25 g of a solute of molar mass 250 g mol^{ | vedclass

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(D) Number of moles of solute $= \frac{25 \ g}{250 \ g \ mol^{-1}} = 0.1 \ mol$. Volume of solution $= 100 \ mL = 0.1 \ L$. Molarity $(M) = \frac{	ex

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$1$ and $1$

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(D) Number of moles of solute $= \frac{25 \ g}{250 \ g \ mol^{-1}} = 0.1 \ mol$. Volume of solution $= 100 \ mL = 0.1 \ L$. Molarity $(M) = \frac{	ext{Moles of solute}}{	ext{Volume of solution in } L} = \frac{0.1 \ mol}{0.1 \ L} = 1 \ M$. Mass of solution $= 	ext{Density} 	imes 	ext{Volume} = 1.25 \ g \ mL^{-1} 	imes 100 \ mL = 125 \ g$. Mass of solvent (water) $= 	ext{Mass of solution} - 	ext{Mass of solute} = 125 \ g - 25 \ g = 100 \ g = 0.1 \ kg$. Molality $(m) = \frac{	ext{Moles of solute}}{	ext{Mass of solvent in } kg} = \frac{0.1 \ mol}{0.1 \ kg} = 1 \ m$. Thus,the molarity and molality are $1 \ M$ and $1 \ m$ respectively.

$25 \ g$ of a solute of molar mass $250 \ g \ mol^{-1}$ is dissolved in $100 \ mL$ of water to obtain a solution whose density is $1.25 \ g \ mL^{-1}$. The molarity and molality of the solution are respectively

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