The density of ' \(x\) ' \(\mathrm{M}\) solution (' \(x\) ' molar) of \(\mathrm{NaOH}\) is \(1.12\mathrm{g}{\mathrm{mL}}…

• Density of solution = 1.12 g/mL

• Molality (m) = 3 mol/kg

• Molar mass of NaOH = 40 g/mol

• Molarity (x) = ?

We are to find the molarity \(x\) (in mol/L) of NaOH.

Molality \(m=\frac{\text{mol of solute}}{\text{mass of solvent in kg}}\)

Given:

\(\text{Molality}=3\text{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{/}\mathrm{k}\mathrm{g}\)

Let’s assume we have 1 kg (1000 g) of water. Then the number of moles of NaOH = 3 mol.

Mass of NaOH =

\(3\text{ }\text{mol}\times 40\text{ }\mathrm{g}\mathrm{/}\mathrm{m}\mathrm{o}\mathrm{l}=120\text{ }\mathrm{g}\)

So total mass of solution =

\(1000\text{ }\mathrm{g}\text{ (water)}+120\text{ }\mathrm{g}\text{ (NaOH)}=1120\text{ }\mathrm{g}\)

Given density = 1.12 g/mL

So for 1120 g of solution:

\(\text{Volume}=\frac{1120\text{ }\mathrm{g}}{1.12\text{ }\mathrm{g}\mathrm{/}\mathrm{m}\mathrm{L}}=1000\text{ }\mathrm{m}\mathrm{L}=1\text{ }\mathrm{L}\)

Molarity \(x=\frac{\text{mol of solute}}{\text{volume of solution in L}}\)

We have 3.0 mol NaOH in 1 L of solution

x = 3.0