If 0.5 mol of a non-volatile solute is dissolved in 1 kg of water, what is the v

If 0.5 mol of a non-volatile solute is dissolved in 1 kg of water, what is the vapor pressure lowering? (Vapor pressure of pure water = 23.76 mmHg)

If 0.5 mol of a non-volatile solute is dissolved in 1 kg of water, what is the vapor pressure lowering? (Vapor pressure of pure water = 23.76 mmHg)

Step 1: Identify the number of moles of solute. In this case, it is given as 0.5 mol.
Step 2: Identify the mass of the solvent (water) given in the problem, which is 1 kg.
Step 3: Convert the mass of water to moles. The molar mass of water (H2O) is approximately 18 g/mol. Since 1 kg = 1000 g, we calculate the moles of water: 1000 g / 18 g/mol = 55.5 mol.
Step 4: Use the formula for vapor pressure lowering: Vapor pressure lowering = (n_solute / n_solvent) * P°_solvent.
Step 5: Substitute the values into the formula: n_solute = 0.5 mol, n_solvent = 55.5 mol, and P°_solvent = 23.76 mmHg.
Step 6: Calculate the vapor pressure lowering: (0.5 / 55.5) * 23.76 = 1.88 mmHg.

Raoult's Law – The vapor pressure of a solvent is directly proportional to the mole fraction of the solvent in the solution.
Mole Calculation – Understanding how to calculate the number of moles of solvent (water) based on its mass.
Vapor Pressure Lowering – The decrease in vapor pressure of a solvent when a non-volatile solute is added.