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AIIMS 201926 May
Exam
AIIMS
Year
2019
Date
26 May
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry

AIIMS 2019 · 26 May · Chemistry · Some Basic Concepts of Chemistry

Calculate molarity of 63 \% w / w HNO_3 solution if density is 1.4...

Calculate molarity of 63%w/wHNO363 \% \mathrm{w} / \mathrm{w} \mathrm{HNO}_3 solution if density is 1.4 g/mL1.4 \mathrm{~g} / \mathrm{mL}.

Options

  1. A14M14 \mathrm{M}
  2. B12M12 \mathrm{M}
  3. C10M10 \mathrm{M}
  4. D8M8\mathrm{M}

Correct answer

A. 14M14 \mathrm{M}

Step-by-step solution

63%63 \% nitric acid by mass means that mass of nitric acid =63 g=63 \mathrm{~g} Mass of solution =100 g=100 \mathrm{~g} Molar mass of HNO3=63 g mol−1\mathrm{HNO}_3=63 \mathrm{~g} \mathrm{~mol}^{-1} Mole of HNO3=63 g63 g mol−1=1 mole\mathrm{HNO}_3=\frac{63 \mathrm{~g}}{63 \mathrm{~g} \mathrm{~mol}^{-1}}=1 \mathrm{~mole} Density of solution =1.4 g mL−1∴=1.4 \mathrm{~g} \mathrm{~mL}^{-1}\therefore \quad Volume of solution =100 g1.4 g mL−1=71.428 mL=0.071428 L=\frac{100 \mathrm{~g}}{1.4 \mathrm{~g} \mathrm{~mL}^{-1}}=71.428 \mathrm{~mL}=0.071428 \mathrm{~L} Molarity of solution = Moles of solute  Volume of solution in L=1.00.071428 mol L−1=14M=\frac{\text{ Moles of solute }}{\text{ Volume of solution in } \mathrm{L}} =\frac{1.0}{0.071428} \mathrm{~mol} \mathrm{~L}^{-1}=14 \mathrm{M}
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