Chapter 13.6, Problem 2PE

Interpretation Introduction

Interpretation:

The boiling point and the freezing point of the solution should be calculated.

Concept introduction

Boiling point is the temperature at which liquid turns into a gas. Example: boiling point of water is $\text{100°C}$. That is water changes from liquid phase to gas phase at temperature ${\text{100}}^{\text{o}}\text{C}$.

${\text{ΔT}}_{\text{b}}{\text{= T}}_{\text{b}}{\text{- T}}_{\text{b}}^{\text{°}}$

Where,

${\text{ΔT}}_{\text{b}}-$ Change in boiling point

${\text{T}}_{\text{b}}-$ Boiling point of the solution

${\text{T}}_{\text{b}}^{\text{°}}-$ Boiling point of pure solvent

The Boiling point elevation${\text{(ΔT}}_{\text{b}}\text{)}$ is indicated as ${\text{ΔT}}_{\text{b}}{\text{= K}}_{\text{b}}\text{m}$

Where,

${\text{ΔT}}_{\text{b}}-$ Change in boiling point

${\text{K}}_{\text{b}}-$ Molal boiling point constant

$m-$ Molality of the solution

Freezing point is the temperature at which liquid turns into solid.

The Freezing point depression ${\text{(ΔT}}_{\text{f}}\text{)}$ is indicated as ${\text{ΔT}}_{\text{f}}{\text{= K}}_{\text{f}}\text{m}$

Where,

${\text{ΔT}}_{\text{f}}-$ Change in freezing point

${\text{K}}_{\text{f}}-$ Molal freezing point constant

$m-$ Molality of the solution

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $\text{12}\text{g}$ of ${}^{\text{12}}\text{C}$.

From given mass of substance moles could be calculated by using the following formula,

$\text{Moles}\text{of}\text{substance }\text{= }\frac{\text{Given}\text{mass}\text{of}\text{substance}}{\text{Molecular}\text{mass}}$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$.