Chapter 13.3, Problem 4CP

Interpretation Introduction

Interpretation:

The molarity of the solution is to be calculated.

Concept Introduction:

The number of grams of solute present in $100g$

of solution is called the percent by mass of the solution.

Percent by mass of a solution is obtained by dividing the mass of solute by the mass of solution (i.e. the mass of the solute and the mass of the solvent) and then multiplying the result by $100\%$. Mathematically, the percent by mass of a substance is written as:

$\text{percent by mass}=\frac{\text{mass of solute}}{\text{mass of solute}+\text{mass of solvent}}\times 100\%$

Molarity is defined as the total number of moles of solute present in one litre of the solution.

Mathematically, it can be expressed as follows:

$\text{Molarity }\left(M\right)=\frac{\text{moles of solute}}{\text{litre of solution }}$

$\text{No}\text{. of moles}=\frac{\text{mass }}{\text{molar mass }}$

The relation between volume and density is expressed as follows:

$d=\frac{\text{m}}{\text{V}}$

Here, $d$ is the density of the solution, m is the mass of the solution and V is the volume of the solution.

Molality is defined as the total number of moles present in one kilogram of the solvent.

Mathematically, it can be expressed as follows;

$\text{molality }\left(m\right)=\frac{\text{moles of solute}}{\text{mass of solvent }\left(\text{in Kg}\right)}$

$\text{No}\text{. of moles}=\frac{\text{mass }}{\text{molar mass }}$