Question

A gas is found to have the formula ${\left( {{{\text{C}}_3}{{\text{O}}_2}} \right)_{\text{n}}}$. Its vapor density is $34$. The value of ${\text{n}}$ will be

Answer 1

Hint : Vapor density is the density of vapor with respect to that of hydrogen. It can also be explained as the mass of ‘n’ number of molecules of certain volume divided by the mass of ‘n’ number of molecules of same volume of hydrogen. It has no unit.

Complete step by step solution :
Vapor density can be defined as molar mass of a gas divided by the molar mass of hydrogen. Molar mass of hydrogen is $2.016{\text{g}}$. Thus it can be concluded that the vapor density is the half of the molar mass of the given gas. Molar mass can be defined as the double of the vapor density.
Here the formula mass of ${\left( {{{\text{C}}_3}{{\text{O}}_2}} \right)_{\text{n}}}$can be calculated by adding the atomic masses.
Atomic mass of carbon=12, atomic mass of oxygen=16
There are three carbons and two oxygens.
Thus the formula mass of {\left( {{{\text{C}}_3}{{\text{O}}_2}} \right)_{\text{n}}}$$$ = 3 \times 12 + 2 \times 16 = 68{\text{g}}$ Vapor density is given as 34. Molar mass of the given compound has to be calculated. Molar mass is the double of vapor density. Molar mass represented by ‘m’ $ = 2 \times {\text{d}}$, {\text{d}}$is the vapor density.${\text{m}} = 2 \times 34 = 68{\text{g}}$Formula mass is calculated as$68{\text{g}}$$$$$$
Therefore by dividing the molar mass and the formula mass, we get the value of n.
Therefore $n = \dfrac{{68{\text{g}}}}{{68{\text{g}}}} = 1$

Additional information:
Vapor density is also defined with respect to air. It is defined as above at the same temperature and pressure. It is an expression of the density of vapor or gas.

Note : If a gas has vapor density less than air, it will generate in air. Some examples are acetylene, methane, hydrogen. If the vapor density is greater than one it will sink in air. Some examples are propane, hydrogen sulfide, ethane, butane, chlorine, sulfur dioxide.