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BITSAT 2013
Exam
BITSAT
Year
2013
Subject
Chemistry
Chapter
Redox Reactions

BITSAT 2013 · Chemistry · Redox Reactions

Volume of 3 \% solution of sodium carbonate necessary to neutralise...

Volume of 3%3 \% solution of sodium carbonate necessary to neutralise a litre of 0.1N0.1 \mathrm{ N} sulphuric acid

Options

  1. A176.66ml176.66 \mathrm{ml}
  2. B156.6ml156.6 \mathrm{ml}
  3. C116.0ml116.0 \mathrm{ml}
  4. D196.1ml196.1 \mathrm{ml}

Correct answer

A. 176.66ml176.66 \mathrm{ml}

Step-by-step solution

Normality of 3%Na2CO3N=3×100053×100=0.566N3 \% \mathrm{Na}_{2} \mathrm{CO}_{3}\mathrm{N}=\frac{3 \times 1000}{53 \times 100}=0.566 \mathrm{ N} For H2SO4\mathrm{H}_{2} \mathrm{SO}_{4} sol. N1=0.1,V1=100mL\mathrm{N}_{1}=0.1, \mathrm{ V}_{1}=100 \mathrm{ mL} For Na2CO3\mathrm{Na}_{2} \mathrm{CO}_{3} sol. N2=0.566\mathrm{N}_{2}=0.566 Now apply N1V1=N2V2V2=N1V1N2=0.1×1000mL0.566=176.66mL\mathrm{N}_{1} \mathrm{ V}_{1}=\mathrm{N}_{2} \mathrm{ V}_{2}\mathrm{V}_{2}=\frac{\mathrm{N}_{1} \mathrm{ V}_{1}}{\mathrm{ N}_{2}}=\frac{0.1 \times 1000 \mathrm{ mL}}{0.566}=176.66 \mathrm{ mL}
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