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(A) The formula for normality $(N)$ is $N = \frac{	ext{mass in } g 	imes 1000}{	ext{Equivalent mass} 	imes 	ext{Volume in } mL}$.First,calculate

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$0.3$

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(A) The formula for normality $(N)$ is $N = \frac{	ext{mass in } g 	imes 1000}{	ext{Equivalent mass} 	imes 	ext{Volume in } mL}$.First,calculate the equivalent mass $(E)$ of $H_3PO_4$. The molar mass of $H_3PO_4$ is $98 \ g/mol$ and its basicity is $3$.$E = \frac{98}{3} \approx 32.67 \ g/eq$.Now,substitute the values into the formula:$N = \frac{4.9 	imes 1000}{32.67 	imes 500} = \frac{4900}{16335} \approx 0.3 \ N$.

What will be the normality of a solution containing $4.9 \ g$ of $H_3PO_4$ dissolved in $500 \ mL$ water (in $N$)?

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