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Q.
Calculate molality of $1$ litre solution of $93\%$ $H_{2}SO_{4}$ by volume. The density of solution is $1.84 \,g\, ml^{-1}$
Solutions
Solution:
Given $H_{2}SO_{4}$ is 93% by volume
wt. of $H_{2}SO_{4}=93\,g$
Volume of solution = 100 ml
$\therefore $ Weight of solution $=100 \times 1.84\,g =184\,g$
wt. of water $= 184 - 93 = 91 \,g$
Molality $=\frac{\text{Moles}}{\text{wt. of water in kg}}$
$=\frac{93\times1000}{98\times91}=10.42$