Question

An unknown gas shows a density of 3 g per litre at ${273}^{\circ }$ and 1140 mm Hg pressure. What is the gram molecular mass of this gas?

Answer 1

Given:
P= 1140 mm Hg
Density = D = 3 g / L
T = 273 0C = 273+273 = 546 K
M = ?

We know that, at STP, the volume of one mole of any gas is 22.4 L
Hence we have to find out the volume of the unknown gas at STP.

First, apply Charle’s law.
We have to find out the volume of one litre of unknown gas at standard temperature 273 K.

$V_1=1L$
$T_1=546K$
$V_2=?$
$T_2=273K$
$\frac{V_1}{T_1}=\frac{V_2}{T_2}$
$V_2=\frac{(V_1\times T_2)}{T_1}=\frac{(1L\times 273K)}{546K}=0.5L$

We have found out the volume at standard temperature. Now we have to find out the volume at standard pressure.

Apply Boyle’s law.
$P_1=1140mmHg$
$V_1=0.5L$
$P_2=760mmHg$
$V_2=?$
$P_1\times V_1=P_2\times V_2$
$V_2=\frac{(P_1\times V_1)}{P_2}=\frac{(1140mmHg\times 0.5L)}{760mmHg}=0.75L$

Now, 22.4 L is the volume of 1 mole of any gas at STP, then 0.75 L is the volume of X moles at STP
$Xmoles=\frac{0.75L}{22.4L}=0.0335moles$
The original mass is 3 g
$n=\frac{m}{M}$
$0.0335moles=\frac{3g}{M}$
$M=\frac{3g}{0.0335moles}$
M= 89.55 g / mole
Hence, the gram molecular mass of the unknown gas is 89.55 g