Question

1) Adrenaline is the hormone that triggers the release of extra glucose molecules in times of stress. A solution of 1.75g of adrenaline in 36g of CHCI3 depresses the solution freezing point by 1.25 °C. Calculate the molar mass (g/mol) of adrenaline. K_f CHCI3 = 4.68 °C/m. Please show work.

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2) What is the boiling point (°C) of an aqueous 0.5 M Al(NO₃)₃ (MM = 212.996 g/mol) solution whose density is 1.005 g/mL at P = 1atm? Assume theoretical i. Please show work.

Answer 1

1) Given,

Mass of adrenaline(solute) = 1.75 g

Mass of CHCl₃ (solvent)= 36 g x ( 1kg/ 1000 g) = 0.036 kg

Freezing point depression(T_f) = 1.25 ^oC

K_f for CHCl₃ = 4.68 ^oC/m

The problem is based on the concept of freezing point depression.

We know, the formula for freezing point depression,

T_f = i x K_f x m

Here, "i" is the van't Hoff factor

"m" is the molality of the solution.

Calculating the molality of the solution from the given data,

1.25 ^oC = 1 x 4.68 ^oC/m x m

molality(m) = 0.2671 m

Now, We know, the formula to calculate molality,

Molality = number of moles of solute/kg of solvent

Substituting the known values and solving for moles of solute,

0.2671 = number of moles of solute / 0.036 kg

Number of moles of adrenaline = 0.009615 mol

Now, the molar mass of adrenaline = Grams /mol

= 1.75 g / 0.009615 mol

= 182 g/mol

Thus, the molar mass of adrenaline is 182 g/mol

2) Given,

The concentration of aqueous Al(NO₃)₃ solution = 0.5 M

The density of solution = 1.005 g/mL

Assume 1 L of solution,

Thus, we have, 0.5 mol of Al(NO₃)₃ in 1 L of solution

Calculating the molality of Al(NO₃)₃ solution,

Now,

Mass of solution = 1000 mL x ( 1.005 g / 1 mL)

Mass of solution = 1005 g

Calculating the mass of Al(NO₃)₃ from the moles,

= 0.5 mol of Al(NO₃)₃ x (212.996 g / 1 mol)

= 106.5 g of Al(NO₃)₃

Now, Mass of solution = Mass of solute + Mass of solvent

1005 g = 106.5 g + Mass of solvent

Mass of solvent = 898.5 g x ( 1 kg /1000 g)

Mass of solvent (water)= 0.8985 kg

Now, the molality of solution = 0.5 mol / 0.8985 kg

Molality = 0.5565 m

Al(NO₃)₃ = Al³⁺ + 3NO₃^-

Vant Hoff factor(i) = 4

K_b for water = 0.512 ^oC/m

T_b = i x K_b x m

T_b = 4 x 0.512 ^oC/mx 0.5565 m

T_b = 1.14 ^oC

Also we know,

T_b = T_solution - T_solvent

1.14 ^oC = T_solution - 100 ^oC

T_solution = 101.14 ^oC