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JEE Main 2019 · Chemistry · Mole Concept and Stoichiometry · easy

A 1.330 g mixture of NaCl and KCl is dissolved in water and treated with excess AgNO3, giving 2.870 g of AgCl. (Na = 23, K = 39, Cl = 35.5, Ag =…

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A 1.330 g1.330\ \mathrm{g} mixture of NaCl and KCl is dissolved in water and treated with excess AgNO3\mathrm{AgNO_3}, giving 2.870 g2.870\ \mathrm{g} of AgCl\mathrm{AgCl}. (Na = 23, K = 39, Cl = 35.5, Ag = 108.) The mass percentage of NaCl in the mixture is nearest to
  1. A. 30.030.0
  2. B. 44.044.0
  3. C. 56.056.0
  4. D. 65.565.5
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Answer: (B) 44.044.0

Molar mass of AgCl = 143.5, so n(Cl−)=2.870/143.5=0.0200n(\mathrm{Cl^-}) = 2.870/143.5 = 0.0200 mol. Let NaCl = xx mol and KCl = yy mol: x+y=0.0200x+y = 0.0200 and 58.5x+74.5y=1.33058.5x + 74.5y = 1.330. Substituting y=0.02−xy = 0.02 - x: 58.5x+1.49−74.5x=1.33058.5x + 1.49 - 74.5x = 1.330, so 16x=0.16016x = 0.160, x=0.0100x = 0.0100. Mass of NaCl =0.585= 0.585 g, hence 0.585/1.330×100=44.0%0.585/1.330 \times 100 = 44.0\%. Check: 0.0100×58.5+0.0100×74.5=1.3300.0100\times58.5+0.0100\times74.5 = 1.330 g.
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