A solution of (Al2(SO4)3) contain 22% salt by weight and density of solution i

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Question -

A solution of $(A l_{2} (S O_{4})_{3})$ contain 22% salt by weight and density of solution is 1.253.The molarity, normality and molality of the solution is

Options:

A . 0.806 M, 4.83 N, 0.825 m

B . 0.825 M, 48.3 N, 0.805 m

C . 4.83 M, 4.83 N, 4.83 m

D . None

Answer: Option A
:
A

Molecular wt. of $A l_{2} (S O_{4})_{3} = 2 \times 27 + 3 \times (32 + 4 \times 16) = 342 g / m o l e$
Equivalent wt. of $A l_{2} (S O_{4})_{3} = E q . w t . o f A l^{3 +} + E q . W t o f S O_{4}^{2 -} = \frac{27}{3} + \frac{96}{2}$
$= 57 g e q$
No. of. equivalent per mole $= \frac{342}{57} = 6$
Let volume of solutions = 1 L
Wt. of solutions = $V \times d e n s i t y = 1000 \times 1.253 = 1253 g$
Wt. of solute = $1253 \times 22 % = 257.66$
Moles. of solute = $\frac{275.66}{342} = 0.806$
Wt. of solvent = $1253 - 257.66 = 977.34$
Molarity = $\frac{m o l e s o f s o l u t e}{V o l u m e o f s o l u t i o n} = 0.806 M$
Normality = $6 \times M o l a l i t y = 6 \times 0.806 = 4.836 N$
Molalit = $\frac{M o l e s o f s o l u t e}{W t . o f s o l v e n t i n k . g} = \frac{0.806}{977.34 / 1000} = 0.825 m$

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