What is the approximate molality of 10 % (w/w | vedclass

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(B) Given,$10 \% (w/w)$ aqueous glucose solution means $10 \ g$ of glucose is present in $100 \ g$ of the solution.Mass of solute (glucose) $= 10 \ g$

Vedclass · Accepted Answer

$0.62$

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(B) Given,$10 \% (w/w)$ aqueous glucose solution means $10 \ g$ of glucose is present in $100 \ g$ of the solution.Mass of solute (glucose) $= 10 \ g$.Mass of solvent (water) $= 100 \ g - 10 \ g = 90 \ g = 0.09 \ kg$.Molar mass of glucose $= 180 \ g \ mol^{-1}$.Number of moles of glucose $= \frac{10 \ g}{180 \ g \ mol^{-1}} = 0.0556 \ mol$.Molality $(m) = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{0.0556 \ mol}{0.09 \ kg} \approx 0.617 \ m$.Rounding to the nearest option,the value is $0.62 \ m$.

What is the approximate molality of $10 \% (w/w)$ aqueous glucose solution (in $m$)? (Molar mass of glucose $= 180 \ g \ mol^{-1}$)

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