Which of the following relations is incorrect for solutions ?

3 N A l_{2} (S O_{4})_{3} = 0.5 M A l_{2} (S O_{4})_{3}

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3 M H_{2} S O_{4} = 6 N H_{2} S O_{4}

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1 M H_{3} P O_{4} = 1 / 3 N H_{3} P O_{4}

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1 M A l_{2} (S O_{4})_{3} = 6 N A l_{2} (S O_{4})_{3}

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Solution

The correct option is A $1 M H_{3} P O_{4} = 1 / 3 N H_{3} P O_{4}$
We know that, $N o r m a l i t y = M o l a r i t y \times n - f a c t o r$
$A . 3 N A l_{2} (S O_{4})_{3} = 0.5 M A l_{2} (S O_{4})_{3}$
n-factor for $A l_{2} (S O_{4})_{3} = 6$
So, $0.5 M \times 6 = 3 N A l_{2} (S O_{4})_{3}$

$B . 3 M H_{2} S O_{4} = 6 N H_{2} S O_{4}$
n-factor for $H_{2} S O_{4} = 2$
So, $3 M \times 2 = 6 N H_{2} S O_{4}$
$(M o l a r i t y \times n - f a c t o r = N o r m a l i t y)$

$D . 1 M A l_{2} (S O_{4})_{3} = 6 N A l_{2} (S O_{4})_{3}$
n-factor for $A l_{2} (S O_{4})_{3} = 6$
So, $1 M \times 6 = 6 N A l_{2} (S O_{4})_{3}$

$C .$ n-factor for $H_{3} P O_{4} = 3$
So, $1 M H_{3} P O_{4} = 1 M \times 3 = 3 N H_{3} P O_{4}$

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