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MolSynapse / Question bank / Equivalent Concept and Titration / Acid-base titrations and double titration (NaOH, Na2CO3, NaHCO3)

JEE Main levelSingle correct optionDifficulty 2 of 3eq046

A $1.25 g$ sample of impure $NaOH$ (impurity is neutral to $HCl$ ) is dissolved in water and neutralised by $36.4 mL$ of $0.500 N$ $HCl$ . The percentage purity of the sample ( $NaOH = 40$ ) is:

A29.1%
B58.2%
C91.0%
D72.8%

Answer: B

Equivalents of

HCl = 0.500 \times 0.0364 = 0.0182

, which equals the equivalents (moles) of

NaOH

. Mass of

NaOH = 0.0182 \times 40 = 0.728 g

. Purity

= 0.728/1.25 \times 100 = 58.2%

.

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