Answer:
36.2 K
Explanation:
Step 1: Given data
- Initial pressure of the gas (P₁): 8.6 atm
- Initial temperature of the gas (T₁): 38°C
- Final pressure of the gas (P₂): 1.0 atm (standard pressure)
- Final temperature of the gas (T₂): ?
Step 2: Convert T₁ to Kelvin
We will use the following expression.
K = °C +273.15
K = 38 °C +273.15 = 311 K
Step 3: Calculate T₂
We will use Gay Lussac's law.
P₁/T₁ = P₂/T₂
T₂ = P₂ × T₁/P₁
T₂ = 1.0 atm × 311 K/8.6 atm = 36.2 K
The answer is 6.4109 grams
Step 1
The osmotic pressure is calculated as follows:

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Step 2
<em>Information provided:</em>
The mass of solute = 13.6 g
Volume of solution = 251 mL
Absolute temperature = T = 298 K
The molar mass of solute = M = 354.5 g/mol
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Step 3
Procedure:
1 L = 1000 mL => Volume = 251 mL x (1 L/1000 mL) = 0.251 L
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C = moles of solute/volume of solution (L)
C = mass of solute/(molar mass x Volume (L))
C = 13.6 g/(354.5 g/mol x 0.251 L)
C = 0.153 mol/L
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π = C x R x T
π = 0.153 mol/L x 0.082 atm L/mol K x 298 K
π = 3.74 atm
Answer: π = 3.74 atm