Question

In Victor Mayer's method, $0.2$ g of a volatile compound on volatilisation gave $56$ ml of vapour at STP. Its molecular weight is :

• A. 40
• B. 60
• C. 80
• D. 120

Answer 1

The correct option is C 80

We can use the Ideal Gas Law to solve this problem

$PV=nRT$

Since $n=\frac{m}{M}$, we can rearrange this equation to get

$PV=\left(\frac{m}{M}\right)RT$

And we can solve this equation to get

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

STP is defined as 1 bar and 0 °C.

In this problem,

m=0.2 g

R=0.083 14 bar⋅L⋅K${}^{-1}$mol${}^{-1}$

T=273.15 K

p=1 bar

V=56 mL=0.056 L

∴ $M=\frac{0.2g\times 0.08314bar\cdot L\cdot K^{-1}mo{l}^{-1}\times 273.15K}{1bar\times 0.056L}=80$ g/mol

∴ The molar mass of the compound is 80 g/mol, so the molecular mass is 80 u.

Note: The answer can have only one significant figure, because that is all you gave for the mass of the compound.

Hence, the correct option is $C$