Question

Find molarity, molality, and normality of ${\mathrm{H}}_2{\mathrm{SO}}_4$ in $98\%$ w/w solution.$(\mathrm{\rho }=1.8{\mathrm{gml}}^{-1})$.

Answer 1

Part-$1$: Given data

1. Density $\left(\mathrm{\rho }\right)=1.8{\mathrm{gml}}^{-1}$
2. $98\%\mathrm{w}/\mathrm{w}\mathrm{solution}$$\Rightarrow$$98\mathrm{g}$ solute is present in $100\mathrm{g}$ solution.

Part-$2$ Calculating molarity$\left(\mathrm{M}\right)$

1. Molarity$\left(\mathrm{M}\right)$ of a solution is defined as the number of moles per litre of solution.
2. The solution contains $98\mathrm{g}{\mathrm{H}}_2{\mathrm{SO}}_4$, so the number of moles$=$$\frac{98}{98}=1mol$
3. Volume of solution$\left(V\right)$ $=$$\frac{m}{\rho }=\frac{100g}{1.8gm{l}^{-1}}=\frac{1000}{18}ml=\frac{1000}{18}\times \frac{1}{1000}=\frac{1}{18}litre.$
4. $M=\frac{numberofmoles}{Volumeofsolution}=\frac{1}{{\displaystyle \frac{1}{18}}}=18M$

Part-$3$ Calculating molality$\left(\mathrm{m}\right)$

1. Molality is defined as the number of moles of solute per kilogram of solvent.
2. In $100\mathrm{g}$ solution $98\mathrm{g}$is the amount of solute. so the amount of solvent $=$Total weight of solution $-$Weight of solute$=2\mathrm{g}$
3. Weight of solvent in Kg$=$$\frac{2}{1000}kg$
4. $m=\frac{numberofmolesofsolute}{Ki\log ramofsolvent}=\frac{1}{{\displaystyle \frac{2}{1000}}}=500m$

Part-$4$: Calculating Normality$\left(\mathrm{N}\right)$:

1. Normality is defined as the gram equivalence of solute per litre of solution.
2. Gram equivalence is calculated based on the number of ions that react in a chemical reaction.
3. In $1\mathrm{mol}{\mathrm{H}}_2{\mathrm{SO}}_4$,$2{\mathrm{H}}^{+}$ ion reacts in the given reaction.
4. $N=\frac{gramequivalenceofsolute}{Volumeofsolution.}=\frac{2}{{\displaystyle \frac{1}{18}}}=36N$