Question

An X molal solution of a compound in benzene has a mole fraction of solution equal to 0.2. The value of X is:
(A) 14
(B) 3.2
(C) 1.4
(D) 2

Answer 1

Hint: To answer this question we should know the formula to calculate molarity. 1 molal solution contains 1 mole of solute for every 1 Kg of solvent, this help us remember the concept and the formula.

Complete answer:
Given: The mole fraction of solute in benzene = 0.2
            The molality = X
            The solvent is benzene.
The molecular mass of benzene = 12 $\times $6 + 1 $\times $6
                                                         =72 + 6
                                                        = 78g/mol
The sum total mole fraction of solvent and solute = 1
Mole fraction of solute + mole fraction of solvent = 1
So, Mole fraction of solvent = 1 - Mole fraction of solute
                                                   = 1 - 0.2
                                                   = 0.8
Mole fraction of solute (${{X}_{1}}$) = $\frac{{{n}_{1}}}{{{n}_{1}}+{{n}_{2}}}$
                                         0.2 = $\frac{{{n}_{1}}}{{{n}_{1}}+{{n}_{2}}}$ ………………..equation 1
Where, ${{n}_{1}}$= number of moles of moles of solute.
              ${{n}_{2}}$= number of moles of moles of solvent
Mole fraction of solvent (${{X}_{2}}$) = $\frac{{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}$
                                            0.8 = $\frac{{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}$………………..equation 2
Now , let's divide equation 1 and 2
\[\frac{0.2}{0.8}=\frac{{{n}_{1}}}{{{n}_{2}}}\]……………..equation 3
Molarity defined as the number of moles in 1Kg of solvent .
Molarity = $\frac{moles\text{ }of\text{ }solute~}{kilogram\text{ }of\text{ }solvent}$
Now, let's calculate the number of moles of solvent.
Moles of solvent = $\frac{weight\text{ }in\text{ }g}{Molecular\text{ }weight}$
We know that 1Kg = 1000g
so, let's consider weight of solvent in g is 1000g
Moles of solvent = $\frac{1000}{78}$
                               = 12.8205
Therefore, ${{n}_{2}}$ = 12.8205
substituting ${{n}_{2}}$ in equation 3
\[\frac{0.2}{0.8}=\frac{{{n}_{1}}}{12.8}\]
${{n}_{1}}$ = 3.2
Moles of solute present in 1 Kg of solvent is 3.2
Therefore, the molarity is 3.2.

Thus, option B is the correct answer.

Note:
$Molarity=\frac{{{X}_{1}}\times 1000}{{{X}_{2}}}\times mol.\text{ }wt.\text{ }of\text{ }solvent$
Where, ${{X}_{1}}$= Mole fraction of solute
              ${{X}_{2}}$= Mole fraction of solvent
This is the actual relationship used, but the defines the same meaning that is moles of solute in 1Kg of solvent. So, the irony from this is it very important to understand the concept.