Question

A solution is prepared by dissolving 3.44 g of sucrose, C12H22O11, in 118 g of water.

a) What is the molality of the resulting solution?

b) What is the mole fraction of water in this solution?

Answer 1

molality=molessugar/kgsolvent
= 3.44/(molmassSucrose*.118)

Mole fraction moles solute/totalmoles
where moles solute=3.44/molmassSucrose
moles water=18/118

and total moles=moles solute+moles water.

Answer 2

Just a slight correction to a typo above.
mols H2O = 118/18

Answer 3

B)
moles of h2o=118g/18(g/mol)=6.5556 mol
moles of c12h22O11= 3.44g/342.3(g/mol)=0.10049 mol
6.5556 mol/(6.5556 mol + 0.10049 mol= 0.984902 mol

A)
M= (0.10049)/(0.00344 kg + 0.118 kg)= 0.82749 m c12h22O11