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(B) The number of equivalents is given by the formula: $	ext{Equivalents} = 	ext{Molarity} 	imes 	ext{Volume} 	imes 	ext{n-factor}$.For $H_2SO_4

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$2.4$

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(B) The number of equivalents is given by the formula: $	ext{Equivalents} = 	ext{Molarity} 	imes 	ext{Volume} 	imes 	ext{n-factor}$.For $H_2SO_4$,the n-factor (basicity) is $2$. The number of equivalents of $H_2SO_4$ is $0.6 \, M 	imes 10 \, mL 	imes 2 = 12 \, 	ext{meq}$.Since $y \, mL$ of $1 \, M \, NaOH$ neutralizes $10 \, mL$ of $0.6 \, M \, H_2SO_4$,the number of equivalents of $NaOH$ must be equal to the number of equivalents of $H_2SO_4$.Equivalents of $NaOH = 1 \, M 	imes y \, mL 	imes 1 = y \, 	ext{meq}$.Equating the two: $y = 12$.Now,for acid $X$,$5 \, mL$ is neutralized by $y \, mL$ $(12 \, mL)$ of $1 \, M \, NaOH$.Using the principle of equivalence: $N_X 	imes V_X = N_{NaOH} 	imes V_{NaOH}$.$N_X 	imes 5 \, mL = 1 \, N 	imes 12 \, mL$.$N_X = \frac{12}{5} = 2.4 \, N$.

$A$ solution $(5 \, mL)$ of an acid $X$ is completely neutralized by $y \, mL$ of $1 \, M \, NaOH$. The same volume $(y \, mL)$ of $1 \, M \, NaOH$ is required to neutralize $10 \, mL$ of $0.6 \, M \, H_2SO_4$ completely. The normality $(N)$ of the acid $X$ is $......$

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