Which one of these solutions has the highest normality?
$6 g$ of $N a O H$ per $100 m L$
$8 g K O H$ per $100 m L$
$0.5 M H_{2} S O_{4}$
$1 N H_{3} P O_{4}$

Question

Which one of these solutions has the highest normality-6 g of NaOH per 100 mL8 g KOH per 100 mL0-5 M H-2-SO-4-1 N H-3-PO-4-

Toppr Admin · Accepted Answer

Normality of-xA0- 8-xA0-g-xA0-KOH per 100-xA0-mL is-N-856-xD7-1000100-1-4 N-Normality of 0-5-xA0-M-xA0-H2SO4N-Molarity-xD7-basisity-0-5-xD7-2-1 N-Normality of 6-xA0-g of NaOH per 100-xA0-mL-N-640-xD7-1000100-1-5 N-Normality of 1-xA0-N-xA0-H3PO4 is 1N-Therefore- the normality of 6-xA0-g of NaOH per 100-xA0-mL is highest-Hence- option C is correct-

Which one of these solutions has the highest normality?6 g of NaOH per 100 mL8 g KOH per 100 mL0.5 M H2SO41 N H3PO4

Normality of 8 g KOH per 100 mL is:N=856×1000100=1.4 N.Normality of 0.5 M H2SO4N=Molarity×basisity=0.5×2=1 N.Normality of 6 g of NaOH per 100 mL.N=640×1000100=1.5 N.Normality of 1 N H3PO4 is 1N.Therefore, the normality of 6 g of NaOH per 100 mL is highest.Hence, option C is correct.

Which one of these solutions has the highest normality?
$6 g$ of $N a O H$ per $100 m L$
$8 g K O H$ per $100 m L$
$0.5 M H_{2} S O_{4}$
$1 N H_{3} P O_{4}$