Question

determine the volume of diluted nitric acid ( density= 1.11 g/ ml, 19% w/v HNO3) that can be prepared by diluting with water 50 ml of conc HNO3 ( d= 1.42 g/ ml, 69.8 % w/v)

Answer 1

As we know:

V₁N₁=V₂N₂  

Here,

N = normality of the solution

V is the volume of the solution

$V_{1}\times \frac{Number of g equivalent }{Volume of solution}=V_{2}\times \frac{Number of g equivalent }{Volume of solution}$  

$V_{1}\times \frac{\frac{given mass}{equivalent mass}}{Volume of solution}=V_{2}\times \frac{\frac{given mass}{equivalent mass}}{Volume of solution}$  

$V_{1}\times \frac{\frac{19}{36.5}}{0.1}=50\times \frac{\frac{69.8}{36.5}}{0.1}$  

V₁=183.69 mL

The volume of diluted nitric acid is 183.69 mL

Answer 2

Answer : The volume of dilute nitric acid is 183.684 ml.

Solution : Given,

19% w/v dilute nitric acid means 19 g of dilute nitric acid present in 100 ml of solution.

Mass of dilute nitric acid = 19 g

Volume of solution = 100 ml

69.8% w/v dilute nitric acid means 69.8 g of dilute nitric acid present in 100 ml of solution.

Mass of conc. nitric acid = 69.8 g

Volume of conc. nitric acid = 50 ml

The formula used for the relation between the volume and molarity is,

$M_1V_1=M_2V_2\\\frac{w_1}{mV}\times V_1=\frac{w_2}{mV}\times V_2\\ w_1V_1=w_2V_2$

where,

$w_1$ = mass of dilute nitric acid

$w_2$ = mass of conc. nitric acid

$V_1$ = volume of dilute nitric acid

$V_2$ = volume of conc. nitric acid

V = volume of solution

m = molar mass of nitric acid

As per the question, the volume of solution and molar mass of nitric acid are same.

Now put all the given values in above formula, we get the volume of dilute nitric acid.

$19g\times V_1=69.8g\times 50ml\\V_1=183.684ml$

Therefore, the volume of dilute nitric acid is 183.684 ml.