A solution contains $15 %$ glucose $[C_{6} H_{12} O_{6}]$ in water and has density $1.5$ g/ml. Find the molarity and molality of this solution $[C = 12, H = 1, O = 16]$ .

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Solution

$15$ % glucose $C_{6} H_{12} O_{6}$ in $H_{2} O$
$15 g$ of glucose in $85 g$ of $H_{2} O$

Solute $⟶$ glucose $⟶ 15 g$

Solvent $⟶ H_{2} O ⟶ 85 g$

No. of moles of solute= $\frac{G i v e n M a s s}{M o l a r M a s s}$

Molar Mass of $C_{6} H_{12} O_{6}$ = $6$ x $12 + 12$ x $1 + 6$ x $16$
= $72 + 12 + 96 = 180$
$n = \frac{15}{180} = 0.0833 D e n s i t y ⟶ 1.5 g / m L 1.5 g ⟶ 1 m L 1 g ⟶ \frac{1}{1.5} m L 100 g ⟶ \frac{1}{1.5} \times 100 m L \frac{100}{1.5} m L o f s o l u t i o n \frac{100}{1.5} \times \frac{1}{1000} L o f s o l u t i o n \frac{1}{15} L o f s o l u t i o n M o l a r i t y = \frac{M o l e s o f s o l u t e}{V o l u m e o f s o l u t i o n i n L} = \frac{15 / 180}{1 / 15} = \frac{15}{180} \times \frac{15}{1} = 1.25 M M o l a l i t y = \frac{M o l e s o f s o l u t e}{M a s s o f s o l v e n t i n k g} = \frac{15 / 180}{85 / 1000} = \frac{15}{180} \times \frac{1000}{85} = \frac{50}{51} = 0.980 m$

Molarity of solution= $1.25 M$

Molality of solution= $0.980 m$

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