1) What is the solubility of neon (in units of grams per liter) in water at 25 °C, when the Ne gas over the solution has a partial pressure of 0.500 atm? kH for Ne at 25 °C is 4.51×10^-4 mol/L·atm.
Answer: ? g/L
2) What is the solubility of butane (in units of grams per liter) in water at 25 °C, when the C4H10 gas over the solution has a partial pressure of 457 mm Hg? kH for C4H10 at 25 °C is 1.13×10^-3 mol/L·atm.
Answer: ? g/L
Gas Solubility Using Henry's Law?
2018-06-13 3:16 pm
回答 (1)
2018-06-13 4:02 pm
1)
Henry's law constant, kH = 4.51 × 10⁻⁴ mol/(L atm)
Partial pressure, P = 0.500 atm
Molar mass of Ne = 20.2 g/mol
Henry's law: C = kH × P
Solubility, C
= kH × P
= [4.51 × 10⁻⁴ mol/(L atm)] × (0.500 atm) × (20.2 g/mol)
= 4.56 × 10⁻³ g/L (to 3 sig. fig.)
====
2)
Henry's law constant, kH = 1.13 × 10⁻³ mol/(L atm)
Partial pressure, P = 457 mmHg = 457/760 atm
Molar mass of C₄H₁₀ = (12.0×4 + 1.0×10) g/mol = 58.0 g/mol
Henry's law: C = kH × P
Solubility, C
= kH × P
= [1.13 × 10⁻³ mol/(L atm)] × (457/760 atm) × (58.0 g/mol)
= 3.94 × 10⁻² g/L (to 3 sig. fig.)
Henry's law constant, kH = 4.51 × 10⁻⁴ mol/(L atm)
Partial pressure, P = 0.500 atm
Molar mass of Ne = 20.2 g/mol
Henry's law: C = kH × P
Solubility, C
= kH × P
= [4.51 × 10⁻⁴ mol/(L atm)] × (0.500 atm) × (20.2 g/mol)
= 4.56 × 10⁻³ g/L (to 3 sig. fig.)
====
2)
Henry's law constant, kH = 1.13 × 10⁻³ mol/(L atm)
Partial pressure, P = 457 mmHg = 457/760 atm
Molar mass of C₄H₁₀ = (12.0×4 + 1.0×10) g/mol = 58.0 g/mol
Henry's law: C = kH × P
Solubility, C
= kH × P
= [1.13 × 10⁻³ mol/(L atm)] × (457/760 atm) × (58.0 g/mol)
= 3.94 × 10⁻² g/L (to 3 sig. fig.)
收錄日期: 2021-05-01 22:24:49
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