(A) Mass of sugar $= 3.42 \times 10^{-2} \ kg = 34.2 \ g$. Mass of sugar syrup $= 234.2 \ g$. Mass of solvent (water) $= 234.2 \ g - 34.2 \ g = 200 \

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(A) Mass of sugar $= 3.42 \times 10^{-2} \ kg = 34.2 \ g$. Mass of sugar syrup $= 234.2 \ g$. Mass of solvent (water) $= 234.2 \ g - 34.2 \ g = 200 \ g = 0.2 \ kg$. Moles of sugar $= \frac{\text{Mass}}{\text{Molar mass}} = \frac{34.2 \ g}{342 \ g \ mol^{-1}} = 0.1 \ mol$. Molality $(m) = \frac{\text{Moles of solute}}{\text{Mass of solvent in } kg} = \frac{0.1 \ mol}{0.2 \ kg} = 0.5 \ mol \ kg^{-1}$.

Class 12 Chemistry Solutions Method of expressing concentration of solution

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$3.42 \times 10^{-2} \ kg$ of sugar (molar mass $342 \ g \ mol^{-1}$) is dissolved in water to produce $234.2 \ g$ of sugar syrup. What is the molality of the sugar syrup?

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A
$0.50 \ mol \ kg^{-1}$
B
$0.76 \ mol \ kg^{-1}$
C
$0.67 \ mol \ kg^{-1}$
D
$0.85 \ mol \ kg^{-1}$

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$3.42 \times 10^{-2} \ kg$ of sugar (molar mass $342 \ g \ mol^{-1}$) is dissolved in water to produce $234.2 \ g$ of sugar syrup. What is the molality of the sugar syrup?

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$A$ solution of $Al_2(SO_4)_3$ $(d = 1.253 \ g/mL)$ contains $22\%$ salt by weight. The molarity,normality,and molality of the solution are:

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