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  • of sulphuric acid solution contain <img alt="96 %" src="https://entrancecorner.onco

20\, \mathrm{ml} of sulphuric acid solution contain 96 % by (s.p.gr=1.84) weight of pure acid. What will be the volume of 3.5 \mathrm{M} \mathrm{NaOH} solution required to neutralise the acid.

Option: 1

108.2


Option: 2

205.7


Option: 3

145.6


Option: 4

224.8


Answers (1)

best_answer

\text { wt of solute }=20 \times 1.84 \times \frac{96}{100} \mathrm{gm}

\text { No. of moles of solute }=\frac{\text { Weight of solute }}{\text { Molecular Weight }}

\text { Molecular weight of sulphuric acid = 98 gm}

\text { No. of moles }=\frac{1.84 \times 20 \times 0.96}{98}

\text { No. of Moles }=0.360

\mathrm{n_{H^{+}}=2 \times 0.360}

\mathrm{n_{H^+}=\frac{3.5 \times V}{1000}}

\mathrm{2 \times 0.360=\frac{3.5 \times V}{1000}}

\mathrm{\frac{1000 \times 2 \times 0.360}{3.5}=V}

\mathrm{V=205.7}

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seema garhwal

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