2.48 g of Na_(2)S_(2)O_(3)xH_(2)O was dissolved per litre of the solution 20mL of tis solution required 10 ml 0.01 m iodin e solution find out the value of x

2.48 g of Na_(2)S_(2)O_(3)xH_(2)O was dissolved per litre of the solut

1.	2.48 g of Na_(2)S_(2)O_(3)xH_(2)O was dissolved per litre of the solution 20mL of tis solution required 10 ml 0.01 m iodin e solution find out the value of x
Answer» Solution :The balanced eqaution for the REDOX REACTION is : `2Na_(2)Cr_(2)O_(3)+I_(2)rarrNa_(2)S_(4)O_(6)+2` NAL Let the molarity of `Na_(2)S_(2)O_(3)xH_(2)O "solution" =M_(1)` Applying molarity equation to the above redox reaction we have `(M_(1)xx20)/(2)(Na_(2)S_(2)O_(3))=(10xx.01)/(1)(I_(2))` `therefore M_(1)=0.01 M` But the actual amount dissolved =2.48g Equating these VALUES we have `(158+18x)xx0.01=2.48 or x=5`