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`
