The v.p.of aqueous solution of cane sugar ( mass 342 ) is 756mmHg at 373k. How many grams of sugar in 1 kg of water ( remember v.p.of pure water at 373k is 1 atm )

{ answer:-10 gm }

To solve the problem of determining how many grams of cane sugar are dissolved in 1 kg of water, we can use Raoult's Law, which relates the vapor pressure of a solution to the vapor pressure of the pure solvent and the mole fraction of the solvent in the solution. Let's break this down step by step.

Understanding Vapor Pressure and Raoult's Law

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid phase. For pure water at 373 K (which is the boiling point), the vapor pressure is 1 atm, equivalent to 760 mmHg. When a non-volatile solute, like cane sugar, is added to water, the vapor pressure of the solution decreases. This is where Raoult's Law comes into play:

P_solution = X_solvent * P°_solvent

Where:

• P_solution is the vapor pressure of the solution.
• X_solvent is the mole fraction of the solvent (water in this case).
• P°_solvent is the vapor pressure of the pure solvent (760 mmHg for water at 373 K).

Calculating the Mole Fraction

Given that the vapor pressure of the aqueous solution of cane sugar is 756 mmHg, we can find the mole fraction of water in the solution:

X_solvent = P_solution / P°_solvent

Substituting the values:

X_solvent = 756 mmHg / 760 mmHg = 0.9947368

Finding the Mole Fraction of Sugar

The mole fraction of sugar (X_sugar) can be calculated as:

X_sugar = 1 - X_solvent

So:

X_sugar = 1 - 0.9947368 = 0.0052632

Relating Mole Fraction to Mass

Now, let's relate the mole fraction of sugar to the mass of sugar in the solution. We know that:

X_sugar = n_sugar / (n_sugar + n_water)

Where:

• n_sugar is the number of moles of sugar.
• n_water is the number of moles of water.

For 1 kg of water, the number of moles of water (n_water) can be calculated as follows:

n_water = mass / molar mass = 1000 g / 18 g/mol = 55.56 moles

Setting Up the Equation

Now we can set up the equation using the mole fraction of sugar:

0.0052632 = n_sugar / (n_sugar + 55.56)

Cross-multiplying gives:

0.0052632 * (n_sugar + 55.56) = n_sugar

Expanding this, we get:

0.0052632 * n_sugar + 0.292 = n_sugar

Rearranging leads to:

n_sugar - 0.0052632 * n_sugar = 0.292

n_sugar * (1 - 0.0052632) = 0.292

n_sugar = 0.292 / 0.9947368 ≈ 0.293 moles

Calculating the Mass of Sugar

Finally, we can find the mass of sugar using its molar mass:

mass_sugar = n_sugar * molar mass_sugar

Substituting the values:

mass_sugar = 0.293 moles * 342 g/mol ≈ 10 g

Thus, the amount of cane sugar dissolved in 1 kg of water is approximately 10 grams. This aligns with the answer provided in your question.