1.

`Na_(2)SO_(4).xH_(2)O` has `50% H_(2)O`. Henxe, `x` is :A. `4`B. `5`C. `6`D. `8`

Answer» Correct Answer - 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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