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# $0.5\;g$ of fluming $H_2SO_4$(oleum) is diluted with water. This solution is completely neutralised by $26.7\;mL$ of $0.4\;N\;NaOH$. The percentage of free $SO_3$ in the sample is

$(a)\;30.6 \% \\ (b)\;40.6\% \\(c)\;20.6 \% \\(d)\;50\%$

Oleum is basically $H_2S_2O_7$, a mixture of $(H_2SO_4+2SO_3 )$
Meq. of $H_2SO_4$ +Meq. of $SO_3$=Meq. of $NaOH$
Suppose the mass of $SO_3$ in Oleum =w
Therefore the mass of $H_2SO_4=0.5\;w$
Eq.wt of $H_2SO_4=49$
and Eq.wt of $SO_3=\large\frac{80}{2}$$=40 \large\frac{(0.5 -w)}{49}$$ \times 1000 + \large\frac{w}{40}$$\times 1000 =26.7 \times 0.4 w=0.103 \therefore \% of SO_3=\large\frac{0.103}{0.5}$$\times 100=20.6 \%$
Hence c is the correct answer.