Question

$1\text{g}$ of a gaseous compound of boron and hydrogen occupies $0.820\text{L}$ at $1\text{atm}$ and $3{}^{\circ }\text{C}$. The compound is:
$(R=0.0820{\text{L atm mol}}^{-1}{\text{K}}^{-1};\text{molar mass of}H=1\text{g},B=10.8\text{g}$

• A. $B{H}_3$
• B. $B_4{H}_{10}$
• C. $B_2{H}_6$
• D. $B_2{H}_{12}$

Answer 1

The correct option is C $B_2{H}_6$

We know,
$PV=nRT$
$\Rightarrow PV=\frac{m}{M}RT$
Where, $m=\text{mass of the compound}=1\text{g}$
$M=\text{molar mass of the compound}$
$T=\text{temperature}=276\text{K}$

$\Rightarrow M=\frac{mRT}{PV}=\frac{1\times 0.082\times 276}{1\times 0.82}=27.6\text{g}$

The compound with molar mass $27.6\text{g}$ among the given compounds is $B_2{H}_6$ .