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JEE Main 201911 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry

JEE Main 2019 · 11 January · Morning Shift · Chemistry · Some Basic Concepts of Chemistry

A 10 mg effervescent tablet containing sodium bicarbonate and...

A 10mg10 \mathrm{mg} effervescent tablet containing sodium bicarbonate and oxalic acid releases 0.25 mLofCO20.25 \mathrm{~mL} of \mathrm{CO}_{2} at T=298.15 K\mathrm{T}=298.15 \mathrm{~K} and P=1\mathrm{P}=1 bar. If molar volume of CO2\mathrm{CO}_{2} is 25.0 L25.0 \mathrm{~L} under such condition, what is the percentage of sodium bicarbonate in each tablet? [Molar mass of NaHCO3=84 g mol−1]\left.\mathrm{NaHCO}_{3}=84 \mathrm{~g} \mathrm{~mol}^{-1}\right]

Options

  1. A0.84
  2. B33.6
  3. C16.8
  4. D8.4

Correct answer

D. 8.4

Step-by-step solution

2 NaHCO3+H2C2O4⟶Na2C2O4+\mathrm{NaHCO}_{3}+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \longrightarrow \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+2 CO2+\mathrm{CO}_{2}+2 H2O\mathrm{H}_{2} \mathrm{O}Moles of CO2\mathrm{CO}_{2} evolved =0.2525×103=10−5=\frac{0.25}{25 \times 10^{3}}=10^{-5}∴ Moles of NaHCO3=10−5∴ Mass of NaHCO3=84×10−5 g=0.84×10−3 g=0.84mg\begin{aligned} \therefore \quad& \text{ Moles of } \mathrm{NaHCO}_{3}=10^{-5} \\ \therefore \quad \text{ Mass of } \mathrm{NaHCO}_{3}&=84 \times10^{-5} \mathrm{~g}\\ &=0.84 \times10^{-3} \mathrm{~g}=0.84 \mathrm{mg} \end{aligned}∴%\therefore \quad \% by weight =×100=8.4%=\times 100=8.4 \%
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