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A 0.407 g sample of a monoprotic acid is dissolved in water and titrated with 0.190 M KOH. What is the molar mass of the acid if 15.5 mL of the KOH solution is required to neutralize the sample?

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Final answer:

To find the molar mass of the monoprotic acid, we calculate the moles of KOH used in the titration and then relate it to the moles of the acid. Dividing the mass of the acid sample by the calculated moles of the acid yields the molar mass, which is 138.2 g/mol in this case.

Step-by-step explanation:

The student is asking to find the molar mass of a monoprotic acid that requires 15.5 mL of 0.190 M KOH to be neutralized. To solve this, we first calculate the number of moles of KOH that reacted:

Number of moles of KOH = Molarity of KOH × Volume of KOH in liters
= 0.190 mol/L × 0.0155 L
= 0.002945 mol of KOH

Since the acid is monoprotic, it donates one proton per molecule; thus, the number of moles of the acid is equal to the number of moles of KOH:

Moles of the acid = 0.002945 mol

To find the molar mass of the acid, we divide the mass of the acid sample by the number of moles of the acid:

Molar mass of the acid = Mass of the acid sample / Moles of the acid
= 0.407 g / 0.002945 mol
= 138.2 g/mol

Therefore, the molar mass of the acid is 138.2 g/mol.

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