An aromatic hydrocarbon with empirical formula `C_(5)H_(4)` on sulphonation gave a monosulphonic acid. `0.104 g` of this acid `0.104 g` of this acid required `10 mL` of `NaOH` for complete neutralization. The molecular formula of the acid is

Options

A. `C_(10)H_(8)`
B. `C_(5)H_(4)`
C. `C_(15)H_(12)`
D. `C_(20)H_(16)`

Answer

A

Explanation

`ArH overset(H_(2)SO_(4))rarr ArSO_(3)H overset(NaOH)rarr ArSO_(3)Na + H_(2)O`
According to the law of equivalence,
`("Milliequi valents")_(ArSO_(3)H) = ("Milliequi valents")_(NaOH)`
= Normality `xx V_(mL)`
`= ((1)/(20))(10) = 0.5`
`:.` Equivalents of `ArSO_(3)H = (0.5)/(1000) = 5 xx 10^(-4)`
But `5 xx 10^(-4)` equivalent's of `ArSO_(3)H` weigh `0.104 g`. Thus, mass of one equivalent of `ArSO_(3)H` will be `208 g`.
Since `ArSO_(3)H` is a monobasic aciud, `1` equivalent `= 1 mol`.
Thus, molecular mass of `ArSO_(3)H` is `208 u`.
Molecular mass of `ArH = ("Molecular mass of" ArSO_(3)H) - ("Formula mass of" SO_(3)H) + ("Atomic mass of "H)`
`= (208)-(81) + (1)`
`= 128 u`
Empirical formula mass of `ArH = (5 xx 12 + 4 xx 1) = 64 u`
`:. n = ("Molecular mass")/("Empirical formula mass") = (128 u)/(64 u) = 2`
Hence, Molecular formula `= n xx` Empirical formula
`= 2 xx (C_(5)H_(4)) = C_(10)H_(8)`