At 30 oC, pure benzene has a vapor pressure of 121.8 mmHg. Dissolving 15.0 g of a non-volatile solute, R , in 250 g of benzene produced a solution having a vapor pressure of 120.2 mmHg. Determine the approximate MW of solute R

At a certain temperature, the vapor pressure of pure benzene (C6H6) is 0.930 atm. A solution was prepared by dissolving 10.0 g of a nondissociating, nonvolatile solute in 78.11 g of benzene at that temperature. The vapor pressure of the solution was found to be 0.900 atm. Assuming the solution behaves ideally, determine the molar mass of the solute.

Part #1:

30.0 g of glucose (C6H12O6, molar mass = 180.16 g/mol) are dissolved in 100. grams of water (molar mass 18.02). What is the mole fraction of water in solution? Note: glucose is a non-volatile solute.

Part #2:

At 25.0 oC, pure water has a vapor pressure of 24.2 torr. What would the vapor pressure of the solution in part one be (in torr)?

Part #3:

A liquid is a mixture of benzene and toluene. The sample contains 0.225 moles of benzene and 0.775 moles of toluene. At a given temperature, pure benzene has a vapor pressure of 38.02 torr, while pure toluene has a vapor pressure of 8.73 torr. What is the total combined vapor pressure of the two liquids (in torr?)?

IDK how to do 2b, 2c, and 2d???...PLEASE HELP