Question

Mole fraction of ${C_3}{H_5}{(OH)_3}$ in a solution of 36 g of water and 46 g glycerine is:
A. 0.46
B. 0.36
C. 0.20
D. 0.40

Answer 1

Hint: Mole fraction of a compound is determined by first calculating the moles of the compound present in the mixture. The moles of the compound can be calculated if the mass of the compound is given by dividing it with the molecular mass of the compound.

Complete step by step answer:
Given,
The mass of water is 36 g.
Mass of glycerine is 46 g
${C_3}{H_5}{(OH)_3}$ is the molecular formula of glycerine.
The formula for calculating the moles is shown below.
$n = \dfrac{m}{M}......(i)$
Where
n is the number of moles of the compound.
m is the mass of the compound.
M is the molar mass of the compound.
The molecular weight of water is 18.
To calculate the moles of water, substitute the values of mass and molecular weight in equation (i).
$n = \dfrac{{36g}}{{18g/mol}}$
$\Rightarrow n = 2mol$
The molecular weight of glycerine is 92 g/mol.
To calculate the moles of glycerine substitute the values in the above equation.
$n = \dfrac{{46g}}{{92g/mol}}$
$\Rightarrow n = 0.5mol$
Mole fraction of a compound is determined by dividing the moles of a compound with the total moles of the compound present in the mixture.
The formula for calculating the mole fraction of a compound is shown below.
${X_A} = \dfrac{{{m_A}}}{{{m_A} + {m_B}}}......(ii)$
Where,
${X_A}$ is the mole fraction of the compound.
${m_A}$ is moles of A.
${m_B}$ is moles of B.
To calculate the mole fraction of glycerine, substitute the values in equation (ii).
\[{X_G} = \dfrac{{0.5}}{{0.5 + 2}}\]
$\Rightarrow {X_G} = 0.2$
Thus, the mole fraction of ${C_3}{H_5}{(OH)_3}$ in a solution of 36 g of water and 46 g glycerine is 0.2.

Therefore, the correct option is C

Note:
${C_3}{H_5}{(OH)_3}$ and ${C_3}{H_8}{O_3}$both are the molecular formula of glycerine. For calculating the mole fraction of water, moles of water is divided by the total moles present in the mixture.