A $0.900 g$ aspirin tablet (acetylsalicylic acid, $C X_{9} H X_{8} O X_{4}$ , molar mass $180$ ) is boiled with $50.0 mL$ of $0.500 M$ $NaOH$ , which neutralises and hydrolyses it (2 mol $NaOH$ per mol aspirin). The excess $NaOH$ needs $30.4 mL$ of $0.500 M$ $HCl$ . The mass percent of aspirin in the tablet is

A500.0%
B196.0%
C98.0%
D49.0%

Answer: C

NaOH taken = 0.0500 x 0.500 = 0.0250 mol; excess = 0.0304 x 0.500 = 0.0152 mol; consumed = 0.0098 mol. Aspirin = 0.0098/2 = 0.0049 mol = 0.882 g. Percent = 0.882/0.900 x 100 = 98.00%.

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A 0.900 g aspirin tablet (acetylsalicylic acid, CX9​HX8​OX4​, molar mass 180) is boiled with 50.0 mL of 0.500 M NaOH, which neutralises and hydrolyses it (2 mol NaOH per mol aspirin). The excess NaOH needs 30.4 mL of 0.500 M HCl. The mass percent of aspirin in the tablet is

Answer: C

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