2.48 g of `Na_(2)S_(2)O_(3). xH_(2)O` is dissolved per litre solution 20 ml of this solution required 10 ml 0.01 M iodine solution. What is value of x ?

Answer

5

Explanation

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`