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Determine the vapor pressure of a solution at 25 0C that contains 76.6 g of glucose...

Determine the vapor pressure of a solution at 25 0C that contains 76.6 g of glucose (C6H12O6) in 250.0 mL of water. The vapor pressure of pure water at 250C is 23.8 torr.

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Concepts and reason

Colligative property:

It is a property that depends upon the mass of a solute present in a solution. It does not depend on the nature of the solute. Lowering vapor pressure, depression of freezing point and elevation of boiling point are the examples of colligative properties.

Vapor pressure:

Vapor pressure is defined as the pressure exerted by vapors above the liquid surface in equilibrium with liquid at that particular temperature. The vapor pressure of the solution is equal to the vapor pressure of the pure component in the solution multiplied by the mole fraction of that component in the solution.

Fundamentals

The relationship between the vapor pressure of the constituent of a solution and the mole fraction of the constituent (solute) is given by Raoult’s Law.

(1-Xoe
P
solution
solvent
or

х.
solution
solvent solvent

Here,

The vapor pressure of a solution is Р.
solution
.

The vapor pressure of pure solvent is solvent
.

The mole fraction of solute is solute
.

The mole fraction of solvent is X
solvent
.

The mole fraction of a solute is defined as the number of moles of the component (solute) divided by the total number of moles of all the components (solvent and solute) in the solution.

• Mathematically, it can be expressed as follows.

Moles of solute
Mole fraction,xolte
solute
Moles of solute +Moles of solution

The density of a substance is the mass of the substance per unit volume. Density is denoted by the symbol .

Mass
Density
Volume

The given information in the problem is written down.

The given vapor pressure of pure water at 25°C
is 23.8 torr
.

The given mass of glucose is 76.6g
.

The given volume of water is 250.0mL
.

The density of water is 1g/mL
. The mass of water is calculated by the density equation as follows.

т%3DPV

Substitute 1g/mL
for and 250.0mL
for .

m = 1g/mLx250.0 mL
250.0g

The mass of water is obtained to be 250.0g
.

The moles of glucose and water can be calculated as follows.

Mass of glucose
Moles of glucoseMolar mass of glucose
76.6g
180.156 g /mol
=0.425 mol

So, the number of moles of glucose is 0.425mol
.

Mass of water
Moles of water Molar mass of water
250.0g
18.01528 g/mol
13.88 mol

Thus, the number of moles of water is 13.88 mol
.

Use the below equation to calculate the mole fraction of glucose.

Moles of glucose
Mole fraction of glucose,x^c
Moles of glucose+Moles of water

Substitute as follows.

0.425mol
Mole fraction of glucose,xghcose 0.425mol +13.88mol
0.0297

Therefore, the mole fraction of glucose is 0.0297
.

The calculated mole fraction of glucose is 0.0297
.

The given vapor pressure of pure water at 25°C
is .

Use the below expression to find the vapor pressure of the solution as follows.

(1-x,
X
glucose
solution
water

Substitute 0.0297
for х.
glucose
and 23.8 torr
for water
.

Paion(1-0.0297) (23.8tor)
=23.09 torr
solution
(Rounded to three significant figures)
=23.1 torr

Hence, the vapor pressure of the solution is 23.1 torr
.

Ans:

The vapor pressure of the given solution at 25°C
is 23.1 torr
.

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