Question

A solution is prepared by dissolving 30 gm of sulphuric acid
in 70gm of water. The solution has density of 1.215 gm cm^-3
Calculate - (i) weight % of H2SO4 (ii) Mole fraction of H2SO4
(iii) Molarity (iv) Motality (v) Normality​

Answer 1

Explanation:

Density of solution ,d$= 1.215 g/cm^3$

Mass of the solution = 30 g+ 70 g = 100 g

Volume of the solution = V

$d=1.215 g/cm^3=\frac{100 g}{V}$

$V = 82.30 cm^3= 82.30 mL$ (=

$1 cm^3= 1 ml$

Weight % of sulfuric acid

$\%=\frac{30 g}{100 g}\times 100=30\%$

Mole fraction of sulfuric acid:

$\chi=\frac{n_s}{n_w+n_s}$

$n_s=\frac{30 g}{98 g/mol}$ = Moles of sulfuric acid

$n_w=\frac{70 g}{18 g/mol}$ =Moles of water

$\chi_s=\frac{\frac{30 g}{98 g/mol}}{\frac{70 g}{18 g/mol}+\frac{30 g}{98 g/mol}}=0.0729$

Molarity of the solution:

$M=\frac{\text{Moles of}H_2SO_4}{\text{Volume of the solution (L)}}$

$M=\frac{30 g}{98 g/mol\times 0.08230 L}=3.7195 mol/L$

Molality of the solution:

$m=\frac{\text{Moles of}H_2SO_4}{\text{Mass of the solvent (kg)}}$

$m=\frac{30 g}{98 g/mol\times 0.070 kg}=4.3731 mol/kg$

Normality of the solution:

Equivalent mass :

$\frac{\text{molar mass of compound}}{\text{removable}(H^+)or(OH^-)}$

Equivalent mass of sulfuric acid =$\frac{98 g/mol}{2}=49 g/mol$

$N=\frac{\text{Mass of}H_2SO_4}{\text{Equivalent mass of}H_2SO_4\times \text{Volume of the solution (L)}}$

$N=\frac{30 g}{49 g/mol\times 0.08230 L}=7.4391 N$