Asked by Kate
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
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
Answers
Answered by
Steve
1st error: 3/300 = 0.01, not 0.10
Answered by
Kate --- Confused
how did you get that number?
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.