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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:

 

Na2CO3. xH2O + 2HCl → 2NaCl + (x + 1)H2O + CO2

 

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

 

Also, n1 M1 V1 =n2 M2 V2 ---(i)

n1 = acidity of Na2CO3 ,  M1= molarity of Na2CO3 solution  and V1= volume of Na2CO3 solution

n2  = basicity of HCl , M2 = molarity of HCl and V2 = volume of HCl

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

2* M1 * 20= 1*0.1*19.8

⇒ M1 = 0.0495M

Molarity= strength / molar mass of solute 

⇒ Molar mass= strength/ molarity

⇒ molar mass of  Na2CO3. xH2O = 7/0.0495

⇒ molar mass of  Na2CO3. xH2O = 141.414g

As, this molar mass= molar mass of Na2CO3 + mass of number of molecules of water in it

Molar mass of anhydrous Na2CO3 = 106

Molar mass of water= 18

 

So, 106+ 18x= 141.414

⇒x= 1.976≈ 2

 

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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.

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