Calculate the molality of a sucrose solution (C12H22O11) that was prepared by mixing 1.450 grams of the sugar with 30.00 mL of water. C12H22O11 = 342.34 g / mol dwater = 1.00 g / mL

Answer 1

Molality= moles sucrose / kg of solvent (water)

moles sucrose = 1.450 g x 1 mol / 342.34 g = 0.0042356 moles sucrose

Now, we must find the kg of water used.

30.00 ml x 1 g / ml x 1 kg/1000 g = 0.03000 kg

Molality = 0.0042356 moles / 0.03000 kg = 0.1412 molal (4 sig. figs.)