Answer:
2.7 °C.kg/mol
Explanation:
Step 1: Calculate the freezing point depression (ΔT)
The normal freezing point of a certain liquid X is-7.30°C and the solution freezes at -9.9°C instead. The freezing point depression is:
ΔT = -7.30 °C - (-9.9 °C) = 2.6 °C
Step 2: Calculate the molality of the solution (b)
We will use the following expression.
b = mass of solute / molar mass of solute × kilograms of solvent
b = 102. g / (162.2 g/mol) × 0.650 kg = 0.967 mol/kg
Step 3: Calculate the molal freezing point depression constant Kf of X
Freezing point depression is a colligative property. It can be calculated using the following expression.
ΔT = Kf × b
Kf = ΔT / b
Kf = 2.6 °C / (0.967 mol/kg) = 2.7 °C.kg/mol
Answer:
1.17 mol
Explanation:
Step 1: Write the balanced equation
2 Al + 6 HCl → 2 AlCl₃ + 3 H₂
Step 2: Calculate the moles corresponding to 85.0 g of HCl
The molar mass of HCl is 36.46 g/mol.
85.0 g × 1 mol/36.46 g = 2.33 mol
Step 3: Calculate the number of moles of H₂ produced from 2.33 moles of HCl
The molar ratio of HCl to H₂ is 6:3.
2.33 mol HCl × 3 mol H₂/6 mol H₂ = 1.17 mol H₂
