1 g of a gaseous compound of boron & oxygen occupies 0.80 L at 1 atm and 303 K. The compound is .
(R=0.0820 $L a t m m o l^{- 1} K^{- 1}$ , Z: H=1.0, B=10.8)

B H_{3}

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B_{2} H_{6}

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B_{3} H_{12}

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Solution

The correct option is B $B_{2} H_{6}$
Ideal gas equation is $P V = n R T$ ------->(1)

$n = \frac{m}{M}$ -------->(2)

applying (2) in (1),

$P V = \frac{m}{M} \times R T$ ------->(3)

Rearranging equation (3)

$M = \frac{m R T}{P V}$

$= \frac{1 \times 0.082 \times 303}{1 \times 0.82}$

$= 30.3 g$

This molar mass corresponds to diborane( $B_{2} H_{6}$ ).

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