Question

What is the normality of lead (II) nitrate if the density of its 26% (w/w) aqueous solution is 3.105 g/mL? Take molar mass of lead (II) nitrate to be 331g/mol.

(a) 2.437 N

(b) 4.878 N

(c) 0.243 N

(d) 0.488 N

This question was posed to me in an online quiz.

The doubt is from Expressing Concentration of Solutions in division Solutions of Chemistry – Class 12

Answer 1

The correct option is (b) 4.878 N

For explanation: Consider 100g of solution. It is made up of 26g lead (II) nitrate and 74g water.

Volume of solution, V = 100g/(3.105g/ml) = 32.2061 ml = 0.0322 L

Equivalent weight of lead nitrate = 331/2 = 165.5 g/eq

Number of equivalents, N = 26g/(165.5g/eq) = 0.1571 eq

Normality = N/V = 4.878 N.