Question

1. Nitrogen gas can be obtained by decomposing ammonium nitrate at high temperatures.  The nitrogen gas is collected over water in a 500 mL (three significant figures) flask at 19℃.  The ambient pressure is 745 mmHg.  (Vapor pressure of water at that temperature is 16.48 mmHg). 
2. What is the partial pressure of nitrogen?
3. How many moles of water are there in the wet gas?
4. How many moles of dry gas are collected?
5. If 0.128 g of Ne are added to the flask at the same temperature what is the partial pressure of neon in the flask?
6. What is the total pressure after Ne is added?

Answer 1

I: Partial pressure of Nitrogen

Ambient pressure is the total pressure = 745mmHg.

Vapour pressure of water = 16.48mmHg

According to Dalton's Law of partial pressure;

P_(Total) = P₁ + P₂ + .......

Hence;

P_(Total) = P_N2 + P_H2O

745mmHg = P_N2 + 16.48mmHg

PN2 = (745 - 16.48) mmHg

= 728.52mmHg

II: Moles of water in wet gas

Given that;

PV = nRT

and;

P = 16.48mmHg

V = 500mL = 500x10^-3L

R = 62.36367LmmHgK^-1mol^-1

T = 19^oC + 273 = 292K

n=?

Therefore;

n$= \dfrac{PV}{RT} = \dfrac{16.48mmHgx 500x10^-3L}{62.36367LmmHgmol^-1K^-1x292K}$ = 4.525x10^-4 mol

III: Moles of dry gas collected

Since; PV=nRT

and;

P= 728.52mmHg

V= 500x10^-3L

T = 292K

n $=\dfrac{PV}{RT} = \dfrac{728.52mmHgx500x10^-3L}{62.36367LmmHgmol^-1K^-1x292K} =0.02mol$

IV: Partial pressure of Neon

Neon added = 0.128g

Moles of neone $\dfrac{mass}{Molar mass} = \dfrac{0.128g}{20.18g/mol}$ = 6.34 x 10^-3 mol

Since PV=nRT

where n=6.34 x 10^-3 mol

R = 62.36367LmmHgK^-1mol^-1

T = 19^oC + 273 = 292K

V= 0.5L

P $=$ $\dfrac{nRT}{V} = \dfrac{6.34x10^-3molx62.36367LmmHgK^-1K^-1x292K}{0.5L}$ = 230.9mmHg

V: Total pressure after Neon is added

P_(total) = P_H2O + P_N2 + P_Ne

= (16.48 + 728.52 + 230.9)mmHg

=975.9mmHg