A mixture of methanol (methyl alcohol) and water contains 50.00% water by mass.

Question

A mixture of methanol (methyl alcohol) and water contains 50.00% water by mass.

answer both parts plz<3

Problem 3.29 Mixing and Volume Additivity A mixture of methanol (methyl alcohol) and water contains 50.00% water by mass. Ideal Specific Gravity The specific gravity of water at 20°C is about 0.9982; the specific gravity of methanol at 20°C is about 0.7917 Assuming volume additivity of the components, estimate the specific gravity of the mixture at 20°C. SG = What volume of this mixture is required to provide 250.0 mol of methanol? liters SHOW HINT LINK TO TEXT By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor. Attempts: 0 of 5 used SAVE FOR LATER SUBMIT ANSWER Actual Specific Gravity Repeat part (a) with the additional information that the specific gravity of the mixture at 20°C is 0.9156 (making it unnecessary to assume volume additivity). What is the real volume of the mixture required to provide 250.0 mol methanol? What percentage error results from the volume-additivity assumption? liters SHOW HINT By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor. Attempts: 0 of 5 used SAVE FOR LATER SUBMIT ANSWER

Explanation / Answer

Part a

Let's take the initial mass of mixture = 100 g

Mass of water = 50 g

Density of water at 20°C = 0.9982 g/cm3

Volume of water = mass/density

= 50g / 0.9982 g/cm3

= 50.0901 cm3

Mass of methanol = 50 g

Density of methanol at 20°C = 0.7917 g/cm3

Volume of methanol = mass/density

= 50g / 0.7917 g/cm3

= 63.1552 cm3

Based on volume additivity

Total volume = 50.0901 + 63.1552 = 113.2453 cm3

Density of mixture = mass of mixture /volume of mixture

= 100/113.2453

= 0.8830 g/cm3

Specific gravity of mixture = 0.8830

Mass of given methanol = moles x molecular weight

= 250 mol x 32.04 g/mol

= 8010 g

50 g methanol consists = 113.2453 cm3 of mixture

8010 g methanol consists =( 113.2453 x 8010/50) cm3 of mixture

= 18141.897 cm3 x 10^-3L/cm3

= 18.141 L

Part b

If specific gravity of mixture = 0.9345

Density of mixture = 0.9156 g/cm3

Volume of mixture = mass of mixture / density of mixture

= 100/0.9156 = 109.2179 cm3

50 g methanol consists = 109.2179 cm3 of mixture

8010 g methanol consists =( 109.2179 x 8010/50) cm3 of mixture

= 17496.72 cm3 x 10^-3L/cm3

= 17.496 L

Percent error in volume

= (ideal value - actual value) * 100 / (actual value)

= (18.141 - 17.496)*100/17.496

= 3.686%

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