6.11% w/v of Cu2+ implies that 6.11 g of Cu2+ is present in 100 ml of the solution
therefore, 250 ml of the solution would have: 250 ml * 6.11 g/100 ml = 15.275 g
# moles of Cu2+ = 15.275 g/63.546 g mole-1 = 0.2404 moles
1 mole of CuCl2 contain 1 mole of Cu2+ ion
Hence, 0.2404 moles of Cu2+ would correspond to 0.2404 moles of CuCl2
Molar mass of CuCl2 = 134.452 g/mole
The mass of CuCl2 required = 0.2404 moles * 134.452 g/mole = 32.32 grams
Answer:
The boiling point elevation is 3.53 °C
Explanation:
∆Tb = Kb × m
∆Tb is the boiling point elevation of the solution
Kb is the molal boiling point elevation constant of CCl4 = 5.03 °C/m
m is the molality of the solution is given by moles of solute (C9H8O) divided by mass of solvent (CCl4) in kilogram
Moles of solute = mass/MW =
mass = 92.7 mg = 92.7/1000 = 0.0927 g
MW = 132 g/mol
Moles of solute = 0.0927/132 = 7.02×10^-4 mol
Mass of solvent = 1 g = 1/1000 = 0.001 kg
m = 7.02×10^-4 mol ÷ 0.001 kg = 0.702 mol/kg
∆Tb = 5.03 × 0.702 = 3.53 °C (to 2 decimal places)
Answer:
A)Trial 1 because the average rate of reaction is lower.
Explanation:
I accidentally gave myself low rating my bad
