An amount of 1.00 g of a gaseous compound of boron and hydrogen occupies 0.820 litre at 1.00 atm and at 3∘C. The compound is (R=0.0820L atm mol−1K−1, Atomic weight: H = 1.0, B = 10.8)
A
BH3
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B
B4H10
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C
B2H6
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D
B3H12
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Solution
The correct option is CB2H6 PV=nRTPV=mMRT⇒M=mRTPV=1×0.082×2761×0.82=27.6g
27.6 g correspond to molar mass of B2H6
Hence correct option is c.