Question

A 0.623 g sample of a monoprotic acid is dissolved in water and titrated with 0.260 M KOH.

Answer 1

Answer:

126. g/mol is the molar mass of the acid .

Explanation:

$HA+KOH\rightarrow KA+H_2O$

Moles of KOH = n

Volume of the solution of KOH = 19.0 mL = 0.019 L

Molarity of the KOH solution = 0.260 M

$Molarity=\frac{Moles}{Volume(L)}$

$0.260M=\frac{n}{0.019 L}$

$n=0.260 M\times 0.019 L=0.00494 mol$

Accrding to recation, 1 mol KOH react with 1 mole of HA ,then 0.00494 mol of KOH will react with :

$\frac{1}{1}\times 0.00494 mol=0.00494 mol$ of HA

Mass of HA = 0.623 g

Molar mass of HA = M

$0.00494 mol=\frac{0.623 g}{M}$

[tex[M=\frac{0.623 g}{0.00494 mol}=126. g/mol[/tex]

126. g/mol is the molar mass of the acid .