The molality and molarity of a solution of gl | vedclass

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(C) Mass by mass percentage is the mass of solute dissolved in $100 \ g$ of the solution.Mass of solution $= 100 \ g$Mass of glucose $= 10 \ g$Mass of

Vedclass · Accepted Answer

$0.617 \ m, 0.67 \ M$

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(C) Mass by mass percentage is the mass of solute dissolved in $100 \ g$ of the solution.Mass of solution $= 100 \ g$Mass of glucose $= 10 \ g$Mass of solvent $= 90 \ g$Molar mass of glucose $(C_6H_{12}O_6) = 12 	imes 6 + 1 	imes 12 + 16 	imes 6 = 180 \ g \ mol^{-1}$Molality $= \frac{	ext{Mass of solute}}{	ext{Molar mass of solute}} 	imes \frac{1000}{	ext{Mass of solvent (in g)}}$$= \frac{10}{180} 	imes \frac{1000}{90} = 0.617 \ m$Volume of solution $= \frac{	ext{Mass of solution}}{	ext{Density of solution}} = \frac{100 \ g}{1.2 \ g \ mL^{-1}} = 83.33 \ mL$Molarity $= \frac{	ext{Mass of solute}}{	ext{Molar mass of solute}} 	imes \frac{1000}{	ext{Volume of solution (in mL)}}$$= \frac{10}{180} 	imes \frac{1000}{83.33} = 0.67 \ M$

The molality and molarity of a solution of glucose in water which is labelled as $10 \% (w/w)$ are respectively (density of solution $= 1.2 \ g \ mL^{-1}$)

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