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Question
Calculate molarity and molality of 6.3% solution of nitric acid having density 1.04 g cm−3
. (H = 1, N = 14, O = 16)
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Solution
Given: Density of 6.3% HNO3 = 1.04 g cm−3
To find: a. Molarity b. Molality
Formulae:
a) Molarity = Number of moles of solute/Volume of solution in L b) Molality = Number of moles of solute/Mass of solvent in kg
Calculation: Density of solution = Mass of solution/Volume of solution
∴ Volume of solution = `100/1.04` = 96.15 cm3 = 96.15 × 10–3 dm3
Molarity = `"Massof solute"/"Molar mass of solute×Volume of solution in L"`
∴ Molarity = `(6.3 xx 10^(-3))/(63 xx 10^(-3) xx 96.15 xx 10^(-3))` = 1.04 mol/dm3
b) 6.3% HNO3 is present means 6.3 g of HNO3 is present in 100 g of solution.
Mass of H2O = 100 − 6.3 = 93.7 g
Molality of HNO3 = Massof HNO3/Molar mass of HNO3 ×Massof solvent in kg
`= (6.3 xx 10^(-3))/(63xx10^(-3) xx 93.7 xx 10^(-3))`
= 1.067 mol/kg
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