Answer:
- 0.99 °C ≅ - 1.0 °C.
Explanation:
- We can solve this problem using the relation:
<em>ΔTf = (Kf)(m),</em>
where, ΔTf is the depression in the freezing point.
Kf is the molal freezing point depression constant of water = -1.86 °C/m,
m is the molality of the solution (m = moles of solute / kg of solvent = (23.5 g / 180.156 g/mol)/(0.245 kg) = 0.53 m.
<em>∴ ΔTf = (Kf)(m)</em> = (-1.86 °C/m)(0.53 m) =<em> - 0.99 °C ≅ - 1.0 °C.</em>
Answer:
4,997.9
Explanation:
just convert fluid ounces to millilters.
As,
Kw = [H+] [OH-]
For water, [H+] = [OH-]
Therefore we can write
Kw = [H+]²
9.311 × 10-14 = [H+]²
[H+] =
3.04 × 10-7 = [OH-]
Ph = - log [H+]
= - log (
3.04 × 10-7)
= 6.52
Thus, Ph = PoH = 6.52
