When 2.46 of hydrated salt `(MSO_(4)x H_(2)O)` is completely dehydrated, 1.20 g of anhydrous salt is obtained. It molecular weight of anhydrous salt is `120 g mol^(-1)`, the value of x is

Options

A. 2
B. 4
C. 5
D. 6

Answer

D

Explanation

`underset(2.46g)(MSO_(4)).xH_(2)O to underset(1.20g)(MSO_(4)) + xH_(2)O`
Mass of water = 2.46 – 1.20 = 1.26 g
`therefore ` No of moles of `H_(2)O = (1.26)/(18) = 0.07`
No.of moles of `MSO_(4)= (1.20)/(120)=0.01`
`therefore ` 0.07 moles of water associated with 0.01 mole of salt.
Thus, there are 7 moles of water per mole of anhydrous salt.