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Consider two solutions, one formed by adding 10 g of glucose (C6H12O6) to 1 L of water and the other formed by adding 10 g of sucrose (C12H22O11) to 1 L of water.1) Calculate the vapor pressure for the first solution at 20 ?C. (The vapor pressure of pure water at this temperature is 17.5 torr.).2) Calculate the vapor pressure for the second solution at 20 ?C. (The vapor pressure of pure water at this temperature is 17.5 torr.).

Respuesta :

Answer:

The vapor pressure for the first solution at 20°C is 17.48 Torr.

The vapor pressure for the second solution at 20°C is 17.49 Torr.

Explanation:

1) mass of solute that is glucose = 10 g

Moles of glucose  ,[tex]n_1=\frac{10 g}{180 g/mol}=0.05555 mol[/tex]

Mass of water = m

Volume of water = V = 1L = 1000 mL

Density of water = 1 g/ml

[tex]m=d\times V=1 g/mL\times 1000 mL=1000 g[/tex]

Moles of water = [tex]n_2=\frac{1000 g}{18 g/mol}=55.55 mol[/tex]

Vapor pressure of the solution = [tex]p[/tex]

Vapor pressure of the pure solvent that is water = [tex]p_o=17.5 Torr[/tex]

Mole fraction of solute= [tex]\chi_1=\frac{n_1}{n_1+n_2}[/tex]

[tex]\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}[/tex]

[tex]\frac{17.5 Torr-p}{17.5 Torr}=\frac{0.05555 mol}{0.05555 mol+55.55 mol}[/tex]

[tex]p=17.48 Torr[/tex]

The vapor pressure for the first solution at 20°C is 17.48 Torr.

2 ) mass of solute that is sucrose= 10 g

Moles of sucrose ,[tex]n_1=\frac{10 g}{342 g/mol}=0.02924 mol[/tex]

Mass of water = m

Volume of water = V = 1L = 1000 mL

Density of water = 1 g/ml

[tex]m=d\times V=1 g/mL\times 1000 mL=1000 g[/tex]

Moles of water = [tex]n_2=\frac{1000 g}{18 g/mol}=55.55 mol[/tex]

Vapor pressure of the solution = [tex]p[/tex]

Vapor pressure of the pure solvent that is water = [tex]p_o=17.5 Torr[/tex]

Mole fraction of solute= [tex]\chi_1=\frac{n_1}{n_1+n_2}[/tex]

[tex]\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}[/tex]

[tex]\frac{17.5 Torr-p}{17.5 Torr}=\frac{0.02924 mol}{0.02924 mol+55.55 mol}[/tex]

[tex]p=17.49 Torr[/tex]

The vapor pressure for the second solution at 20°C is 17.49 Torr.

pstnonsonjoku

The vapor pressure is lowered in the presence of a solute.

Now we know that 1L of water = 1Kg of water

Number of moles of water in each case = 1000 g/18 g/mol =55.6 moles

For glucose;

Number of moles of glucose = 10g/180 g/mol = 0.056 moles

Vapor pressure =

17.5 - P/17.5 = 0.056 moles/55.6 moles + 0.056 moles

17.5 - P/17.5 =

17.5 - P = 0.018

P = 17.5 - 0.018

P = 17.482 torr

For the sucrose solution;

Number of moles of sucrose = 10 g/342 g/mol = 0.029 moles

Hence vapor pressure;

17.5 - P/17.5 = 0.029 moles/55.6 moles + 0.029 moles

17.5 - P = 0.0091

P = 17.5 - 0.0091

P = 17.49 torr

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