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Class 10 Class 12

Ethylene glycol (molar mass = 62 g mol^–1) is a common automobile auto freeze. Calculate the freezing point of a solution containing 12.4 g of this substance in 100 g of water. Would it be advisable to keep the substance in the car radiator during summer?

Answer:

W_b = 12.4
n_B= 62g/mol
W_A = 100g

k_{f} for water = 1.86 K kg / mol k_{b} for water = 0.512 K kg / mol

∆ T_{f} = \frac{k_{f} \times 1000 \times W_{B}}{W_{A} \times n_{B}}

∆ T_{f} = \frac{1.86 \times 1000 \times 12.4}{100 \times 62} = 3.72 K

Freezing point of the solution
= 273.15 - 3.72 = 269.43K.

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A peeled egg when dipped in water swells, while in saturated salt solution it shrinks. Why?

Answer:

In the first case endo-osmosis occurs because solution inside egg is concentrated one. In later case exo-osmosis occurs because solution outside egg is concentrated one.

552 Views

State the law co-relating the pressure of a gas and its solubility in a liquid. State an application of this law.

Answer:

Henry’s law states that the solubility of a gas in a liquid is directly proportional to pressure of the gas keeping temperature constant.
Mathematically, P = K_H x
x mole fraction of gas in solution
P is partial pressure of gas,
K_H is Henry's constant.

Applications: Henry's law finds various applications in industry and enables us to explain and understand some biological phenomena. The some of important applications are: CO₂solubility in soft drinks, beverages, soda water etc. is increased by applying high pressure and bottles are sealed under high pressure.

153 Views

Explain why solvent-solvent, solute-solute and solute-solvent interactions are important in determining the extent to which a solute dissolves in a solvent.

Answer:

The process of a solute dissolving in a solvent is called dissolution. Dissolution depend on the temperature because increasing temperture dissolution is also increase.
Dissolution also depend on the nature of the solute and solvent.

In the dissolution process, solute-solute, solvent-solvent interactions are established. The solubility of a solute thus, depends on the strength of new solute-solvent interactions relative to the strengths of solute-solute and solvent-solvent interactions.

165 Views

Calculate the temperature at which a solution containing 54 g of glucose, C₃H₁₂O₆, in 250 g of water will freeze. [K_ffor water = 1.86 K kg mol^–1]

Answer:

by applying the formula

∆ T_{f} = \frac{K_{f} \times W_{B} \times 1000}{W_{A} \times M_{B}} W_{A} = 250 g, W_{B} = 54 g, M_{B} = 180 g K_{f} = 1.86 K kg {mol}^{- 1}

Substituting the values

∆ T_{f} = \frac{1.86 \times 54 \times 1000}{250 \times 180} = 2.232 K

∴

Freezing point of solution
= 273 - 2.232 = 270.768 K.

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State the law co-relating the pressure of a gas and its solubility in a liquid. State an application of this law.