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)=6.548 mol
moles of c12h22O11= 3.44g/342.3(g/mol)=0.010 mol
--->6.5556 mol/(6.5556 mol + 0.010 mol= 0.998 mol
A)
M= (0.010)/(0.00344 kg + 0.118 kg)= 0.823 m c12h22O11
I think you have missed the point for A.
For B you haven't been consistent with significant figures and you have different numbers (but close) for the same operation.
mols H2O = 118/18.0 = 6.556 which I would round to 6.56
mols C12H22O11 = 3.44/342 = 0.010
total mols = 6.56+0.01 = 6.57
6.56/6.57 = 0.998
For a.
m stands for molality. M is for molarity.
m = mols solute/kg solvent
m = 0.01/0.118 = ?
To find the molality (m) of the resulting solution, you need to calculate the moles of solute (C12H22O11) and the mass of the solvent (water).
To find the moles of water (H2O), you divide the mass of water (118 g) by the molar mass of water (18 g/mol):
moles of H2O = 118 g / 18 g/mol = 6.548 mol
To find the moles of sucrose (C12H22O11), you divide the mass of sucrose (3.44 g) by the molar mass of sucrose (342.3 g/mol):
moles of C12H22O11 = 3.44 g / 342.3 g/mol = 0.010 mol
To find the mole fraction of water in the solution, you divide the moles of water by the total moles of both water and sucrose:
mole fraction of H2O = moles of H2O / (moles of H2O + moles of C12H22O11) = 6.548 mol / (6.548 mol + 0.010 mol) = 0.998 mol
Finally, to find the molality (m) of the solution, you divide the moles of C12H22O11 by the total mass of the solution (solute + solvent) in kilograms:
molality (m) = moles of C12H22O11 / (mass C12H22O11 + mass H2O) = 0.010 mol / (0.00344 kg + 0.118 kg) = 0.823 m C12H22O11