What is the normal boiling point in ∘ C ^\circ {\rm C} of a solution prepared by dissolving 1.50 g {\rm g} of aspirin (acetylsalicylic acid, C 9 H 8 O 4 {\rm C}_{9}{\rm H}_{8}{\rm O}_{4}) in 75.00 g {\rm g} of chloroform (CHCl 3 ) ({\rm CHCl}_{3})? The normal boiling point of chloroform is 61.7 ∘ C ^\circ {\rm C}, and K b {\rm K}_{{\rm b}} for chloroform is 3.63 ( ∘ C⋅kg)/mol \rm (^\circ C \cdot kg)/mol.

If I can sort through this maze of hieroglyphics

mols aspirin = grams/molar mass
Substitute and solve for mols.

m = mol/kg solvent
Substitute and solve for m

delta T = Kf*m
Substitute and solve for delta T.

Then add delta T to normal boiling point to find the new boiling point.

To find the boiling point of the solution, we need to consider the change in boiling point that occurs due to the presence of a non-volatile solute (aspirin) in the solvent (chloroform). This change can be determined using the formula:

ΔTb = Kbm

where ΔTb is the change in boiling point, Kb is the molal boiling point elevation constant for the solvent, and m is the molality of the solute.

First, we need to calculate the molality (m) of the aspirin in the chloroform solution. Molality is defined as the number of moles of solute per kilogram of solvent.

Molar mass of aspirin (C9H8O4):
C = 12.01 g/mol
H = 1.01 g/mol
O = 16.00 g/mol

Molar mass of C9H8O4 = 12.01*9 + 1.01*8 + 16.00*4 = 180.16 g/mol

Number of moles of aspirin (n):
n = mass/molar mass = 1.50 g / 180.16 g/mol

The mass of chloroform (solvent) is given as 75.00 g.

Molality (m) = n / mass of solvent in kg = n / (mass of solvent in g / 1000)

mass of solvent in kg = 75.00 g / 1000 = 0.075 kg

m = (1.50 g / 180.16 g/mol) / 0.075 kg

Now, we can calculate the change in boiling point (ΔTb) using the formula mentioned earlier:

ΔTb = Kb * m

ΔTb = (3.63 (°C⋅kg)/mol) * [(1.50 g / 180.16 g/mol) / 0.075 kg]

Finally, we can find the boiling point of the solution:

Boiling point of solution = normal boiling point of solvent + ΔTb

Boiling point of solution = 61.7 °C + ΔTb

Therefore, the boiling point of the solution prepared by dissolving 1.50 g of aspirin in 75.00 g of chloroform is approximately equal to 61.7 °C + ΔTb.

To find the normal boiling point of the solution, we can use the equation:

ΔT = K_b * m

where ΔT is the boiling point elevation, K_b is the molal boiling point constant, and m is the molality of the solute.

First, we need to calculate the molality (m) of the solute. We can use the equation:

m = (moles of solute) / (mass of solvent in kg)

To calculate the moles of solute, we need to know the molar mass of acetylsalicylic acid (C9H8O4).

The molar mass of each element is as follows:
C (carbon): 12.01 g/mol
H (hydrogen): 1.01 g/mol
O (oxygen): 16.00 g/mol

Molar mass of acetylsalicylic acid = (9 * 12.01 g/mol) + (8 * 1.01 g/mol) + (4 * 16.00 g/mol)

Next, we need to calculate the mass of solvent (chloroform) in kg. We can convert the given mass of chloroform (75.00 g) to kg by dividing by 1000.

Now we have all the necessary values to find the molality (m) of the solute and can substitute it back into the first equation to find ΔT:

ΔT = K_b * m

Finally, we can add the boiling point elevation (ΔT) to the normal boiling point of the solvent (chloroform) to get the normal boiling point of the solution.