Chapter 12, Problem 12.23QP

Interpretation Introduction

Interpretation:

To calculate the molarity and molality for given amount of $N{H}_3$ solution.

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 kilogram of solvent. Molality is estimation of moles in relationship with solvent in the solution.

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

Molarity (M): The concentration for solutions is expressed in terms of molarity. Molarity is number of moles of the solute present in liter of the solution.

$\text{Molarity = }\frac{\text{No}\text{. of moles of solute}}{\text{Volume of solution in L}}$

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 $kg$.