Determine the vapor pressure lowering associated with 1.32 m C 6 H 12 O 6 solution (MW:180.156 g/mol) at 25°C. The vapor pressure of pure water at 25°C is 23.8 torr.

Determine the vapor pressure lowering associated with 1.32 molle of glucose solution. Here, the molar weight of glucose is 180.156 grams per molle at 25 degrees Celsius. And then here the vapor pressure of pure water at 25 degrees Celsius is 23.8 2 hours. Alright. So here we're going to say that 1.32 mole is molality. Molality is moles of solute over kilograms of solvent. So what it really means is 1.32 moles of glucose divided by 1 kilogram of our solvent which is water. Now here they're talking about the big pressure of pure water so that tells us also it's water. Alright. So here we need the pressure of our solution, so that's what we need to find. So we need to say that pressure of our solution, we need the moles of solvent first. 1 kilogram of water is the same thing as 1,000 grams of water and one mole of water is 18.016 grams. So this equals 55 0.506 moles for water. So we're gonna take that and plug it in for our moles of solvent divided by I. Glucose is covalent, so it doesn't break up into ions, so I is 1, times the moles of our glucose, which is already given to us from the molality, plus the moles of water again, and then this times the pressure of our pure solvent which is 23.8 tors. So when I do that, I'm gonna get as my answer, 0.976 6 tours as my final answer. So here, that give me option d as the correct choice.

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14. Solutions

Vapor Pressure Lowering (Raoult's Law)

General Chemistry

14. Solutions

Vapor Pressure Lowering (Raoult's Law)

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Multiple Choice

Determine the vapor pressure lowering associated with 1.32 m C₆H₁₂O₆ solution (MW:180.156 g/mol) at 25°C. The vapor pressure of pure water at 25°C is 23.8 torr.

0.553 torr

27.6 torr

23.2 torr

0.976 torr

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Verified step by step guidance

Start by understanding the concept of vapor pressure lowering, which is a colligative property. It depends on the number of solute particles in a solution, not their identity.

Use Raoult's Law for vapor pressure lowering: ΔP = X_solute * P°_solvent, where ΔP is the vapor pressure lowering, X_solute is the mole fraction of the solute, and P°_solvent is the vapor pressure of the pure solvent.

Calculate the mole fraction of the solute (C6H12O6). First, find the moles of solute using the molality (m = 1.32 mol/kg) and assume 1 kg of water as the solvent. Moles of solute = 1.32 mol.

Calculate the moles of water. The molar mass of water is approximately 18.015 g/mol, so 1 kg of water is 1000 g, which is 1000 g / 18.015 g/mol = 55.51 mol.

Determine the mole fraction of the solute: X_solute = moles of solute / (moles of solute + moles of solvent). Substitute the values to find X_solute, then use Raoult's Law to calculate ΔP = X_solute * 23.8 torr.