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
1st error: 3/300 = 0.01, not 0.10
how did you get that number?
To find the molality of the resulting solution (part a), you need to calculate the number of moles of sucrose (C12H22O11) and water (H2O) in the solution.
1. Calculate the number of moles of water (H2O):
Mass of water = 118 g
Molar mass of water (H2O) = 18 g/mol
Number of moles of water (H2O) = Mass of water / Molar mass of water = 118 g / 18 g/mol = 6.5556 mol
2. Calculate the number of moles of sucrose (C12H22O11):
Mass of sucrose (C12H22O11) = 3.44 g
Molar mass of sucrose (C12H22O11) = 342.3 g/mol
Number of moles of sucrose (C12H22O11) = Mass of sucrose / Molar mass of sucrose = 3.44 g / 342.3 g/mol = 0.10049 mol
3. Calculate the molality (m):
Molality (m) = Moles of solute / Mass of solvent in kg
Molality (m) of sucrose (C12H22O11) = Number of moles of sucrose / (Mass of sucrose + Mass of water)
Molality (m) = 0.10049 mol / (0.00344 kg + 0.118 kg) = 0.82749 m
So, the molality of the resulting solution is 0.82749 m.
To find the mole fraction of water in the solution (part b), you need to calculate the mole fraction of water.
1. Calculate the total number of moles in the solution:
Total moles in the solution = Moles of water + Moles of sucrose = 6.5556 mol + 0.10049 mol = 6.65609 mol
2. Calculate the mole fraction of water:
Mole fraction of water = Moles of water / Total moles in the solution
Mole fraction of water = 6.5556 mol / 6.65609 mol = 0.984902
So, the mole fraction of water in the solution is 0.984902.