Commercially available H_2SO_4 is 98 g by we | vedclass

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(A) Given: Weight of $H_2SO_4 = 98 \ g$, Molar mass of $H_2SO_4 = 98 \ g \ mol^{-1}$, Weight of solvent (water) $= 2 \ g$. Molality $(m)$ is defined a

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$500$

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(A) Given: Weight of $H_2SO_4 = 98 \ g$, Molar mass of $H_2SO_4 = 98 \ g \ mol^{-1}$, Weight of solvent (water) $= 2 \ g$. Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent. Moles of $H_2SO_4 = \frac{	ext{Given mass}}{	ext{Molar mass}} = \frac{98 \ g}{98 \ g \ mol^{-1}} = 1 \ mol$. Weight of solvent in $kg = \frac{2 \ g}{1000} = 0.002 \ kg$. Molality $(m) = \frac{	ext{Moles of solute}}{	ext{Mass of solvent in } kg} = \frac{1 \ mol}{0.002 \ kg} = 500 \ m$.

Commercially available $H_2SO_4$ is $98 \ g$ by weight of $H_2SO_4$ and $2 \ g$ by weight of water. Its density is $1.38 \ g \ cm^{-3}$. Calculate the molality $(m)$ of $H_2SO_4$ (molar mass of $H_2SO_4$ is $98 \ g \ mol^{-1}$). (in $m$)

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