need help with chem question please

Question

"Determine the vapor pressure and the vapor pressure lowering of a solution made from 460.0mL of water and 218.2g of the solute, glucose (molar mass = 180.157g•mol^-1), at a temperature of 30.0°C.  The vapor pressure of pure water at 30°C is given in the table of water properties in Appendix E of the text (p. 1202). Assume the density of the solution is 0.9956g•mL^-1."

Appendix E:

Water Vapor Pressure at Different Temperatures (°C)

Temp Vapor Pressure (torr) Vapor Pressure (Pa)

0 4.6 613.2812

4 6.1 813.2642

10 9.2 1226.562

15 12.8 1706.522

20 17.5 2333.135

22 19.8 2639.776

25 23.8 3173.064

30 31.8 4239.64

35 42.2 5626.188

40 55.3 7372.707

45 71.9 9585.852

50 92.5 12332.29

55 118.0 15732

Answer 1

Raoults law and Daltons law of partial pressure

moles H2O = 460.0 ml x 1 g / ml = 460 g x 1 mol / 18 g = 25.56 mols

moles glucose = 218.2 g x 1 mol / 180.16 g = 1.211 mols

Total moles = 26.77 moles

Mole fraction water (X_H2O) = 25.56 / 26.77 = 0.9548

Vapor pressure of solution = 0.9548 x 31.8 torr = 30.4 torr

Vapor pressure lowering = 1.4 torr