Chapter 13.6, Problem 3PE

Interpretation Introduction

Interpretation:

The molality of solution in which $0.85g$ of organic compound dissolved in $100g$ of benzene and also the molar mass of the solute should be calculated.

Concept introduction

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 $kg$.

Molality$\left(m\right)$: The concentration for solutions is expressed in terms of molality. Molality is the number of moles of solute present in one kilogram of solvent. Molality is estimation of moles in relationship with solvent in the solution.

$\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{kilograms}\text{of}\text{solvent}}$