A 5.2 g sample of polystyrene was dissolved in enough toluene to form 100 mL of solution. The osmotic pressure of this solution was found to be 0.120 atm at 25 °C. What is the molar mass of the polystyrene sample?
Please show work, I tried solving it using π=MRT but I keep getting the wrong answer.
Chemistry Help?
2016-06-24 1:35 am
回答 (2)
2016-06-24 1:49 am
Π = 0.120 atm
V = 100 mL = 0.1 L
R = 0.0821 atm L (mol K)
T = (273 + 25) K = 298 K
Π = MRT
Π = (n/V)RT
n = ΠV/(RT)
No. of moles of polystyrene, n = 0.120 × 0.1 / (0.0821 × 298) mol
Molar mass of polystyrene = (5.2 g) / [0.120 × 0.1 / (0.0821 × 298) mol] = 10600 g/mol
V = 100 mL = 0.1 L
R = 0.0821 atm L (mol K)
T = (273 + 25) K = 298 K
Π = MRT
Π = (n/V)RT
n = ΠV/(RT)
No. of moles of polystyrene, n = 0.120 × 0.1 / (0.0821 × 298) mol
Molar mass of polystyrene = (5.2 g) / [0.120 × 0.1 / (0.0821 × 298) mol] = 10600 g/mol
2016-06-24 1:52 am
π=MRT
0.120 = M x 0.082057 x 298 K
M = 0.004907
0.004907 = (5.2 g / MW) / 0.100 L
0.0004907 = 5.2 g / MW
MW = 5.2 g / 0.0004907 = 10596
ANSWER: 10600 g/mol (11000; 2 significant figures)
0.120 = M x 0.082057 x 298 K
M = 0.004907
0.004907 = (5.2 g / MW) / 0.100 L
0.0004907 = 5.2 g / MW
MW = 5.2 g / 0.0004907 = 10596
ANSWER: 10600 g/mol (11000; 2 significant figures)
收錄日期: 2021-04-18 15:12:37
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