Question

8 g of na2co3.xh2o are present in 500 ml of solution. if this solution is diluted by water such that volume of solution become 1l. 100 ml of dilute solution is completely neutralised by 10 ml of 1 m hcl. the value of x is

Answer 1

Explanation:

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 0.7gofNa2CO3.XH2O 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 Na2CO3

      M1-> molarity of Na2CO3

      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 get

2×M1×20=1×0.1×19.8

Therefore M1=40019.8=0.0495M

Now Molarity=molarmassofsoluteStrength