Question

A molal solution is one that contains one mole of the solute in
A.1000 g of the solvent
B.One litre of the solvent
C.One litre of the solution
D.22.4 litre of the solvent

Answer 1

We know molality is the moles of solute per specified amount of mass of the solvent. Concentration is expressed as molality and its unit is $mol/kg$ (or) $m$. $Mol/kg$ is one molal solution.

Complete step by step answer: We can define molality as the number of solutes dissolved in 1 kg of solvent. We can calculate the molality of a solution as,
${\text{Molality}}\,{\text{(m) = }}\dfrac{{{\text{Mass}}\,{\text{of}}\,{\text{solute(in mol)}}}}{{{\text{Mass of}}\,{\text{solvent(in kg)}}}}$
One kg of solvent is 1000 g of solvent.
$\therefore$ Option (A) is correct.
Example:
Calculate the molality of copper (II) bromide whose mass percentage is $0.50\%$.
The mass percentage of the solution is $0.50\%$.
Consider the total mass of the solution as 100g.
The mass of water is calculated as,
Mass of water=Total mass of the solution-Mass of copper (II) bromide
Mass of water=$100g - 0.50g\,CuB{r_2} = 99.5g$
The moles of copper (II) bromide are calculated from its molar mass.
Moles of copper (II) bromide=$0.50g \times \dfrac{{1mol}}{{223.37g}} = 0.00223mol$
Mass of water (in kg)=$99.5g \times \dfrac{{1kg}}{{1000g}} = 0.0995\,kg$
The molality of copper (II) bromide is given as,

${\text{Molality}}\,{\text{(m) = }}\dfrac{{{\text{Mass}}\,{\text{of}}\,{\text{solute(in mol)}}}}{{{\text{Mass of}}\,{\text{solvent(in kg)}}}} \\\ Molality\,(m) = \dfrac{{0.00223\,mol}}{{0.0995kg}} \\\ Molality\,(m) = 0.02241m \\\$
The molality of copper (II) bromide is $0.02241\,m$.
One mole of the solute in one litre of the solution is molarity. We can write the formula as,
${\text{Molarity = }}\dfrac{{{\text{Mass}}\,{\text{of}}\,{\text{solute(in}}\,{\text{moles)}}}}{{{\text{Volume}}\,{\text{of}}\,{\text{solution}}\,{\text{(in}}\,{\text{litres)}}}}$
One molar solution is one mole of solute in 1L of water.
Option (C) is incorrect.
We can define molar volume as the volume occupied by one mole of any substance at standard temperature and pressure. One mole of gas at standard temperature and pressure contains 22.4 Litres of gas and not solvent.
Option (D) is incorrect.
$\therefore$Option A is correct.

Note:
We can also calculate molality using the freezing point depression formula.
$m = \dfrac{{\Delta {T_f}}}{{{K_f}}}$
Here,
$\Delta {T_f} =$Freezing point of the solution
$m =$Molal concentration
${k_f} =$Freezing point depression constant (depends on the solvent used)
For example, let us calculate the molality of hydrochloric acid in benzene solution as,
Given,
Moles of hydrochloric acid =$0.030\,mol$
Mass of benzene =$100.0{\text{ }}g$
Freezing point of the solution $\left( {{T_f}} \right)$ =${4.04^o}C$
Freezing point of benzene $\left( {{T_o}} \right)$ =${5.51^o}C$
Freezing point depression constant of benzene =${4.90^o}C/m$
The change in freezing point is calculated as,
$\Delta {T_f} = {T_o} - {T_f} \\\ \Delta {T_f} = {5.51^o}C - {4.04^o}C \\\ \Delta {T_f} = {1.47^o}C \\\$
The molality of the solution is calculated as,
$m = \dfrac{{\Delta {T_f}}}{{{K_f}}} \\\ m = \dfrac{{{{1.47}^o}C}}{{{{4.90}^o}C/m}} \\\ m = 0.3\,m \\\$
The molality of hydrochloric acid in benzene solution is $0.3\,m$.
We can also calculate the molarity, mass percentage and mole fraction from molality. Molality depends on the masses of solvent and solute, and is not affected by temperature and pressure.