my question first hor i usedp = 101000pa

V=(4.6 X 10^ -5) m^3

m=0.16g

R=8.314

T=373K

my answer i get like 106.7 but the answer is 106.4

???

Your working is ok. It's a silly issue about significant figures.

Student answers are slightly different when you use

1) 3 sig figs throughout

2) 4 sig figs for intermediate working and 3 sig figs for final answer

3) stored-in-memory calculator's 10 sig fig values for all calculations throughout.

Furthermore, in this context of gas laws,

R = 8.314472 (pressure in Pa, volume in m3)

R = 0.0820574587 (pressure in atm, volume in dm3)

But when you use 3 sig fig values for (arguably "intemediate") calculations, of R = 8.31 or 0.0821, you end up with a slightly different 3 sig fig answer.

In the actual 'A' level exams, Cambridge should not penalize students for this issue in the mark scheme (as this is not an issue of Chemistry, but mathematical 'nomenclature'), as long as the student doesn't try to annoy the marker by spamming 10 sig fig values on the answer booklet.

Nonetheless, with regards to gas constant R, since it's included in the Data Booklet, use the value of R = 8.31 as stated in the Data Booklet.

Similarly, although 1 atm = 1.01325 x 10^5 Pa, but if you use the 3 sig fig value of 1.01 x 10^5, you will end up with a slightly different answer.

(You should be as consistent as possible; Bigcable22 you used 4 sig fig for gas constant R but used 3 sig fig for converting atm to Pa; either use 4 sig fig each, or 3 sig fig each.)