Question

0.7 g of Na₂Co₃ .XH₂O were dissolved in water and the volume was made to 100mL , 20 mL of this solution required 19.8 mL of N/10 HCl for complete neutralisation, the value of 'x' is

A) 7

B) 3

C) 2

D) 5

Answer 1

Na₂CO₃. XH₂O + 2HCl → 2NaCl + (X + 1)H₂O + CO₂

The strength of a solution is defined as the amount of solute in grams, present in one litre of solution. It is expressed in gL^-1. We are given that 0.7 g of Na₂CO₃. XH₂O is dissolved 100 ml of solution. The strength of the solution is therefore = 0.7/[1000/100] = 7 gL^-1

The molarity equation is written as
n₁M₁V₁ = n₂M₂V₂ (1)

where n₁ is the acidity of Na₂CO₃, M₁ is its molarity and V₁ is the volume of sodium carbonate solution used. n₂ is the basicity of HCl, M₂ is its molarity and HCl is the volume of HCl used .

Since HCl is monobasic, therefore its molarity is the same as its normality. Substituting the given values in equation (1), we get
2 × M₁ × 20 = 1 × 0.1× 19.8

Therefore
M₁ = 19.8/400 = 0.0495 M

Now Molarity = Strength in gL^-1/ molar mass of solute

Therefore Molar mass = Strength / Molarity

Thus, molar mass of Na₂CO₃. xH₂O = 7/0.0495 = 141.414 g

But molar mass of Na₂CO₃. XH₂O = Mass of anhydrous Na₂CO₃ + mass of x molecules of water = 106 + 18x

Therefore 106 + 18x = 141.414 which gives x = 35.414/18 = 1.976

Hence the value of x here can be rounded off to 2.