Question

What is the normality of lead (II) nitrate if the density of its 26% (w/w) aqueous solution is 3.105 g/mL?

Answer 1

Explanation :

Step 1: Assumptions 

Assuming 100 g of solution.

26% (w/w) lead (II) nitrate = 26 gm

74% (w/w) water = 74 gm

Step 2: Calculate the volume of solution 

Volume of solution,

V = mass ÷ density 

V = 100 gm ÷ (3.105 gm /ml)

    = 32.2061 ml

    = 0.0322 L

Step 3: Calculate equi. weight 

Equivalent weight of lead nitrate

    = 331/2

    = 165.5 gm /eq

Step 4: Calculate Normality 

Number of equivalents,

N = 26 gm ÷ (165.5 gm /eq)

    = 0.1571 eq

Normality = N/V = 4.878 N.

Normality is 4 .878 N