Question

$20c{m}^3$ of 0.25 N HCl, $20c{m}^3$ N HCl and $90c{m}^3$ of 0.1 N $HN{O}_3$ were mixed and diluted to a litre. The normality of the resulting solution is.

• A. 0.8 N
• B. 1.05 N
• C. 0.105 N
• D. 0.16 N

Answer 1

The correct option is C 0.105 N

$20\times 0.25+20\times 0.5+90\times 1=130\times N$
$\therefore N=\frac{5+10+90}{130}=\frac{105}{130}$
After dilution $130\times \left(\frac{105}{130}\right)=x\times 1000$
$x=0.105N$