Chapter 13.3, Problem 3PE

Interpretation Introduction

Interpretation:

The molality of $\text{44}\text{.6}$ percent solution of $\text{NaCl}$ should be calculated.

Concept Introduction:

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.

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

Percent by mass: It is the ratio of solute mass to the solution mass which finally multiplied by $\text{100}$. The formula for percent by mass calculation is as follows,

$\begin{array}{l}\text{Percent mass =}\frac{\text{Mass}\text{of}\text{solute}}{\text{Mass of solution}}\text{×100\%}\\ \text{Mass of solution = Mass of solute + Mass of solvent}\end{array}$

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 $kg$.