Question

How many mL of conc. HNO, of specific gravity 1.41 containing 69% by mass are required to prepare 500 ml of 0.5 N HNO3? ​

Answer 1

Answer:

16.2mL of 69% HNO₃ required to prepare 500mL of 0.5N HNO₃  

Explanation:

Given,

   Specific gravity, d = 1.41g/mL

  % weight of HNO₃ = 69%

Since HNO₃ gives only one H⁺, its equivalent weight is equal to molecular weight.

    Equivalent weight of HNO₃,EW =  63g

Normality of 69% HNO₃ = (%weight×10×d) / EW

⇒                                       = (69×10×1.41) / 63 = 15.443N

now, we have to prepare 500mL(V₂) 0.5N(N₂) HNO₃ using 15.443N(N₁) HNO₃

                N₁V₁ = N₂V₂

⇒    V₁× 15.443 =  0.5×500

⇒                  V₁ = 250/15.443 = 16.2mL

Hence, the volume of 69% HNO₃ required is 16.2mL