Question

Which amongst the following has the highest normality ?

• A. $8.0 \,g$ of $KOH/100\, mL$ of solution
• B. $0.5$ molar $H_2SO_4$
• C. $1\, N\, H_3PO_4$
• D. $6\, g$ of $NaOH/100\, g$ of water

Answer 1

Answer: (D)

$6\, g$ of $NaOH/100\, g$ of water

Answer 2

$8 \,g \,KOH / 100cc=\frac{8}{56}\times\frac{1000}{100}=\frac{10}{7}N$ $=1.4\,N$, $0.5 \,M\, H_{2}SO_{4}=1\,N \, H_{2}SO_{4}$ $6\, g \,NaOH/ 100cc=\frac{6}{40}\times\frac{1000}{100}$ $=1.5\,N$