0.7g of a sample of Na2CO3.xH2O were dissoled in water and the volume made up to 100ml . 20ml of this solution
required 19.8ml of 0.1 N HCl for complete neutralization . Find the value of x .
ans = ~2
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 X M1 X 20 = 1 X 0.1 X 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.