Question

A gaseous mixture enclosed in a vessel of volume ${}^{\prime }{V}^{\prime }$ consists of $2$ gram mole of gas $A$ and $4$ gram mole of gas $B$ at a certain temperature $T$. The gram molecular weights of the gases $A$ and $B$ are $4$ and $16$ respectively. The molecular weight of the mixture will be

• A. $12\text{g}$
• B. $6\text{g}$
• C. $4\text{g}$
• D. $32\text{g}$

Answer 1

The correct option is A $12\text{g}$

Given,
Molecular weight of gas $A,\left(M_A\right)=4\text{g}$
Molecular weight of gas $B,\left(M_B\right)=16\text{g}$
No. of moles of gas $A$ in mixture $\left(n_A\right)=2$
No. of moles of gas $B$ in mixture $\left(n_B\right)=4$
Molecular weight of the mixture will be given by
$M=\frac{n_A{M}_A+n_B{M}_B}{n_A+n_B}=\frac{2\left(4\right)+4\left(16\right)}{2+4}$
$=12\text{g}$