Answer:
232.5 g C2H6O2
Explanation:
The equation you need to use here is ΔTf = i Kf m
Since pure water freezes at 0 C, your ΔTf is just 4.46 C
i = 1 (ethylene glycol is a weak electrolyte)
Kf = molal freezing constant, which for water is 1.86 C/m
m = molality = x mols C2H6O2 / 1.15 kg H2O (don't know the moles of ethylene glycol we're dissolving yet)
Than,
4.46 C = 1.86 C/m (x mol C2H6O2 / 1.15 kg H2O)
Solve for x, you should get x = 2.75 mol C2H6O2
3.75 mol C2H6O2 (62 g C2H6O2 / 1 mol C2H6O2) = 232.5 g C2H6O2
Answer:
(A) 4.616 * 10⁻⁶ M
(B) 0.576 mg CuSO₄·5H₂O
Explanation:
- The molar weight of CuSO₄·5H₂O is:
63.55 + 32 + 16*4 + 5*(2+16) = 249.55 g/mol
- The molarity of the first solution is:
(0.096 gCuSO₄·5H₂O ÷ 249.55 g/mol) / (0.5 L) = 3.847 * 10⁻⁴ M
The molarity of CuSO₄·5H₂O is the same as the molarity of just CuSO₄.
- Now we use the dilution factor in order to calculate the molarity in the second solution:
(A) 3.847 * 10⁻⁴ M * 6mL/500mL = 4.616 * 10⁻⁶ M
To answer (B), we can calculate the moles of CuSO₄·5H₂O contained in 500 mL of a solution with a concentration of 4.616 * 10⁻⁶ M:
- 4.616 * 10⁻⁶ M * 500 mL = 2.308 * 10⁻³ mmol CuSO₄·5H₂O
- 2.308 * 10⁻³ mmol CuSO₄·5H₂O * 249.55 mg/mmol = 0.576 mg CuSO₄·5H₂O
No it means that the liquid is way different then water<span />
