Question

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%.

Answer 1

Given, density of $HN{ O }_{ 3 } = 1.41 g/mL$ and mass percent of $HN{ O }_{ 3 }$ is 69%

i) Calculate the mass of $HN{ O }_{ 3 }$ in the solution  

$=\quad 1.41\quad { g }/{ ml }\quad \times \quad$ 69%

$=\quad 0.9729\quad { g }/{ ml }$

ii) Convert the above got answer in 1 to mole quantity by dividing it with the molar mass of $HN{ O }_{ 3 }$ which is 63 g/mol  

$=\quad \frac { 0.9729\quad { g }/{ ml } }{ 63\quad { mol }/{ ml } }$

$=\quad 0.01544\quad { mol }/{ ml }$

iii) The molar quantity is expressed as mole/Litre, we have to multiply the above answer with 1000 (1L = 1000ml)  

$=\quad 0.01544\quad { mol }/{ ml }\quad \times \quad 1000$

$=\quad 15.44\quad M$

Answer 2

Concentration of HNO3 = 15.44 mol/L

Explanation:

Mass Percent of HNO3 given in the Sample = 69 %

Thus, 100 g of HNO3 contains 69 g of HNO3 by mass.

Molar mass of HNO3:

= { 1 + 14 + 3(16)} g mol ^ - 1

= 1 + 14 + 48

= 63g mol^{-1}

No. of moles in 69 g of HNO3:

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

= 1.095 mol es

 

Volume of 100g HNO3 solution:

= $\frac{MassofSolution}{DensityofSolution}$

= $\frac{100g}{1.41g mL^-1}$

= 70.92mL

= 70.92 * $10^{-3}$ L

= 15.44mol/L

Therefore, Concentration of HNO3 = 15.44 mol/L