Calculate the vapor pressure of a solution made by dissolving 550 grams of glucose (molar mass = 180.2 g/mol) in 1020.0 ml of water at 25°C. The vapor pressure of pure water at 25°C is 23.76 mm Hg. Assume the density of the solution is 1.00 g/ml. (760 torr = 760 mmHg = 1 atm)

22.55 mmHg (0.03 atm)

Explanation:

According to Raoult's law, the vapor pressure (Psolution) of a solution is given by:

Psolution = Xsolvent x Psolvent

Where Xsolvent is the mole fraction of the solvent in the solution and Psolvent is the vapor pressure of the pure solvent.

From the data, we have: Psolvent = 25.76 mmHg

We have to calculate Xsolvent, which is equal to the moles of solvent divided into the total number of moles.

The solution is composed of the solute (glucose) dissolved in the solvent (water). So, the total number of moles is calculated from the moles of solute and solvent.

To calculate the moles of solute (glucose), we divide the mass of glucose into its molar mass:

moles of glucose = mass/molar mass = 550 g/(180.2 g/mol) = 3.05 mol

The same for the moles of solvent (water). The mass of water is obtained from the product of the volume and density:

mass of water = volume x density = 1020.0 mL x 1.00 g/mL = 1020.0 g

molar mass H₂O = (1 g/mol x 2) + 16 g/mol = 18 g/mol

moles of water = mass water/molar masss = 1020.0 g/(18 g/mol) = 56.67 mol

Now, we can calculate Xsolvent:

Xsolvent = moles of water/total moles

total moles = moles glucose + moles water = 3.05 mol + 56.67 mol = 59.72 mol

⇒ Xsolvent = 56.67 mol/(59.72 mol) = 0.9489

Finally, we calculate the vapor pressure of the solution:

Psolution = 0.9489 x 23.76 mmHg = 22.55 mmHg

22.55 mmHg x 1 atm/760 mmHg = 0.03 atm

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