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If `13.1g` of `Na_(2) SO_(4) XH_(2)O` contains `6g` of `H_(2)O`, the value of `X` isA. `10`B. `5`C. `3`D. `7` |
Answer» Correct Answer - D According to formula, 1 mol `Na_(2) SO_(4) (142g)` always combines with `X` moles `H_(2)O (18X g)`. According to data, mass of `Na_(2) SO_(4) = (13.1 - 6)g = 7.1g`. In a compound the ratio of mass is always fixed. Thus, `(.^(m)Na_(2)SO_(4))/(.^(m)H_(2)O) = (142g)/(18Xg) = (7.1)/(6g)` `:. X = ((42xx6))/((18xx7.1)) = (852)/(127.8) = 6.66 ~~ 7` |
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