Question

Molality: It is defined as the number of moles of the solute present in 1 kg of the solvent. It is denoted by m
\[molality(m)=\dfrac{Number\ \text{of}\ \text{moles}\ \text{of}\ \text{solute}}{Number\ \text{of}\ \text{kilograms}\ \text{of}\ \text{the}\ \text{solvent}}\]
Let ${{w}_{A}}$ grams of the solute of molecular mass ${{m}_{A}}$ be present in ${{w}_{B}}$ grams of the solvent, then
$molality(m)=\dfrac{{{w}_{A}}}{{{m}_{A}}\times {{w}_{B}}}\times 1000$
Relation between mole fraction and molality:
${{X}_{A}}=\dfrac{n}{n+N}$and ${{X}_{B}}=\dfrac{n}{n+N}$
$\dfrac{{{X}_{A}}}{{{X}_{B}}}=\dfrac{n}{N}=\dfrac{Moles\ \text{of}\ \text{solute}}{Moles\ \text{of}\ \text{solvent}}=\dfrac{{{w}_{A}}\times {{m}_{B}}}{{{w}_{B}}\times {{m}_{A}}}$
$\dfrac{{{X}_{A}}\times 1000}{{{X}_{B}}\times 1000}=\dfrac{{{w}_{A}}\times 1000}{{{w}_{B}}\times 1000}=m$or $\dfrac{{{X}_{A}}\times 1000}{(1-{{X}_{A}}){{m}_{B}}}=m$
If the ratio of the mole fraction of a solute is changed from $\dfrac{1}{3}$to $\dfrac{1}{2}$in the 800g of solvent then, the ratio of molality will be:
A. 1:3
B. 3:1
C. 4:3
D. 1:2

Answer 1

Hint: Molality is also known by the name molal concentration. It generally measures the solute concentration in a solution. The solution is composed of two components called solute and solvent. There are many different ways to express the concentration of solutions like molarity, molality, normality, formality, volume percentage, weight percentage and part per million. Mole fraction represents the number of molecules of a particular component in a mixture divided by the total number of moles in the given mixture. This is a general way of expressing the concentration of a solution.

Complete Step by step solution: The molar fraction can be represented by X. If the solution consists of components A and B then the mole fraction is given by the
$Mole\ \text{fraction}\ \text{of}\ \text{solute, }{{\text{X}}_{\text{A}}}\text{=}\dfrac{Moles\ \text{of}\ \text{solute}}{Moles\ \text{of}\ \text{solute+Moles}\ \text{of}\ \text{solvent}}=\dfrac{n}{N+n}$
And mole fraction of solvent is given by
$Mole\ \text{fraction}\ \text{of}\ \text{solvent, }{{\text{X}}_{\text{B}}}\text{=}\dfrac{Moles\ \text{of}\ \text{solute}}{Moles\ \text{of}\ \text{solute+Moles}\ \text{of}\ \text{solvent}}=\dfrac{N}{N+n}$
$\dfrac{{{X}_{A}}}{{{X}_{B}}}=\dfrac{n}{N}=\dfrac{Moles\ \text{of}\ \text{solute}}{Moles\ \text{of}\ \text{solvent}}=\dfrac{{{w}_{A}}\times {{m}_{B}}}{{{w}_{B}}\times {{m}_{A}}}$
$\dfrac{{{X}_{A}}\times 1000}{{{X}_{B}}\times 1000}=\dfrac{{{w}_{A}}\times 1000}{{{w}_{B}}\times 1000}=m$
\[\dfrac{{{X}_{A}}\times 1000}{{{X}_{B}}\times {{m}_{B}}}=\dfrac{{{w}_{A}}}{{{w}_{B}}}\times \dfrac{1000}{{{m}_{A}}}=molality\]
Or $\dfrac{{{X}_{A}}\times 1000}{(1-{{X}_{A}})}=m$
Initially ${{X}_{A}}=\dfrac{1}{3}$
Therefore, ${{m}_{1}}=\dfrac{\dfrac{1}{3}\times 1000}{1-\dfrac{1}{3}}=\dfrac{1000}{2}=500$
When ${{X}_{A}}=\dfrac{1}{2}$
Then ${{m}_{2}}=\dfrac{\dfrac{1}{2}\times 1000}{1-\dfrac{1}{2}}=1000$
Therefore, $\dfrac{{{m}_{1}}}{{{m}_{2}}}=\dfrac{500}{1000}=\dfrac{1}{2}$= 1:2

Hence option D is the correct answer.

Note: It is an important point to note that mole fraction represents a fraction of molecules so we can say that different molecules have different masses therefore the mole fraction is also different in their cases it is also a different term from the mass fraction.