$0.63 g$ of dibasic acid was dissolved in water. The volume of the solution was made $100 m L . 20 m L$ of this acid solution required $10 m L \frac{N}{5} N a O H$ solution. What are the equivalent mass and molecular mass of the acid?

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Solution

$N_{1} V_{1} (A c i d) \equiv N_{2} V_{2} (N a O H)$
$N_{1} \times 20 = \frac{1}{5} \times 10$
$N_{1} = \frac{1}{5} \times \frac{10}{20} = \frac{1}{10}$
Strength of the acid solution $=$ Eq. mass of the acid $\times N o r m a l i t y$
$= E \times \frac{1}{10} = \frac{E}{10} g / L$
Mass of acid in $100 m L$ of the solution $= 0.63$ (given)
So, $\frac{E}{100} = 0.63$ or $E = 63$
Mol. mass $= B a s i c i t y \times E q . m a s s$
$= 2 \times 63 = 126$ .

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0.63 g of dibasic acid was dissolved in water. The volume of the solution was made 100 mL. 20 mL of this acid solution required 10 mL N5NaOH solution. What are the equivalent mass and molecular mass of the acid?

N1V1(Acid)≡N2V2(NaOH)N1×20=15×10N1=15×1020=110Strength of the acid solution = Eq. mass of the acid ×Normality=E×110=E10g/LMass of acid in 100 mL of the solution =0.63 (given)So, E100=0.63 or E=63Mol. mass =Basicity×Eq.mass=2×63=126.

$0.63 g$ of dibasic acid was dissolved in water. The volume of the solution was made $100 m L . 20 m L$ of this acid solution required $10 m L \frac{N}{5} N a O H$ solution. What are the equivalent mass and molecular mass of the acid?