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03/01/2022 2:59 pm
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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 mL-1, then what shall be the molarity of the solution?
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03/01/2022 3:06 pm
Molar mass of ethylene glycol [C2H4OH)2]
= 2 × 12 + 6 × 1 + 2 ×16
= 62 g mol-1
Number of moles of ethylene glycol = \(\frac{222.6g}{62\;gmol^{-1}}\)
= 3.59 mol
Therefore, molality of the solution = \(\frac{3.59\;mol}{0.200\;kg}\)
= 17.95 m
Total mass of the solution = (222.6 + 200) g
= 422.6 g
Given,
Density of the solution = 1.072 g mL-1
∴ Volume of the solution = \(\frac{422.6\;g}{1.072\;g\;mL^{-1}}\)
= 394.22 mL
= 0.3942 × 10-3 L
Molarity of the solution = \(\frac{3.59\;mol}{0.39422 \times 10^{-3}L}\)
= 9.11 M