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Na(2)SO(4).xH(2)O has 50% H(2)O. Henxe, ...

`Na_(2)SO_(4).xH_(2)O` has `50% H_(2)O`. Henxe, `x` is :

A

`4`

B

`5`

C

`6`

D

`8`

Text Solution

Verified by Experts

The correct Answer is:
D
`%` by `wt`. Of `H_(2)O`
`=(wt. "of" H_(2)O)/("Total wt".) xx100`
`50=(18x)/(142+18x)xx100`
`71+9x=18x`
`x=71//9=7.88~~8`
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