If i remember correctly it should be a wedge
<span>43.83g of nacl are required to </span><span>make 500.0 ml of a 1.500 m solution.</span>
It is a mixture. it can easily be separated.
<u>Given:</u>
Mass of solvent water = 4.50 kg
Freezing point of the solution = -11 C
Freezing point depression constant = 1.86 C/m
<u>To determine:</u>
Moles of methanol to be added
<u>Explanation:</u>
The freezing point depression ΔTf is related to the molality m through the constant kf, as follows:
ΔTf = kf*m
where ΔTf = Freezing point of pure solvent (water) - Freezing pt of solution
ΔTf = 0 C - (-11.0 C) = 11.0 C
m = molality = moles of methanol/kg of water = moles of methanol/4.50 kg
11.0 = 1.86 * moles of methanol/4.50
moles of methanol = 26.613 moles
Ans: Thus around 26.6 moles of methanol should be added to 4.50 kg of water.
