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

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(D) $1$. Calculate moles of solute: $n = \frac{25 \ g}{250 \ g \ mol^{-1}} = 0.1 \ mol$. $2$. Calculate total mass of solution: $Mass_{solution} = Mas

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$1 \ M$ and $1 \ m$

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(D) $1$. Calculate moles of solute: $n = \frac{25 \ g}{250 \ g \ mol^{-1}} = 0.1 \ mol$. $2$. Calculate total mass of solution: $Mass_{solution} = Mass_{solute} + Mass_{solvent} = 25 \ g + 100 \ g = 125 \ g$. $3$. Calculate volume of solution: $V = \frac{Mass_{solution}}{Density} = \frac{125 \ g}{1.25 \ g \ mL^{-1}} = 100 \ mL = 0.1 \ L$. $4$. Calculate molarity $(M)$: $M = \frac{n}{V(L)} = \frac{0.1 \ mol}{0.1 \ L} = 1 \ M$. $5$. Calculate molality $(m)$: $m = \frac{n}{Mass_{solvent}(kg)} = \frac{0.1 \ mol}{0.1 \ kg} = 1 \ m$. 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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