Question #77505

How many grams of glocose are needed to lower the vapor pressre of water by 100.0 torr?

Other question:

A solution is prepared by mixing 23.5mL of carbon tetrachloride and 35.8 mL of 2,5-dichlorohexane. What is the expected vapor pressure of this solution? What is the moele fraction of carbon tetrachloride in the vapor above the solution? The density of carbon tetrachloride is 1.594 g/mL and its vapor pressure is 91 mmHg. The density of 2,5-dichlorohexane is 1.0g/mL and its vapor pressure is 2.3 mmHg.

Other question:

A solution is prepared by mixing 23.5mL of carbon tetrachloride and 35.8 mL of 2,5-dichlorohexane. What is the expected vapor pressure of this solution? What is the moele fraction of carbon tetrachloride in the vapor above the solution? The density of carbon tetrachloride is 1.594 g/mL and its vapor pressure is 91 mmHg. The density of 2,5-dichlorohexane is 1.0g/mL and its vapor pressure is 2.3 mmHg.

Expert's answer

At 54C vapor pressure of water is 112.5 torr.

P(solution) = x(solvent)*P(solvent)

12.5 = x*112.5

x = 0.11

n(C6H12O6) = 0.88 mol

m(C6H12O6) = 0.88mol*180g/mol = 158.4 g

m(CCl4) = 23.5mL*1.594g/mL = 37.459g

m(C6H12Cl2) = 35.8g

n(CCl4)= 37.459g/154g/mol = 0.24mol

x(CCl4)=0.24/0.24+0.23 = 0.51

P(solution) = 0.51*91 = 46.41mmHg

P(solution) = x(solvent)*P(solvent)

12.5 = x*112.5

x = 0.11

n(C6H12O6) = 0.88 mol

m(C6H12O6) = 0.88mol*180g/mol = 158.4 g

m(CCl4) = 23.5mL*1.594g/mL = 37.459g

m(C6H12Cl2) = 35.8g

n(CCl4)= 37.459g/154g/mol = 0.24mol

x(CCl4)=0.24/0.24+0.23 = 0.51

P(solution) = 0.51*91 = 46.41mmHg

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