JEE Main 2016 · Chemistry · Mole Concept and Stoichiometry · easy

32 g of crystalline sodium carbonate, Na2CO3· xH2O, needs 40.0 mL of 0.500 M HCl for complete neutralisation to CO2. Taking Na=23, C=12, O=16, H=1…

Single correct option

2.32\ \mathrm{g}

of crystalline sodium carbonate,

\mathrm{Na_2CO_3\cdot xH_2O}

, needs

40.0\ \mathrm{mL}

0.500\ \mathrm{M}

\mathrm{HCl}

for complete neutralisation to

\mathrm{CO_2}

. Taking

\mathrm{Na}=23

\mathrm{C}=12

\mathrm{O}=16

\mathrm{H}=1

, the value of

x

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Answer: (A) 7

\mathrm{Na_2CO_3+2HCl\to2NaCl+H_2O+CO_2}

, so 2 mol HCl react per mole of the salt. HCl used

=0.0400\times0.500=0.0200

mol, hence salt

=0.0100

mol. Molar mass of the hydrate

=2.32/0.0100=232\ \mathrm{g\,mol^{-1}}

. Anhydrous

\mathrm{Na_2CO_3}=106

, so

18x=232-106=126

and

x=7

. A common slip is to use a 1:1 mole ratio, which gives

M=116

, lower than the anhydrous mass, a sign that the stoichiometry was wrong.

Original question written in the JEE Main 2016 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake