Ncert Exercises & Solutions

1.6 Calculate the concentration of nitric acid in moles per liter in a sample that has a density, 1.41 g m/L, and the mass percent of nitric acid in it being 69%.

The mass percent of nitric acid in the sample = 69 %
Thus, 100 g of nitric acid contains 69 g of nitric acid by mass.
Molar mass of nitric acid (HNO3)
= {1 + 14 + 3(16)} g mol–1
= 1 + 14 + 48
= 63 g mol–1
$∴$ Number of moles in 69 g of HNO3

$= \frac{69 g}{63 g {mol}^{- 1}} = 1.095 mol$

Volume of 100g of nitric acid solution

$= \frac{Massof solution}{density of solution}$
$= \frac{100 g}{1.41 g {mL}^{- 1}}$
$= 70.952 mL = 70.92 \times 10^{- 3} L$

Concentration of nitric acid

$= \frac{1.095 mole}{70.92 x 10^{- 3} L}$
$= 15.44 mol / L$

$∴$ Concentration of nitric acid = 15.44 mol/L

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