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

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Solution

Solution contains $15$ % glucose in water means $15 g$ of glucose in $100 g$ of solution.
mass of $C_{6} H_{12} O_{6} = 15 g$
mass of $H_{2} O = 85 g$
molar mass of $C_{6} H_{12} O_{6} = 6 \times 12 + 12 \times 1 + 6 \times 16 = 72 + 12 + 96 = 180 g / m o l$
molar mass of $H_{2} O = 2 \times 1 + 1 \times 6 = 18 g / m o l$
Molarity of the solution $= \frac{n o . o f m o l e s o f s o l u t e (C_{6} H_{12} O_{6})}{V o l . o f S o l u t i o n i n L}$
Given $S$ of solution $= 1.5 g / m l$
$S = \frac{M}{V} V = \frac{100 g}{1.5 g} m l = 66.066 m l = 66.066 \times 10^{- 3} l$
$M o l a r i t y = \frac{15 / 180}{66.066 \times 10^{- 3}} = \frac{15 \times 1000}{180 \times 66.066} = 0.00126 \times 10^{3} = 1.26 M M o l a l i t y = \frac{15}{180} \times \frac{1000}{100} = 0.833 m$ .

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