Question

A 48.4-g sample of glucose (a nondissociated, nonvolatile solute with the formula c6h12o6) is dissolved in 151.2 g of water. what is the vapor pressure of this solution at 100°c?

Answer 1

Answer:

Vapor pressure of solution is 736mmHg

Explanation:

Vapor pressure of an ideal solution follows Raoult's law:

$P_{solution} = X_{solvent}P_{solvent}$

Where P is vapor pressure and X is mole fraction

Moles of glucose are:

48.4g × (1mol / 180.156g) = 0.2687mol glucose

Moles of water:

151.2g × (1mol / 18.1g) = 8.354 mol water

Thus, mole fraction of water (Solvent) is:

8.354 mol / (8.354 mol + 0.2687mol) = 0.9688

Vapor pressure of a solvent at boiling point is equal to atmospheric pressure (760mmHg). Replacing in Raoult's law:

$P_{solution} = 0.9688*760mmHg$

Vapor pressure of solution is 736mmHg

Answer 2

The vapor pressure of the solution will be "736 mmHg".

Vapor pressure

According to the question,

Sample of glucose = 48.4 g

Mass of water = 151.2 g

Temperature = 100°C

Now,

Moles of glucose will be:

= 48.4 × ($\frac{1 \ mol}{180.156}$)

= 0.2687 mol

Moles of water will be:

= 151.2 × ($\frac{1 \ mol}{18.1}$)

= 8.354 mol

Water's moles fraction will be:

= $\frac{Moles \ of \ water}{Moles \ of \ water +Glucose}$

By substituting the values,

= $\frac{8.354}{8.354+0.2687}$

= 0.9688

hence,

By using Raoult's Law,

→ $P_{solution}$ = $X_{solvent} P_{solvent}$

By substituting the values,

                = 0.9688 × 760

                = 736 mmHg

Thus the above solution is correct.55

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