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 $3^{0}C$. The compound is (R=0.0820 liter atm $mol{e}^{-10}{K}^{-1};$ at. wt:H=1.0,B=10.8)

• A. $B{H}_3$
• B. $B_4{H}_{10}$
• C. $B_2{H}_6$
• D. $B_3{H}_{12}$

Answer 1

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

Given,

volume - $0.820l$

mass - $1.00g$

pressure - $1.00atm$

temperature - $3^{O}C$

Using ideal gas equation - $PV=nRT$

$n$ - number of mole = $\frac{m}{M}$

$m$ - given mass

$M$ - molar mass

$PV=\frac{m}{M}RT$

here, find the molar mass of the compound

$M=\frac{mRT}{PV}$

Firstly, change the temperature in Kelvin.

$3^{O}C+273=276K$

Now, put the value in the formula.

$M=\frac{1\times 0.0820\times 276}{1\times 0.820}$

= $27.6g/mole$

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

(a) $B{H}_3$ - $10.8+1\times 3=13.8$

(b) $B_4{H}_{10}$ - $4\times 10.8+1\times 10=53.2$

(c) $B_2{H}_6$ - $2\times 10.8+6\times 1=27.6$

(d) $B_3{H}_{12}$ - $3\times 10.8+12\times 1=44.4$

So, option (c) is the correct answer.