Question

1) A student filled a 250 mL container with an unknown gas until a pressure of 760 torr was obtained. The sample of gas was found to weigh 0.164 g. Calculate the molecular mass of the gas if the temperature in the laboratory was 25 C

2) A student collected 35.0 mL of O2 over water at 25 C and a total pressure of 745 torr from the decomposition of a 0.2500 g sample known to contain a mixture of KCl and KClO3. The reaction that produced the oxygen was KClO3 --> 2KCl + 3O2.

a) How many moles of O2 were collected?

b) How many grams of KClO3 were decomposed?

c) What percentage by mass of the same was 2KClO3 ? ( Take H2O vapor pressure at this temperature to be 23.8 torr).

Answer 1

Solution1: Volumeof the container=250ml

Pressure=760 torr

R(gas constant)=62.363 L torr/K mol

Mass of the gas. = 0.164 g

Ideal Gas equation

PV = nRT

760×250×10^-3= 0.164×62.363 ×298/molecular mass

Molecular mass =0.164×62.363×298/760×250×10^-3

= 16.04

Solution 2:(a). 2KClO3----------->2KCl +3O2

No. Of moles of O2 were collected=?

Total Pressure= Pressure of O2+Pressure of water

= 7.55 +23.8 =31.30 torr

No. Of moles of O2 collected= PV/RT

. = 31. 30×35×10^-3/62.363×298

= 0.058 moles

(b) From the equation: hi

KClO3------------>KCl+3/2O2

Gmm of KClO3=( 39+35.5+48)=122.5

Mass of KClO3= no.of moles of O2×1mol of KClO3×molecular mass of KClO3/1.5 moles of O2

0.058×122.5/1.5 =4.73 g

(C) % of 2KClO3 by mass=2×4.630×100/122.5

= 7.55%