QUESTION 2.8

An antifreeze solution is prepared from 222.6 g of ethylene glycol $(C_{2} H_{6} O_{2})$ and 200 g of water. Calculate the molality of the solution. If the density of the solution is $1.072 g m L^{- 1}$ , then what shall be the molarity of the solution?

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Solution

Step I: Calculation of molality of the solution
Mass of ethylene glycol = 222.6 g
Molar mass of ethylene glycol $(C_{2} H_{6} O_{2})$
$= (12 \times 2) + (1 \times 6) + (16 \times 2)$
$= 24 + 6 + 32 = 62 g m o l^{- 1}$
Mass of solution = 200g +222.6 g = 422.6 g = 0.4226kg
Molality (m) $= \frac{\frac{M a s s o f e t h y l e n e g l y c o l}{M o l a r m a s s}}{M a s s o f s o l v e n t i n k g}$
= $\frac{\frac{(222.6 g)}{(62 g m o l^{- 1})}}{0.4226 k g}$ $= 8.49 m o l k g^{- 1}$

Step II: Calculation of molarity of the solution
Density of the solution $= 1.072 g m L^{- 1}$
Mass of solution = Mass of solute + Mass of solution
=222.6g + 200g = 422.6g
Volume $= \frac{M a s s}{D e n s i t y} = \frac{422.6 g}{1.072 g m L^{- 1}} = 394.2 m L$
$= 0.3942 L$
Molarity (M) $= \frac{\frac{M a s s o f e t h y l e n e g l y c o l}{M o l a r m a s s}}{V o l u m e i n l i t r e s}$
$= \frac{\frac{(222.6 g)}{(62 g m o l^{- 1})}}{0.3942 L}$
$= 9.10 m o l L^{- 1} = 9.10 M$

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Molality

Standard XII Chemistry

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QUESTION 2.8 An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g m L−1, then what shall be the molarity of the solution?

QUESTION 2.8

An antifreeze solution is prepared from 222.6 g of ethylene glycol $(C_{2} H_{6} O_{2})$ and 200 g of water. Calculate the molality of the solution. If the density of the solution is $1.072 g m L^{- 1}$ , then what shall be the molarity of the solution?