A 181 mg sample of a nonelectrolyte isolated from throat lozenges was dissolved in enough water to make 10.0 mL of solution at 25°C.?
2019-02-24 10:59 pm
The osmotic pressure of the resulting solution was 4.95 atm. Calculate the molar mass of the compound.
回答 (1)
2019-02-24 11:26 pm
Osmotic pressure, Π = 4.95 atm
Molarity, M = ? mol/L
Gas constant, R = 0.08206 L atm / (mol K)
Temperature, T = (273 + 25) K = 298 K
Π = MRT
4.95 = M × 0.08206 × 298
Molarity, M = 4.95 / (0.08206 × 298) mol/L = 0.2024 mol/L
Molar mass of the protein = (181/1000 g) × (10.0/1000 L) / (0.2024 mol/L) = 8940 g/mol
(to 3 sig. fig.)
Molarity, M = ? mol/L
Gas constant, R = 0.08206 L atm / (mol K)
Temperature, T = (273 + 25) K = 298 K
Π = MRT
4.95 = M × 0.08206 × 298
Molarity, M = 4.95 / (0.08206 × 298) mol/L = 0.2024 mol/L
Molar mass of the protein = (181/1000 g) × (10.0/1000 L) / (0.2024 mol/L) = 8940 g/mol
(to 3 sig. fig.)
收錄日期: 2021-04-24 07:35:22
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20190224145928AAzwJfM