Answer to Question #302214 in General Chemistry for Vavine Ruku GI

Question

When you react 3.9367 grams of Na2CO3.nH20 with excess HCl.0.6039 grams of a gas is given off. What is the number of water molecules bonded to the salt?

Answer 1

Question

0.7 g of Na2​CO3​.xH2​O is dissolved in 100 ml,20 ml of which required 19.8 ml of 0.1 N HCl. The value of x is:

A

4

B

3

C

2

D

1

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Solution

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Correct option is C)

Na2​CO3​.XH2​O+2HCl→2NaCl+(X+1)H2​O+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 0.7gofNa2​CO3​.XH2​O is dissolved 100 ml of solution. 
The strength of the solution is therefore =[100/1000]
0.7​=7gL−1
The molarity equation is written as n1×M1×V1=n2×M2×V2....(1)where n1-> acidity of Na2​CO3​    M1-> molarity of Na2​CO3​    V1-> volume of sodium carbonate solution used    n2-> basicity of HCl    M2-> molarity of HCl    V2-> 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 get2×M1×20=1×0.1×19.8
Therefore M1=400
19.8​=0.0495M
Now Molarity=molarmassofsolute
StrengthingL−1​
Therefore, molar mass of Na2​CO3​.xH2​O =0.0495
7​=141.414 g/mol 
But molar mass of Na2​CO3​.xH2​O= Mass of anhydrous Na2​CO3​+ mass of x molecules of water =106+18x
Therefore 106 + 18x = 141.414 , which gives x=18
35.414​=1.976
Hence the value of x here can be rounded off to 2.