Question

Calculate the molarity and Molality of 9.8 %(w/w) solution of sulphuric acid . Given the density of solution is 1.02 g/ml​

Answer 1

Given

Mass of solute = 9.8 g

Mass of solution = 100 g

Mass of solvent = $(9.8-100)\ \text{g}$

Density of solution = 1.02 g/mL

To find

Molarity and Molality of the solution

Solution

Density is given by

$\rho=\dfrac{m}{V}\\\Rightarrow V=\dfrac{m}{\rho}\\\Rightarrow V=\dfrac{100}{1.02}\times 10^{-3}\\\Rightarrow V=98.04\times 10^{-3}\ \text{L}$

Molar mass of sulphuric acid = 98 g/mol

Moles of solute

$\dfrac{9.8}{98}=0.1\ \text{mol}$

Molarity is given by

$M=\dfrac{\text{Moles of solute}}{V}\\\Rightarrow M=\dfrac{0.1}{98.04\times 10^{-3}}\\\Rightarrow M=1.02\ \text{M}$

Molarity of the solution is $1.02\ \text{M}$.

Molality is given by

$\dfrac{\text{moles of solute}}{\text{mass of solvent}}=\dfrac{0.1}{(100-9.8)\times 10^{-3}}=1.11\ \text{mol/kg}$

Molality of the solution is $1.11\ \text{mol/kg}$.