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) =$\frac{Numberofmolesofsolute}{Numberofkilogramsofthesolvent}$
Let $w_A$ grams of the solute of molecular mass $m_A$ be present in $w_B$ grams of the solvent, then
Molality (m) = $\frac{w_A}{m_A\times w_B}\times 1000$
Relation between mole fraction and Molality :
$X_A=\frac{n}{N+n}and{X}_B=\frac{N}{N+n}$
$\frac{X_A}{X_B}=\frac{n}{N}=\frac{Molesofsolute}{Molesofsolvent}=\frac{W_A\times m_B}{w_B\times m_A}$
$\frac{X_A\times 1000}{X_B\times m_B}=\frac{w_A\times 1000}{W_B\times m_A}=m$, or, $\frac{X_A\times 1000}{(1-X_A)m_B}=m$

If the ratio of the mole fraction of a solute is changed from $\frac{1}{3}$ to $\frac{1}{2}$ in the $800$ g of solvent then, the ratio of molality will be :

• A. 1:3
• B. 3:1
• C. 4:3
• D. 1:2

Answer 1

The correct option is D 1:2

The expression for the molality is $\frac{X_A\times 1000}{(1-X_A)m_B}=m$.
For two different mole fractions of A, i.e. 1/2 and 1/3, the expression becomes,
$\frac{\frac{1}{2}\times 1000}{(1-(\frac{1}{2}\left)\right)m_B}=m_1$ ......(1)
$\frac{\left(\frac{1}{3}\right)\times 1000}{(1-(\frac{1}{3}\left)\right)m_B}=m_2$......(2)
Divide equation (2) with equation (1).
$\frac{\frac{\left(\frac{1}{3}\right)\times 1000}{(1-(\frac{1}{3}\left)\right)m_B}}{\frac{\frac{1}{2}\times 1000}{(1-(\frac{1}{2}\left)\right)m_B}}=\frac{m_2}{m_1}$
Hence, $\frac{m_2}{m_1}=\frac{1}{2}$.