Chapter 14, Problem 111A

Interpretation Introduction

Interpretation:

Molality of $\mathrm{Ca}{\left(N{O}_3\right)}_2$ solution have to be calculated.

Concept introduction:

Molality: Molality is defined as the number of moles of solute per kg of the solvent.

In expression, $\mathrm{Molality}ofthesolution=\frac{Numberofmolesof the solute}{ Mass of the solvent(inKg)}$

Again $\mathrm{Number}ofmolesofsolute=\frac{Massofthesolute}{Molecularweightofthesolute}$

Answer to Problem 111A

The molality of $\mathrm{Ca}{\left(N{O}_3\right)}_2$ solution is 0.501 mol/kg.

Explanation of Solution

Given information:

Mass of $\mathrm{Ca}{\left(N{O}_3\right)}_2$ = 246.14 gm

Number of moles of $\mathrm{Ca}{\left(N{O}_3\right)}_2$ = 1.5 mole

Volume of the solution = 3.0 L

Calculate the mass of solution.

$\begin{array}{c}\mathrm{mass}ofsolution=3.0L\times 1.08\mathrm{kg}/L\\ =3.24\mathrm{kg}\end{array}$

Calculate the mass of $\mathrm{Ca}{\left(N{O}_3\right)}_2$.

$\begin{array}{c}\mathrm{Mass}\mathrm{of}\mathrm{Ca}{\left(N{O}_3\right)}_2=246.14\mathrm{gm}\\ =0.246\mathrm{kg}\end{array}$

Calculate the mass of solvent.

$\begin{array}{c}\mathrm{Mass}ofsolvent=3.24\mathrm{kg}solution-0.246\mathrm{kg}solute\\ =2.994\mathrm{kg}\end{array}$

Write the expression for molality of the solution.

$\mathrm{Molality}ofthesolution=\frac{Numberofmolesof the solute}{ Mass of the solvent(inKg)}$

Substitute the values in the above equation.

$\begin{array}{l}MolalityofCa{\left(N{O}_3\right)}_{2}solution=\frac{Numberofmolesof Ca{\left(N{O}_3\right)}_2}{ Mass of the solvent(inKg)}\\ =\frac{1.5mole}{ 2.994kg}=0.501m\end{array}$

The molality of $\mathrm{Ca}{\left(N{O}_3\right)}_2$ solution is $0.501\mathrm{mol}/kg$.