Question

7.35 g of a dibasic acid was dissolved in water and diluted to 250 mL. If 25 mL of
this acid solution was neutralized with 15 mL of 1 N NaOH solution. Determine
the molecular weight of acid.

Answer 1

Answer:

n1v1=n2n2

so we get the normality of acid.

then we put the formula of normality.

my ans is 980gm. is it right or wrong

Answer 2

Answer: 61.25 g/mol

Explanation:-

Normality is defined as the number of gram equivalents dissolved per liter of the solution.

$\text{number of gram equivalents}}=Normality\times {\text {Volume in L}}$

$\text{number of gram equivalents}}=1N\times 0.015L=0.015$

As 1 gram equivalent of acid is neutralized by 1 gram equivalent of base, number of gram equivalents of acid are 0.015.

$Normality=\frac{\text{no of gram equivalents} \times 1000}{\text{Volume in ml}}=\frac{0.015\times 1000}{25ml}=0.6N$

According to dilution law,

$N_1V_1=N_2V_2$

where,

$N_1$ = normality of stock solution = ?

$V_1$ = volume of stock solution = 250 ml

$N_2$ =  normality of dilute solution = 0.6 N

$V_2$ = volume of dilute solution = 25 ml

$N_1\times 250=0.6\times 25$

$N_1=0.06N$

$Molarity=\frac{Normality}{valency}$

$Molarity=\frac{0.06}{2}=0.03M$

$Molarity=\frac{n\times 1000}{V_s}$

where,

n= moles of solute  

$V_s$ = volume of solution in ml = 250 ml

${\text {moles of solute}=\frac{\text {given mass}}{\text {molar mass}}=\frac{7.35g}{Mg/mol}$

Now put all the given values in the formula of molarity, we get

$0.03M=\frac{7.35\times 250}{M\times 1000ml}$

$M=61.25g/mol$

Thus molecular weight of acid is 61.25 g/mol.