Question

A gaseous mixture contains $C{H_4}$ and ${C_2}{H_6}$ in equimolar proportion. The weight of $1.12L$ of this mixture STP is:
(A) $2.3g$
(B) $0.8g$
(C) $1.15g$
(D) $11.5g$

Answer 1

Molecules are the combination of atoms. Given molecules are methane and ethane both are alkanes. Gaseous mixture consists of $C{H_4}$ and ${C_2}{H_6}$ in equimolar proportions. The weight of each molecule can be calculated from the number of moles and molar mass. The sum of the weight of the two molecules gives the weight of the mixture.

Complete answer:
The mixture consists of the molecule’s methane $\left( {C{H_4}} \right)$ and ethane $\left( {{C_2}{H_6}} \right)$ in equimolar proportion.
STP means standard conditions of pressure, temperature and volume.
At STP, one mole of a gas occupies $22.4L$
Given mixture consists of $1.12L$
Thus, $1.12L$ of mixture consists of $\dfrac{{1.12}}{{22.4}} = 0.05moles$
From the above calculation, the mixture consists of $0.05$ moles of methane and ethane in equimolar proportions.
The number of moles of methane will be $0.025$ moles
The number of moles of ethane will be $0.025$ moles
The molar mass of methane is $16$ . The molar mass of ethane is $30$ .
The amount of methane present in mixture is $0.025 \times 16 = 0.4g$
The amount of ethane present in the mixture is $0.025 \times 30 = 0.75g$
The weight of the mixture can be calculated by the sum of the weight of the methane and ethane which is equal to $0.4 + 0.75 = 1.15g$
Thus, the weight of the mixture is $1.15g$ .
Option C is the correct one.

Note:
The number of moles is the ratio of the amount of the substance to the molar mass of that substance. The molar mass has the units of $gm{\left( {mol} \right)^{ - 1}}$ and the number of moles has moles. Thus, the amount of substance will have units of grams.