The molality of 15% by wt solution of H2SO4 is

Question

The molality of 15% by wt solution of H2SO4 is (A) 18 (B) 2.6 (C) 1.2 (D) 1.8

Tardigrade Admin · Accepted Answer

15% w/w solution weight of solute = 15 g Weight of solvent = 85 g m=(wA × 1000/MA × wB) =(15 × 1000/98 × 85)=1.8 [A arrow S o l u t e , B arrow S o l v e n t ]

Q. The molality of 15% by wt solution of $H_{2} S O_{4}$ is

18

2.6

1.2

1.8

15% $w / w$ solution weight of solute = 15 g

Weight of solvent = 85 g

$m = \frac{w _{A} \times 1000}{M _{A} \times w _{B}}$

$= \frac{15 \times 1000}{98 \times 85} = 1.8$

$[A \to S o l u t e, B \to S o l v e n t]$

Q. The molality of 15% by wt solution of H2​SO4​ is

18

2.6

1.2

1.8

15% w/w solution weight of solute = 15 g Weight of solvent = 85 g m=MA​×wB​wA​×1000​ =98×8515×1000​=1.8 [A→Solute,B→Solvent]

Q. The molality of 15% by wt solution of $H_{2} S O_{4}$ is

15% $w / w$ solution weight of solute = 15 g

Weight of solvent = 85 g

$m = \frac{w _{A} \times 1000}{M _{A} \times w _{B}}$

$= \frac{15 \times 1000}{98 \times 85} = 1.8$

$[A \to S o l u t e, B \to S o l v e n t]$