The correct answer is slow
Answer:
We can't see the options so we don't know what we can put
Explanation:
Answer:
Mole Fraction (H₂O) = 0.6303
Mole Fraction (C₂H₅OH) = 0.3697
Explanation:
(Step 1)
Calculate the mole value of each substance using their molar masses.
Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol
Molar Mass (H₂O): 18.014 g/mol
200.0 g H₂O 1 mole
--------------------- x ------------------ = 11.10 moles H₂O
18.014 g
Molar Mass (C₂H₅OH): 2(12.011 g/mol) + 6(1.008 g/mol) + 15.998 g/mol
Molar Mass (C₂H₅OH): 46.068 g/mol
300.0 g C₂H₅OH 1 mole
---------------------------- x -------------------- = 6.512 moles C₂H₅OH
46.068 g
(Step 2)
Using the mole fraction ratio, calculate the mole fraction of each substance.
moles solute
Mole Fraction = ------------------------------------------------
moles solute + moles solvent
11.10 moles H₂O
Mole Fraction = -------------------------------------------------------------
11.10 moles H₂O + 6.512 moles C₂H₅OH
Mole Fraction (H₂O) = 0.6303
6.512 moles C₂H₅OH
Mole Fraction = -------------------------------------------------------------
11.10 moles H₂O + 6.512 moles C₂H₅OH
Mole Fraction (C₂H₅OH) = 0.3697
Refer to the diagram shown below.
The piston supports the same load W at both temperatures.
The ideal gas law is

where
p = pressure
V = volume
n = moles
T = temperature
R = gas constant
State 1:
T₁ = 20 C = 20+273 = 293 K
d₁ = 25 cm piston diameter
State 2:
T₂ = 150 C = 423 K
d₂ = piston diameter
Because V, n, and R remain the same between the two temperatures, therefore

If the supported load is W kg, then

Similarly,


Because p₁/p₂ = T₁/T₂, therefore

The minimum piston diameter at 150 C is 20.8 cm.
Answer: 20.8 cm diameter