Which amongst the following has the highest normality ?

Question

Which amongst the following has the highest normality ? (A) 8.0 g of KOH/100 mL of solution (B) 0.5 molar H2SO4 (C) 1 N H3PO4 (D) 6 g of NaOH/100 g of water

Tardigrade Admin · Accepted Answer

8 g KOH / 100cc=(8/56)×(1000/100)=(10/7)N =1.4 N, 0.5 M H2SO4=1 N H2SO4 6 g NaOH/ 100cc=(6/40)×(1000/100) =1.5 N

Q. Which amongst the following has the highest normality ?

$8.0 g$ of $K O H /100 m L$ of solution

$0.5$ molar $H_{2} S O_{4}$

$1 N H_{3} P O_{4}$

$6 g$ of $N a O H /100 g$ of water

$8 g K O H /100 cc = \frac{8}{56} \times \frac{1000}{100} = \frac{10}{7} N$
$= 1.4 N$ ,
$0.5 M H_{2} S O_{4} = 1 N H_{2} S O_{4}$
$6 g N a O H /100 cc = \frac{6}{40} \times \frac{1000}{100}$
$= 1.5 N$

Q. Which amongst the following has the highest normality ?

8.0g of KOH/100mL of solution

0.5 molar H2​SO4​

1NH3​PO4​

6g of NaOH/100g of water

8gKOH/100cc=568​×1001000​=710​N =1.4N, 0.5MH2​SO4​=1NH2​SO4​ 6gNaOH/100cc=406​×1001000​ =1.5N

$8.0 g$ of $K O H /100 m L$ of solution

$0.5$ molar $H_{2} S O_{4}$

$1 N H_{3} P O_{4}$

$6 g$ of $N a O H /100 g$ of water

$8 g K O H /100 cc = \frac{8}{56} \times \frac{1000}{100} = \frac{10}{7} N$
$= 1.4 N$ ,
$0.5 M H_{2} S O_{4} = 1 N H_{2} S O_{4}$
$6 g N a O H /100 cc = \frac{6}{40} \times \frac{1000}{100}$
$= 1.5 N$