18 g glucose (C6H12O6) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is:

The molar masses of glucose and water are 180 g/mol and 18 g/mol respectively.
Number of moles of glucose = 18/180 = 0.1
Number of moles of water = 178.2/18 = 9.9
Mole fraction of glucose = 0.1/(0.1 + 9.9) = 0.01
Relative lowering in vapor pressure is equal to the mole fraction of glucose.
760 - P/760 = 0.01
760 - P = 7.6
P = 760 - 7.6 = 752.4 torr.
Hence, the correct option is 752.4