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?
a) What is the molality of the resulting solution?
b) What is the mole fraction of water in this solution?
Answers
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.
= 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.
Just a slight correction to a typo above.
mols H2O = 118/18
mols H2O = 118/18
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
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
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