a)
number of moles of Sucrose = mass/molar mass
                                                         Â
= 50.4 g / 342.296 g
                                                         Â
= 0.147 mol
Volume = 0.355 L
Molarity = number of moles / Volume
                 Â
= 0.147 mol / 0.355 L
                 Â
= 0.415 M
b)
Molality = number of moles / mass of solvent in Kg
                 Â
= 0.147 mol / 0.332 Kg
                 Â
= 0.443 molal
c)
percent by mass = mass of solute *100 / total mass
                                 Â
= 50.4 *100 / (50.4 + 332)
                                 Â
= 13.2 %
d)
number of moles of water = mass/ molar mass = 332/18 =18.44
moles
Mole fraction = number of moles of solute / total
moles
                            Â
= 0.147 / (0.147 + 18.44)
                            Â
= 7.91*10^-3