Calculate normality, molarity and molality of the solution containing 22 %( w / w ) salt Al 2( SO 4)3 d =1.253 gm / ml by weight is

Question

Calculate normality, molarity and molality of the solution containing 22 %( w / w ) salt Al 2( SO 4)3 d =1.253 gm / ml by weight is (A) 4.83 N, 8.25 M, 8.25 m (B) 48.3 N, 0.825 M, 0.825 m (C) 4.83 N, 4.83 M, 4.83 m (D) 4.83 N, 0.805 M,0.825 m

Tardigrade Admin · Accepted Answer

Normality =(w / w % × text10 × d/E2) =( text22 × text10 × text1.253/( text342)6)= text4.83 N Molarity =(w / w % × text10 × d/M2)=( text22 × text10 × text1.253/ text342)= text0.805 M E2= eq.mass of solute M2= molar ass of solute Molality =((w / w %) × 1000/M2 (100 - w / w %)) Molality =( text22 × text1000/ text342 (. text100 - text22 .))= text0.825 m

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

$4.83 N, 8.25 M, 8.25 m$

$48.3 N, 0.825 M, 0.825 m$

$4.83 N, 4.83 M, 4.83 m$

$4.83 N, 0.805 M, 0.825 m$

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

4.83N,8.25M,8.25m

48.3N,0.825M,0.825m

4.83N,4.83M,4.83m

4.83N,0.805M,0.825m

Normality =E2​w/w%×10×d​ =6342​22×10×1.253​=4.83 N Molarity =M2​w/w%×10×d​=34222×10×1.253​=0.805 M E2​= eq.mass of solute M2​= molar ass of solute Molality=M2​(100−w/w%)(w/w%)×1000​ Molality =342(100−22)22×1000​=0.825 m

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

$4.83 N, 8.25 M, 8.25 m$

$48.3 N, 0.825 M, 0.825 m$

$4.83 N, 4.83 M, 4.83 m$

$4.83 N, 0.805 M, 0.825 m$