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\; mL^{-1} then what shall be the molarity of the solution?

Answers (1)

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

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

Volume of solution

                                        = \frac{422.6}{1.072}= 394.22\ mL

So molarity :- 

                         = \frac{3.59}{394.22}\times1000= 9.11\ M

 

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