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(C) To find the highest normality,we calculate the normality $(N)$ for each option:$(A)$ $8 \, g$ of $KOH / 	ext{litre}$: Molar mass of $KOH = 56 \,

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

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(C) To find the highest normality,we calculate the normality $(N)$ for each option:$(A)$ $8 \, g$ of $KOH / 	ext{litre}$: Molar mass of $KOH = 56 \, g/mol$. $N = \frac{8}{56} 	imes 1 = 0.143 \, N$.$(B)$ $1 \, N$ phosphoric acid: Given as $1 \, N$.$(C)$ $6 \, g$ of $NaOH / 100 \, mL$: Molar mass of $NaOH = 40 \, g/mol$. $N = \frac{6 	imes 1000}{40 	imes 100} = 1.5 \, N$.$(D)$ $0.5 \, M \, H_2SO_4$: For $H_2SO_4$,$n$-factor = $2$. $N = M 	imes n	ext{-factor} = 0.5 	imes 2 = 1.0 \, N$.Comparing the values: $0.143 \, N Thus,the solution with $6 \, g$ of $NaOH / 100 \, mL$ has the highest normality.

Which of the following solutions has the highest normality?

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