A 3.664 -g sample of a monoprotic acid was dissolved in water. It took \(20.27 \mathrm{~mL}\) of a \(0.1578 \mathrm{M}\) \(\mathrm{NaOH}\) solution to neutralize the acid. Calculate the molar mass of the acid.

Short Answer

Expert verified
The molar mass of the acid is approximately 1145 g/mol

Step by step solution

01

Calculate the moles of NaOH

To calculate the number of moles of NaOH use the formula: \(n = c \times V\). Here, \(c = 0.1578 \ M\), and \(V = 20.27 \ mL = 0.02027 \ L\). Then \(n_{NaOH} = 0.1578 \times 0.02027 = 0.0032 \ mol\)
02

Calculate the moles of the acid

Given that the sample is monoprotic, it means that one mole of acid will neutralize one mole of NaOH. So, the moles of the acid are also \(0.0032 \ mol\)
03

Calculate the molar mass

With the moles of acid and the mass, the molar mass can be calculated using the formula: \(Molar \ mass = mass \div moles\). Given that mass = 3.664 g and moles = \(0.0032 \ mol\), the molar mass is \( \frac{3.664}{0.0032} = 1145 \ g/mol\)

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Most popular questions from this chapter

The molar mass of a certain metal carbonate, \(\mathrm{MCO}_{3}\), can be determined by adding an excess of \(\mathrm{HCl}\) acid to react with all the carbonate and then "back titrating" the remaining acid with a \(\mathrm{NaOH}\) solution. (a) Write an equation for these reactions. (b) In a certain experiment, \(20.00 \mathrm{~mL}\) of \(0.0800 \mathrm{M} \mathrm{HCl}\) were added to a 0.1022 -g sample of \(\mathrm{MCO}_{3}\). The excess HCl required \(5.64 \mathrm{~mL}\) of \(0.1000 \mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the molar mass of the carbonate and identify \(\mathrm{M}\).

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