A solution of Al2(SO4)3 d=1.253 gm/ml contain 22 % salt by weight. The molarity, normality and molality of the solution is

Question

A solution of Al2(SO4)3 d=1.253 gm/ml contain 22 % salt by weight. The molarity, normality and molality of the solution is (A) 0.825 M, 48.3 N, 0.805 M (B) 0.805 M, 4.83 N, 0.825 M (C) 4.83 M, 4.83 N, 4.83 M (D) None

Tardigrade Admin · Accepted Answer

Molarity = ( % ×10× d/GMM)=(22×10×1.253/342)=0.805 M Normality = ( % ×10× d/GEM)=(22×10×1.253/342 6)=4.83 N Molality =(22×1000/342(100-22))=0.825 m

Q. A solution of $A l_{2} (S O_{4})_{3}$ { $d = 1.253 g m / m l$ } contain $22%$ salt by weight. The molarity, normality and molality of the solution is

$0.825 M, 48.3 N, 0.805 M$

$0.805 M, 4.83 N, 0.825 M$

$4.83 M, 4.83 N, 4.83 M$

None

Molarity $= \frac{% \times 10 \times d}{GMM} = \frac{22 \times 10 \times 1.253}{342} = 0.805 M$
Normality $= \frac{% \times 10 \times d}{GEM} = \frac{22 \times 10 \times 1.253}{3426} = 4.83 N$
Molality $= \frac{22 \times 1000}{342 ( 100 - 22 )} = 0.825 m$

Q. A solution of Al2​(SO4​)3​ {d=1.253gm/ml} contain 22% salt by weight. The molarity, normality and molality of the solution is

0.825M,48.3N,0.805M

0.805M,4.83N,0.825M

4.83M,4.83N,4.83M