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JEE Main 2017 · Chemistry · Mole Concept and Stoichiometry · medium

A 5.00 g sample containing only NaHCO3 and Na2CO3 loses 0.62 g on strong heating (the carbonate is stable). The unheated 5.00 g sample is dissolved…

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A 5.00 g5.00\ \mathrm{g} sample containing only NaHCO3\mathrm{NaHCO_3} and Na2CO3\mathrm{Na_2CO_3} loses 0.62 g0.62\ \mathrm{g} on strong heating (the carbonate is stable). The unheated 5.00 g5.00\ \mathrm{g} sample is dissolved in water and titrated to complete neutralisation (evolution of CO2\mathrm{CO_2}) with 0.50 M HCl0.50\ \mathrm{M\ HCl}. Volume of HCl required is nearly (molar masses: NaHCO3=84\mathrm{NaHCO_3}=84, Na2CO3=106\mathrm{Na_2CO_3}=106, H2O=18\mathrm{H_2O}=18, CO2=44\mathrm{CO_2}=44)
  1. A. 165 mL165\ \mathrm{mL}
  2. B. 140 mL140\ \mathrm{mL}
  3. C. 200 mL200\ \mathrm{mL}
  4. D. 100 mL100\ \mathrm{mL}
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Answer: (A) 165 mL165\ \mathrm{mL}

2NaHCO3→Na2CO3+CO2+H2O2\mathrm{NaHCO_3}\to \mathrm{Na_2CO_3}+\mathrm{CO_2}+\mathrm{H_2O}: mass lost is 6262 g per 168168 g of bicarbonate. So NaHCO3=0.62×168/62=1.68\mathrm{NaHCO_3}=0.62\times168/62=1.68 g =0.0200=0.0200 mol, and Na2CO3=5.00−1.68=3.32\mathrm{Na_2CO_3}=5.00-1.68=3.32 g =0.0313=0.0313 mol. HCl needed: 0.0200×1+0.0313×2=0.08260.0200\times1+0.0313\times2=0.0826 mol. Volume =0.0826/0.50=0.165=0.0826/0.50=0.165 L =165=165 mL. Ignoring the carbonate already present gives 8080 mL; treating carbonate as 1:1 gives 102102 mL.
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