Question

A sample of nitrogen gas, n2, is collected in a100 ml container at a pressure of 688 mm hg and a temperature of 565 °c. how many grams of nitrogen gas are present in this sample?

Answer 1

V= 100 mL = 0.100 L
P=688 mm Hg
T=565°C =565+273=838 K
M(N2)= 2*14.0 g/mol
R=62.36 L·mmHg·K⁻¹·mol⁻¹

$PV= \frac{m}{M} *RT$

$m= \frac{PVM}{RT}$$= \frac{688*0.100*14.0}{62.36*838} =0.0184 g$

Answer 2

Answer: The mass of nitrogen gas present in the given sample is 0.037 g

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

$PV=nRT$

Or,

$PV=\frac{m}{M}RT$

where,

P = Pressure of the gas = 688 mmHg

V = Volume of gas = 100 mL = 0.1 L    (Conversion factor:  1 L = 1000 mL)

m = Mass of nitrogen gas = ?

M = Molar mass of nitrogen gas = 28 g/mol

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

T = Temperature of the gas = $565^oC=[565+273]=838K$

Putting values in above equation, we get:

$688mmHg\times 0.1L=\frac{m}{28g/mol}\times 62.3637\text{L. mmHg }mol^{-1}K^{-1}\times 838K\\\\m=0.037g$

Hence, the mass of nitrogen gas present in the given sample is 0.037 g