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

We will get the chemical equation to be:

Na_{2}CO_{3}. xH_{2}O + 2HCl → 2NaCl + (x + 1)H_{2}O + CO_{2}

We calculate the strength of the given solution as 0.7/(100/1000) g/L = 7 g/L

Also, n_{1} M_{1} V_{1} =n_{2} M_{2} V_{2} ---(i)

n_{1} = acidity of Na_{2}CO_{3} , M_{1}= molarity of Na_{2}CO_{3} solution and V_{1}= volume of Na_{2}CO_{3 }solution

n_{2} = basicity of HCl , M_{2} = molarity of HCl and V_{2} = volume of HCl

As we know for monobasic substances, molarity= normality, so equation (i) becomes

2* M_{1} * 20= 1*0.1*19.8

⇒ M_{1} = 0.0495M

Molarity= strength / molar mass of solute

⇒ Molar mass= strength/ molarity

⇒ molar mass of Na_{2}CO_{3}. xH_{2}O = 7/0.0495

⇒ molar mass of Na_{2}CO_{3}. xH_{2}O = 141.414g

As, this molar mass= molar mass of Na_{2}CO_{3} + mass of number of molecules of water in it

Molar mass of anhydrous Na_{2}CO_{3} = 106

Molar mass of water= 18

So, 106+ 18x= 141.414

⇒x= 1.976≈ 2

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