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?
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
2 answers
1st error: 3/300 = 0.01, not 0.10
how did you get that number?
1 answer