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10 April, 00:43

A 0.2688 g sample of a monoprotic acid neutralized 16.4 mL of 0.08133 M KOH solution. Calculate the molar mass of the acid

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  1. 10 April, 02:13
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    202 g/mol

    Explanation:

    Let's consider the neutralization between a generic monoprotic acid and KOH.

    HA + KOH → KA + H₂O

    The moles of KOH that reacted are:

    0.0164 L * 0.08133 mol/L = 1.33 * 10⁻³ mol

    The molar ratio of HA to KOH is 1:1. Then, the moles of HA that reacted are 1.33 * 10⁻³ moles.

    1.33 * 10⁻³ moles of HA have a mass of 0.2688 g. The molar mass of the acid is:

    0.2688 g/1.33 * 10⁻³ mol = 202 g/mol
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