Determine the percentage by mass of chromium, sulphate radical and water of crystallization in chrome alum `[K_(2)SO_(4)*Cr_(2)(SO_(4))_(3)*24H_(2)O].(Cr=52)`.

Explanation

Gram-formula mass of `K_(2)SO_(4)*Cr_(2)(SO_(4))_(3)*24H_(2)O`
`=[(2xx39+96)+(2xx52+3xx96)+24xx18]g=998g`
`therefore`Amount of Cr in chrome alum`=(2xx52)/(998)xx100=10.42%`
amount of sulphate radicals
`=(4xx"gram-formula mass of "SO_(4)^(2-))/(998)xx100=38.48%`
`therefore`Amount of `H_(2)O=(24xx18)/(98)xx100=43.29%`.