2. Consider 21.23 g of naphthalene (a nonvolatile, nonionizing compound) in 250.
ID: 568213 • Letter: 2
Question
2. Consider 21.23 g of naphthalene (a nonvolatile, nonionizing compound) in 250. gof liquid benzene? (boiling elevation constant Kbp 2.53'C/m for benzene and the normal boiling point of benzene is 80.1 °C). The molar mass of naphthalene is 128.1705 g/mol and the molar mass of benzene is 78.11 g/mol (a) (5pt) Determine the molality (mol/kg) of naphthalene in the above solution (b) (5pt) Determine the mole fraction of naphthalene in the above solution (c) (4pt) Determine the boiling point of the above naphthalene/benzene solution. (d) (4pt) If we dissolve 21.23 g of an unknown compound (a nonvolatile, nonionizing compound) in 250. g of liquid benzene and the vapor pressure of the unknown compound/benzene solution is lower than the vapor pressure of the above solution of 21.23 g of naphthalene in 250. g of benzene, then the molar mass of the unknown compound should be (greater than, equal to, less than) the molar mass of naphthalene. Explain your choice or show your calculation to receive partial credits. (e) (4pt) In question (d), which solution has a greater osmotic pressure: (21.23 g of the unknown compound in 250. g of liquid benzene, 21.23 g of naphthalene in 250.g of liquid benzene, or same for both solutions). Explain your choice or show your calculation to receive partial credits.Explanation / Answer
2)
a) molaity = (mass of solute/molar mass) / mass of solvent in kg
= [21.23/ 128.1705] mol/ 0.25kg
=0.6625 mol/kg
b) mole fraction = moles of naphthalene/ [moles of nalhthalene + moles of benzene]
= (21.23/128.1705) /[21.23/128.1705 + 250/78.11]
=0.04920
c) We know Elevation in boiling point , Delta Tb = kf x molaity
= 2.53C/m x 0.6625m
= 1.6761 C
Thus boiling point of solution = Normal boiling point + delta Tb
=80.1 + 1.6761C
= 81.7761 C
d) If the unknown solution has lower vapor pressure,than naphthalene solution, it means its mole fraction is greater than mole fraction of naphthalene and thus the molar mass of of is less than the molar mass of naphthalene.
we know relative lowering of vapor pressure = mole fraction of solute.
thus
P0 -P unknown = X unknown x P0
P0 -P napthalene = X naphthalene x P0
Dividing
P0 -P unknown / P0 -P napthalene = X unknown /X naphthalene
asPunknown is less than P naphthalene X unknown > X naphthalene
For dilute solutions mole fraction ratio is almost same as mole ratio
As the mass of both are same molar mass of unknown is less than that of naphthalene.
The colligative property is inversely proportional to the molar mass.
e) All colligative properties are inversely proportional to molar mass,
the osmotic pressure of naphthalene solution is greater than the osmotic pressure of unknown compound.
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