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  

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.

the correct answer would be d. hope this !