Q13-115CP

Question

An aqueous solution is 10% glucose by mass $(d = 1.039 g / m L a t 20^{o} C)$ . Calculate its freezing point, boiling point at 1atm, and osmotic pressure.

Step-by-Step Solution

Verified

Answer

Freezing point of solution is $- 7 . 8^{\circ} C$ .

Boiling point of solution is $2 . 1^{\circ} C$ .

Osmotic pressure of solution is 100.98atm.

1Step 1: Molality of solution

Molality is the amount of a substance dissolved in certain mass of solvent. The moles of a solute per kilogramme of a solvent are how it is defined.

Here, an aqueous solution of 10% glucose means 10 g of glucose dissolved in 100 mL (0.1kg) of water.

molar mass of glucose(C₆H₁₂O₆)

$M = 6 \times 12 + 12 \times 1 + 6 \times 16$ =180g/mol.

Mass of solvent $= d e n s i t y \times v o l u m e = 1.039 g / m L \times 100 m L = 103.9 g$

Hence, the moles of solute $= \frac{(m a s s o f s o l u t e)}{(m o l a r m a s s o f s o l u t e)}$

$= \frac{10 g}{(180 g / m o l)} = 0.055 m o l e s$

Now, molality of solution $= \frac{(m o l e s o f s o l u t e)}{(m a s s o f s o l v e n t (k g)} = 0.055 m o l / 0.01039 k g = 4.20 m o l / k g$

2Step 2: freezing point

Freezing point of a substance is the temperature at which the vapor pressure of a substance in its liquid phase is equal the vapor pressure in solid phase.

To calculate it, use the formula below.

$T_{f} = K_{f} \times m$

Where T_f is freezing point depression, $K_{f} (- 1 . 86^{\circ} C / m)$ is freezing point depression constant and m is molality.

Hence,

$T_{f} = - 1 . 86^{\circ} C / m \times 4.20 m o l / k g = - 7 . 8^{\circ} C$

Therefore, the freezing point of solution is $7 . 8^{\circ} C$ .

3Step 3: boiling point

Boiling point is the temperature at which the pressure exerted by surroundings upon a liquid is equaled by the pressure exerted by the vapor of the liquid.

To calculate it, use the formula below.

$T_{b} = K_{b} \times m$

Where T_bis the boiling point elevation, K_b (0.512C/m) is boiling point elevation constant and m is molality.

$T_{b} = 0 . 512^{\circ} C / m \times 4.20 m o l / k g = 2 . 1^{\circ} C$

Therefore, the solution's boiling point is 2.1^. C.

4Step 4: osmotic pressure

Osmotic pressure is defined as the minimum pressure that must be applied to halt the flow of solvent molecules through a semipermeable membrane. Utilizing the formula n=MRT, it may be calculated. Where M is the molarity of solution, R gas constant and T temperature.

Given temperature $= 20^{\circ} C = 20 + 273 = 293 K$

Value of R is $0.082 L . a t m / m o l . K$

Hence, osmotic pressure

$Π = 4.20 m o l / k g \times 0.082 L . a t m / m o l . K \times 293 K = 100.90 a t m$