### Question 1.6

NCERT Class XI Chemistry

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%.

### Video Explanation

(detailed solution given after this video)

Please watch this youtube video for a quick explanation of the solution:

### Solution in Brief

(or jump to detailed one after this)

Take 1 milli-liter of solution and use the mass density formula to obtain the mass of the solution. The mass of nitric acid can be obtained because it is given as 69%.

Next use the molar mass of nitric acid to obtain the moles of nitric acid.

Lastly, multiply by 1000 to obtain the concentration per liter.

A better approach is to take 1000 ml of solution. It will simplify calculation. It has been explained in the video.

### Solution in Detail

Atomic masses of H = 1.008u, N = 14.00u, O = 16.00u.

Molar mass of HNO3 = 1.008 + 14.00 + (16.00 x 3)

$\displaystyle \therefore \text{Molar mass of HNO}_3 = 63.00 \text{ g/mol}$

Take 1 mL of the solution.

$\displaystyle \therefore \text{mass of soln = vol. x }\rho = 1.41 \text{gm}$

but mass of nitric acid has been given as 69%.

$\displaystyle \therefore \text{HNO}_3 = (69\% \times 1.41) \text{ gm}$

$\displaystyle \text{mols of HNO}_3 = \frac{\text{mass of HNO3}}{\text{molar mass of HNO3}} $

$\displaystyle \therefore = \frac{(69\% \times 1.41)}{63.00} \text{ mol}$

$\displaystyle \text{conc. of HNO}_3 = \frac{\text{no. of moles}}{\text{vol. in liters}}$

$\displaystyle = \frac{\bigg (\displaystyle \frac{69\% \times 1.41}{63.00}\bigg) \text{ mol}}{(1/1000) \text{ L}}$

$\displaystyle = 15.4 \text{ M (rounded to 3 S.F.)} \:\underline{Ans}$

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