A mixture of methanol (methyl alcohol) and water contains 60.0% water by mass. A

Question

A mixture of methanol (methyl alcohol) and water contains 60.0% water by mass. Assuming volume additivity of the components, estimate the specific gravity of the mixture at 20 degree C. What volume (in liters) of this mixture is required to provide 150 mol of methanol? Repeat Part (a) with the additional information that the specific gravity of the mixture at 20 degree C is 0.9345 (making it unnecessary to assume volume additivity). What percentage error results from the volume additivity assumption?

Explanation / Answer

Ans. A. Density of methanol at 200C = 0.7920 g/ mL

(Ref: http://www.viscopedia.com/viscosity-tables/substances/methanol/ )

Density of water at 200C = 0.9982 g/mL

Let the mass of the mixture = 100.0 g

Mass of water = 60 % (m/m) of 100.0 g = 60.0 g

Mass of methanol = Total mass- mass of water = 100.0 g- 60.0 g = 40.0 g

Now,

Total volume of the mixture = Volume of 60.0 g water + Volume of 40.0 g methanol

                                    = (Mass/density) of 60.0 g water + (Mass/density) of 40.0 g methanol

                                    = (60.0 g/ 0.9982 g mL-1) + (40.0 g/ 0.7920 g mL-1)

                                    = 60.108195 mL + 50.505051 mL

                                    = 110.613246 mL

Density of the mixture = Mass of the mixture / Volume of the mixture

                                    = 100.0 g / 110.613246 mL

                                    = 0.904051 g/mL

Specific gravity of the mixture = Density of mixture / Density of reference material (water)

                                    = (0.904051 g/mL) / (0.9982 g/mL)

                                    = 0.905681

Therefore, specific gravity of the mixture = 0.905681

Now,

Volume of methanol required:

Mass of 150 mol methanol = 150 mol x molar mass of methanol

                                                = 150 mol x (32.04216 g mol-1)

                                                = 4806.324 g

Volume of 150 mol methanol = volume of 4806.324 g methanol

                                                = (Mass / density) of 4806.324 g methanol

                                                = 6068.591 mL

            Therefore, the required volume = 6068.591 mL

Ans. B. Given, actual specific gravity of the mixture at 200C = 0.9345

            Error = Actual value – calculated value

                        = 0.9345 - 0.905681 = 0.028819

% error = (error / actual value) x 100

            = (0.028819 - 0.9345) x 100

            = 3.084 %

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