The vapour pressure of pure water at 760 mm at 25 degree celcius. The vapour of solution containing 1m solution of glucose will be.

To calculate the vapor pressure of a solution containing a 1M solution of glucose, we can use Raoult's Law. According to Raoult's Law, the vapor pressure of a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent.

In this case, the pure solvent is water, and the solute is glucose.

1. Convert the given atmospheric pressure to units of Torr:
760 mm Hg × (1 Torr/1.33 mm Hg) = 570.68 Torr

2. Calculate the mole fraction of water (solvent) in the solution:
Since we have a 1M solution of glucose, we know that the concentration of water is 55.5 M (since glucose is a non-electrolyte and dissociates into the same number of particles as water).

The mole fraction of water (Xw) can be calculated using the equation:
Xw = (moles of water) / (moles of water + moles of glucose)

Considering 1 mole of glucose in 1L, the number of moles of water can be determined by:
Moles of water = 55.5 mol/L - (1 mol/L) = 54.5 mol/L

Xw = 54.5 mol/L / (54.5 mol/L + 1 mol/L) = 0.982

3. Apply Raoult's Law to calculate the vapor pressure of the solution:
Pvap solution = Xw * Pvap pure water

Pvap solution = 0.982 * 570.68 Torr = 560.14 Torr

Therefore, the vapor pressure of the solution containing a 1M solution of glucose would be approximately 560.14 Torr at 25 degrees Celsius.

To determine the vapor pressure of a solution containing 1M glucose, we need to use Raoult's law, which states that the vapor pressure of a solvent above a solution is proportional to the mole fraction of the solvent present.

In this case, we have a 1M solution of glucose in water. To find the mole fraction of water (which is our solvent), we need to know the number of moles of glucose and water present.

Since it is a 1M solution of glucose, we know that there is 1 mole of glucose in 1 liter of solution. However, we need to convert this to the number of moles of water present.

To do this, we need to know the density of the solution. Let's assume the density is 1 g/mL. Since the molar mass of glucose is 180 g/mol, we can calculate the number of moles of glucose as follows:

1 mol glucose = 1 L solution * (1 mol glucose / 1 L solution) = 1 mol glucose

Next, we need to calculate the number of moles of water in 1 L of solution. We know the density of the solution is 1 g/mL, so 1 L of solution would be 1000 g. Assuming the density of water is 1 g/mL, we have:

1000 g solution - 180 g glucose = 820 g water

Now, we can calculate the moles of water:

1 L solution * (820 g water / 1 L solution) * (1 mol water / 18 g water) = 45.6 mol water

So, in our solution, we have 45.6 moles of water and 1 mole of glucose.

Now we can calculate the mole fraction of water:

Mole fraction of water = moles of water / (moles of water + moles of glucose)
= 45.6 mol water / (45.6 mol water + 1 mol glucose)
= 45.6 / 46.6
≈ 0.979

Since the mole fraction of water is approximately 0.979, the mole fraction of glucose would be:

Mole fraction of glucose = 1 - Mole fraction of water
= 1 - 0.979
≈ 0.021

Using Raoult's Law, we can now calculate the vapor pressure of the solution containing 1M glucose.

Vapor pressure of solution = Mole fraction of water * Vapor pressure of pure water at the same temperature

Vapor pressure of solution ≈ 0.979 * 760 mmHg
≈ 743 mmHg

Therefore, the vapor pressure of the solution containing 1M glucose would be approximately 743 mmHg.

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Do you mean 1m or 1 M. Do you know the difference? I will assume you mean 1 M. Or is that 1 mol?

psolution = XH2O*PoH2O

If this is 1 m we can do it as is. If it 1M, we will the need the density of the solution.