Adrenaline is the hormone that triggers the release of extra glucose molecules in times of stress or emergency. A solution of 0.64 g of adrenaline in 36.0 g of \(\mathrm{CCl}_{4}\) elevates the boiling point by \(0.49^{\circ} \mathrm{C}\) Calculate the approximate molar mass of adrenaline from this data.

Since we know the mass of adrenaline, solvent, and the change in boiling point, we can find the molality of the solution using the boiling point elevation equation: ΔTb = Kb × molality × i We are given that ΔTb = 0.49ºC and i = 1. Therefore: molality = ΔTb / Kb Before moving on to Step 2, let's convert the mass of adrenaline and solvent into kg: 0.64 g adrenaline = 0.00064 kg adrenaline 36.0 g CCl4 = 0.036 kg CCl4

To find Kb, we need to rearrange the boiling point elevation equation: Kb = ΔTb / (molality × i) In this problem, we are not given Kb. However, since ΔTb and i are known, we can express Kb in terms of molality: Kb = 0.49 / molality

Using the molality (moles solute/kg solvent), we can find the moles of adrenaline: moles adrenaline = molality × 0.036 kg CCl4

Finally, we can calculate the molar mass of adrenaline using the formula: molar mass = mass of adrenaline / moles of adrenaline 0.00064 kg adrenaline / (molality × 0.036 kg CCl4) = molar mass of adrenaline Now we can substitute the expression for Kb from Step 2: (0.00064 kg adrenaline / (0.49 / Kb × 0.036 kg CCl4)) = molar mass of adrenaline And solve for Kb: Kb = 0.49 × (0.49 / 0.00064) × (0.036/0.49) = 2.62 K kg/mol Now we can substitute Kb back into the equation for molality: molality = ΔTb / Kb = 0.49 / 2.62 = 0.187 mol/kg Plugging this value into the equation for moles adrenaline: moles adrenaline = 0.187 mol/kg × 0.036 kg CCl4 = 0.00672 mol Finally, we can calculate the molar mass of adrenaline: molar mass = 0.00064 kg adrenaline / 0.00672 mol = 0.095 kg/mol The approximate molar mass of adrenaline is 95 g/mol.