An amount of 1.00 g of a gaseous compound of boron and hydrogen occupies $0.820 L$ at 1.00 atm and at $3^{\circ} C$ . The compound is : $(R = 0.0820 L {atm mole}^{- 1} K^{- 1}$ ; at. wt : $H = 1.0, B = 10.8)$

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${BH}_{3}$

$B_{2} H_{6}$

$B_{4} H_{10}$

$B_{3} H_{12}$

answer is C.

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Detailed Solution

$\begin{array}{l} PV = nRT \\ PV = \frac{m}{M R T} \\ \Rightarrow M = \frac{mRT}{PV} = \frac{1 \times 0.082 \times 276}{1 \times 0.82} = 27.6 g \end{array}$

Its molecular formula is $B_{2} H_{6}$

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