The molality of 648 g of pure water is ..... | vedclass

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(B) Molality is defined as the number of moles of solute per kilogram of solvent.For pure water,the density is $1 \ g \ mL^{-1}$.In $648 \ g$ of water

Vedclass · Accepted Answer

$55.5$

Vedclass · Answer

(B) Molality is defined as the number of moles of solute per kilogram of solvent.For pure water,the density is $1 \ g \ mL^{-1}$.In $648 \ g$ of water,the mass of the solvent (water) is $0.648 \ kg$.The number of moles of water $(H_2O)$ in $648 \ g$ is calculated as: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{648 \ g}{18 \ g \ mol^{-1}} = 36 \ mol$.Molality $(m)$ $= \frac{	ext{moles of solute}}{	ext{mass of solvent in } kg} = \frac{36 \ mol}{0.648 \ kg} \approx 55.55 \ m$.

The molality of $648 \ g$ of pure water is ........... $m$.

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