A 2.567-g sample of a monoprotic acid was dissolved in water. It took 15.24 mL of a 0.1578 M NaOH solution to neutralize the acid. Calculate the molar mass of the acid.

The molar mass of the acid is 1067.42 g/mol  

Explanation:

A monotropic acid can donate one proton only in a acd base reaction thus

Monotropic acid + NaOH → NaSalt + H₂O

From the above reaction, one mole of NaOH reacts with one mole of monotropc acid

however there are 0.1578 M NaOH from 15.24 mL hence we have

Number of moles of NaOH = 0.1578 M/L × (15.24/1000) L = 0.002404872 M of NaOH

The number of moles of the monotropic acid = number of moles of NaOH = 0.0024 M. However Number of moles = mass/(molar mass)

Therefore molar mass = mass/number of moles = 2.567(2.4×10⁻³) = 1067.42 g/mol  

the temperature of the gas, in k, not celsius.

the answer to this question is 3.