Question

0.24g of a volatile substance upon vapourisation. gives 45 ml vapour .what will be the vapour density of the substance?

Answer 1

First find out molar mass of the substance
Mw= 22400×0.24/45= 119.46
VD= Mw/2= 119.46/2 = 59.73

Answer 2

The substance's vapor density is 59.75

Explanation:

According to the ideal gas equation,

$P V=n R T \rightarrow(1)$

Where,

P = Pressure

T = Temperature

V = Volume

From the given,

Pressure, P = 1 atm

Volume, V = $45 \ \mathrm{ml} =\frac{45}{1000}$

Number of moles, n = $\frac{\text {Given mass}}{\text {Mass}}=\frac{0.24}{M}$

Gas constant R = 0.082 atm.L/mol.K

Temperature = 273 K

Substitute the all values in equation (1)

$1 \times \frac{45}{1000}=\frac{0.24}{M} \times 0.082 \times 273$

$M=119.392$

$M=2 \times \text{Vapor density}$

$\text{Vapor density}=\frac{119.392}{2}=59.75$

Therefore, the vapour density of substance is 59.75