Question

0.24gm of a volatile substance upon vaporization gives 45ml vapours at STP. What will be the vapour density of the substance?

Answer 1

Hint: We know that vapour density is the density of a vapour in accordance to that of hydrogen. We can calculate the vapour density by dividing the molecular weight by two. We can calculate the molecular weight using the ideal gas equation.
We can calculate the vapour density by using the formula,
Vapour density$ = \dfrac{{{\text{Molecular weight}}}}{2}$

Complete step by step answer:
Given data contains,
Mass of volatile substance is $0.24gm$.
Volume of the vapour is $45ml$.
We have to know that vapour density is the weight of volume of pure gas compared to equal volume of dry air at the same pressure and temperature. We can determine the vapour pressure by dividing the molecular weight of the vapour by the average molecular weight of air. So it is unit less.
Let us now calculate the vapour density using the mass of the volatile substance and volume of vapour at STP. Since, they have mentioned the volume at standard pressure and temperature. We can take the pressure as one atmosphere and the temperature as $273K$.
We know the equation for ideal gas as,
$PV = nRT$
Where
P indicates the pressure of the gas.
V indicates the volume of the gas.
n indicates the number of moles of the gas.
R indicates the universal gas constant.
T indicates the temperature in Kelvin.
We can write the given volume in liters as \[\dfrac{{45}}{{1000}}\].
We know that the number of moles is given by the equation,
$n = \dfrac{{{\text{Given mass}}}}{{{\text{Molar mass}}\left( M \right)}}$
The number of moles is written as $\dfrac{{0.24}}{M}$.
The value of gas constant R is written as $R = 0.082\dfrac{{atm \cdot L}}{{mol \cdot K}}$.
The value of temperature is $273K$.
Let us now substitute these values in the ideal gas law equation.
$PV = nRT$
$1atm \times \dfrac{{45}}{{1000}}L = \dfrac{{0.24}}{{mol}} \times 0.082\dfrac{{atm \cdot L}}{{mol \cdot K}} \times 273K$
$M = 119.4g$
The molecular weight is $119.4g$.
From the molecular weight, we can calculate the vapour density using the formula
$M = 2 \times {\text{Vapour density}}$
On rearranging, the above equation, we get,
Vapour density$ = \dfrac{{{\text{Molecular weight}}}}{2}$
Let us now substitute the value of molecular weight to calculate vapour density.
Vapour density =$ \dfrac{{119.4g}}{2}$
Vapour density = $59.7$
The vapour density is $59.7$.

Note:
We can also calculate the vapour density from the molar volume.
We know that 1mole of any gas occupies $22.4L$ at standard pressure and temperature.
$0.24gm$ of a volatile substance would give $45ml$ vapours at standard pressure and temperature.
$22400ml$ at standard temperature and pressure would give $\dfrac{{0.24g}}{{45ml}} \times 22400ml = 119.4g$
The molecular weight is $119.4g$.
We know that vapour density is half of the molecular weight.
Vapour density$ = \dfrac{{{\text{Molecular weight}}}}{2}$
Let us now substitute the value of molecular weight to calculate vapour density.
Vapour density$ = \dfrac{{119.4g}}{2}$
Vapour density = $59.7$
The vapour density is $59.7$.