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Question

An amount of 1.00 g of a gaseous compound of borbon and hydrogen occupies 0.820 liter at 1.00 atm and at 30C. The compound is (R=0.0820 liter atm mole−10K−1; at. wt: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 C B2H6

Given,

volume - 0.820 l

mass - 1.00 g

pressure - 1.00 atm

temperature - 3OC

Using ideal gas equation - PV=nRT

n - number of mole = mM

m - given mass

M - molar mass

PV=mMRT

here, find the molar mass of the compound

M=mRTPV

Firstly, change the temperature in Kelvin.

3OC+273=276K

Now, put the value in the formula.

M=1×0.0820×2761×0.820

= 27.6 g/mole

write each compound molar mass according to the given value and match it with this value.

(a) BH3 - 10.8+1×3=13.8

(b) B4H10 - 4×10.8+1×10=53.2

(c) B2H6 - 2×10.8+6×1=27.6

(d) B3H12 - 3×10.8+12×1=44.4

So, option (c) is the correct answer.


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