What will be the normality of a solution containing $4.9 g H_{3} P O_{4}$ dissolved in $500 m L$ water?

0.3

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Solution

The correct option is A $0.3$
$H_{3} P O_{4} dissociates into 3 H^{+} a n d P O_{4}^{³ -}$

$s o, n - factor of H_{3} P O_{4} = 3$

$mass of H_{3} P O_{4} = 4.9 g / m o l$

$molar mass of H_{3} P O_{4} = 3 \times 1 + 31 + 4 \times 16 = 3 + 31 + 64 = 98 g / m o l$

$so, no of moles of H_{3} P O_{4} = 4.9 / 98 = 0.1 / 2 = 0.05 m o l$

$now , molarity = \frac{no of moles of H_{3} P O_{4}}{volume of solution in Litre}$

$= \frac{0.05}{0.5}$

$= 0.1 M$

$w e k n o w,$

$n o r m a l i t y = n - f a c t o r \times m o l a r i t y$

$s o, n o r m a l i t y = 3 \times 0.1 = 0.3 N$

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