Question

If 20.0 mLmL of a 0.100 NN acid solution is needed to reach the end point in titration of 32.5 mLmL of a base solution, what is the normality of the base solution

Answer 1

Answer:

The correct answer is "0.051 N".

Explanation:

Given

Volume,

$V_1=20.0 \ mL$

$V_2=32.5 \ mL$

Normality,

$N_1=0.1 \ N$

$N_2=?$

As we know,

⇒ $N_1V_1=N_2V_2$

or,

⇒ $N_2=\frac{N_1V_1}{V_2}$

On putting the values, we get

⇒      $=\frac{0.1\times 20.0}{32.5}$

⇒      $=\frac{2}{32.5}$

⇒      $=0.051 \ N$