If the partition is removed the average molar mass of the sample will be (Assume ideal behaviour) H2 D2 16.42 L 16.42 L 300 K 300 K 3 atm 6 atm

Question

If the partition is removed the average molar mass of the sample will be (Assume ideal behaviour) H2 D2 16.42 L 16.42 L 300 K 300 K 3 atm 6 atm (A) (1/2) g / m o l (B) (10/3) g / mol (C) (3/2) g / mol (D) (5/3) g / mol

Tardigrade Admin · Accepted Answer

Moles of H 2=(3 × 16.42/0.0821 × 300)=2 Moles of D 2=(6 × 16.42/0.0821 × 300)=4 Average molecular weight =(2 × 2+4 × 4/4+2) =(10/3)

$H_{2}$	$D_{2}$
$16.42 L$	$16.42 L$
$300 K$	$300 K$
$3 a t m$	$6 a t m$

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

$H_{2}$ $D_{2}$

$16.42 L$ $16.42 L$

$300 K$ $300 K$

$3 a t m$ $6 a t m$

$\frac{1}{2} g / m o l$

$\frac{10}{3} g / m o l$

$\frac{3}{2} g / m o l$

$\frac{5}{3} g / m o l$

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

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

$= \frac{10}{3}$

Q. If the partition is removed the average molar mass of the sample will be (Assume ideal behaviour) H2​ D2​ 16.42L 16.42L 300K 300K 3atm 6atm

21​g/mol

310​g/mol

23​g/mol

35​g/mol