Question

A student dissolves 12.g of sucrose C12H22O11 in 300.mL of a solvent with a density of 1.01/gmL . The student notices that the volume of the solvent does not change when the sucrose dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to 2 significant digits. molarity = molality =

Answer 1

molarity(M) = mole of solute/volume of solvent in L

molecular weight of sucrose = 342.30 g/mol, moles of sucrose in the given question = 12/342.30 = 3.5*10-2 mol

molarity = 3.5*10-2/300*10-3 = 0.12M answer

molality(m) = mole of solute/solvent in Kg

density(d) = m/v so mass of solvent in Kg (m) = d*v = 1.01*300*1000 = 3.03*105

so molality = 3.5*10-2/3.03*105 = 1.2*10-7mol/Kg answer
will this be correct because i don't know if I'm right on molality or everything overall

Answer 2

I realize this was posted 3 months ago but...
Molarity = moles of solute/liter of solution

so:
C12H22O11 (Table Sugar) has 342.29 g/mol

12g. x (1 mole of sugar/342.29g of sugar) = 0.035 moles of sugar.
0.035 moles of sugar/ .3 L solution = 0.12 Molarity

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Molality is moles of solute/kg of solvent

So: the density of the solution is 1.01g/mL and there are 300 mL

300mL solution (1.01g/mL) = 303 g of solution
soluition-solute=solvent mass
303g solution - 12g. of solute (sugar) = 291g of solvent (.291kg of solvent)

so:
0.035 Moles of Sugar / .291 kg solvent = 0.12 Molality