Question

10 ml of 10M H2SO4 is diluted to 250 ml, the strength of the diluted solution is
a) 0.80 N b) 0.40 N c) 1.0 N d) 0.60 N

Answer 1

Answer:

c. 1 .................

Answer 2

The strength of the diluted solution is a) 0.80 N.

Explanation:

Given:

• Initial concentration, $M_1 = 10M$
• Initial Volume, $V_1 = 10 mL$
• Final Volume, $V_2 = 250 mL$

To find: Strength of final solution

Formula used:

$M_1V_1 = M_2V_2$ and Normality, $N = n\times M$ where

n is called the n-factor and has a value of 2 for sulphuric acid$(H_2SO_4)$

Solution:

$M_1V_1 = M_2V_2$

$10 \times 10 = M_2\times 250$

$M_2 = \frac{100}{250}$

$M_2= 0.4M$

Now, the molarity of the resulting solution is 0.4 M

But, the question asks to find the strength in terms of normality(N)

Thus,  $N = n\times M = 2\times 0.4 = 0.8N$

Hence, the answer is a) 0.80 N