Chapter 14, Problem 120A

Interpretation Introduction

Interpretation:

The required volume of the stock solution and theboiling pointof the new solution have to be calculated.

Concept introduction:

For electrolyte solutes, the elevation of the boiling point is directly proportional to molality of the solution.

Boiling point elevation $\Delta {\mathrm{T}}_b=i\times K_b\times m$

Where,

∆ T_b = Elevation of boiling point = the boiling point difference between a solution and the pure solvent

i = Van’t Hoff factor = for strong electrolyte ‘i’ is equal to number of dissociate ions.

K_b = the molal boiling point elevation constant

m = molality of the solution

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

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

Molarity: Molarity is defined as the number of moles of solute present in one litre of the solution In expression, $\mathrm{Molarity}ofthesolution=\frac{Numberofmolesof the solute}{ Volume of the solution(inlitre)}$

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

Answer to Problem 120A

The required volume of the stock solution is 302 mL and theboiling pointof the new solutionis 100.1 °C.

Explanation of Solution

$\mathrm{Molarity}ofthesolution=\frac{Numberofmolesof the solute}{ Volume of the solution(inlitre)}$

Data given: Mass of KBr = 135.2 gm

Volume of the solution = 2.3 L

$\begin{array}{l}NumberofmolesofKBr=\frac{MassofKBr}{MolecularweightofKBr}\\ =\frac{135.2gm}{119gm/mol}=1.14mol\end{array}$

$\begin{array}{l}MolarityofKBrsolution=\frac{Numberofmolesof KBr}{ Volume of the solution(inlitre)}\\ =\frac{1.14mole}{ 2.3litre}=0.496M\end{array}$

$\begin{array}{l} M_1{V}_1=M_2{V}_2\hfill \\ Stocksolution=Dilutedsolution\hfill \end{array}$

Data given:

M₁ = molarity of the stock solution = 0.496 M

M₂ = molarity of the diluted solution = 0.1 M

V₂ = volume of diluted solution = 1.5 L = 1500 mL

$\begin{array}{l}Volumeofthestocksolution=V_1=\frac{M_2{V}_2}{M_1}\\ V_1=\frac{0.1M\times 1500mL}{0.496M}=302mL\end{array}$

$\Delta {\mathrm{T}}_b=i\times K_b\times m$

Molality of the new solution = m = 0.10

As, 1 L of water = 1 kg of water

‘i’ for KBr solution = 2

$\begin{array}{l}\Delta {\mathrm{T}}_b=2\times 0.512°C/m\times 0.1m\\ =0.1°C\end{array}$

So, theboiling pointof the new solution = 100 °C + 0.1 °C = 100.1°C.