Question

Question 1.6 Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL–1 and the mass per cent of nitric acid in it being 69%.

Class XI Some Basic Concepts of Chemistry Page 22

Answer 1

we know,
$molarity = \frac{d \times mass \: percentage \: of \: solute \times 10}{m}$
where d is density of solution
m is molar mass of solute.

Here, Given,
d =1.41 g/mL
mass % of nitric acid ( solute ) = 69 %
molar mass of nitric acid ( HNO3) = 63 g/mol

now,
molarity = 1.41 × 69 × 10/63 = 15.4 M

Answer 2

Answer

Mass of nitric acid solution = 69 g (because 69% means 69 g dissolved in 100 ml of solvent)

Density of nitric acid solution = 1.41 g/ml

Volume of nitric acid solution = $\frac{mass}{density}$ = $\frac{100}{1.41}$ = 70.92 ml

No of moles of nitric acid = $\frac{mass of nitric acid}{molar mass of nitric acid}$ = $\frac{69}{63}$ = 1.09

Volume of the solution = 70.92

Molarity of the solution = $\frac{moles of nitric acid}{volume of solution in litres}$

= $\frac{1.09*1000}{70.92}$ = 15.37 moles/liter

∴Concentration of nitric acid = 15.37 mol/L