Question

If the partition is removed the average molar mass of the sample will be (Assume ideal behaviour)

$H_2$	$D_2$
$16.42 \,L$	$16.42 \,L$
$300\,K$	$300\,K$
$3\,atm$	$6\,atm$

• A. $\frac{1}{2} g / m o l$
• B. $\frac{10}{3} g / mol$
• C. $\frac{3}{2} g / mol$
• D. $\frac{5}{3} g / mol$

Answer 1

Answer: (B)

Moles of $H _{2}=\frac{3 \times 16.42}{0.0821 \times 300}=2$

Moles of $D _{2}=\frac{6 \times 16.42}{0.0821 \times 300}=4$

Average molecular weight

$=\frac{2 \times 2+4 \times 4}{4+2}$

$=\frac{10}{3}$