Calculate normality of the mixture obtained by mixing 100 ml of 0.1 N HCl and 50 ml of 0.25 N NaOH solution.

Meq. of HCl $=100\times 0.1=10$

Meq. of NaOH $=50\times 0.25=12.5$

$\therefore$ HCl and NaOH neutralize each other with equal equilents. 

Meq. of NaOH left = 12.5 - 10 = 2.5

Volume of new solution 

                          = 100 + 50 = 150 ml.

$N_{NaOH}$ left $=\frac{2.5}{150}=0.0167N$

Hence normality of the mixture obtained is 0.0167 N