An antifreeze solution is prepared from 222.6 gram of ethylene glycol $(C_{2} H_{6} O_{2})$ and 200 gram 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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m = 17.95 M = 9.11

m = 10 M = 5

m = 23.3 M = 19

m =9.11 M =17.95

answer is A.

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Detailed Solution

Mass of the solute $Molarity of the solution = \frac{number of moles of solute}{volume of solution in litre}$ $C_{2} H_{6} O_{2} = 222.6 g r a m$

Molar mass of $C_{2} H_{6} O_{2} = 62$ gram per mole

Number of moles of solute 222.6/62 = 3.59

Mass of solvent 200 g = 0.200 kg

Molality of the solution(m) : number of moles/weight of solvent in kg

$m = 3.59 / 0.2 = 17.95$

Total mass of the solution : 222.6+200 = 422.6 gram

Volume of the solution = mass of solution/density of solution

$= 422.6 / 1.072 = 394.2 m L = 0.3942 L$

$= 3.59 / 0.3942 = 9.11 M$

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