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

For finding molality we need to find the moles of ethylene glycol.

$\text{Moles of ethylene glycol }= \frac{222.6}{62}= 3.59\ mol$

We know that :

$Molality = \frac{Moles\ of\ ethylene\ glycol}{Mass\ of\ water}\times100$

$= \frac{3.59}{200}\times100 = 17.95\ m$

Now for molarity :-

Total mass of solution = 200 + 222.6 = 422.6 g

$\text{Volume of solution}= \frac{422.6}{1.072}= 394.22\ mL$

$\text{So molarity }= \frac{3.59}{394.22}\times1000= 9.11\ M$

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