Explanation:
Formula to calculate osmotic pressure is as follows.
Osmotic pressure = concentration × gas constant × temperature( in K)
Temperature =
= (25 + 273) K
= 298.15 K
Osmotic pressure = 531 mm Hg or 0.698 atm (as 1 mm Hg = 0.00131)
Putting the given values into the above formula as follows.
0.698 = 
C = 0.0285
This also means that,
= 0.0285
So, moles = 0.0285 × volume (in L)
= 0.0285 × 0.100
= 
Now, let us assume that mass of
= x grams
And, mass of
= (1.00 - x)
So, moles of
=
Now, moles of
=
=
= x = 0.346
Therefore, we can conclude that amount of
present is 0.346 g and amount of
present is (1 - 0.346) g = 0.654 g.
Answer:
Δ S = 93.8 J/mol-K
Explanation:
Given,
Boiling point of chloroform = 61.7 °C
= 273 + 61.7 = 334.7 K.
Enthalapy of vapourization = 31.4 kJ/mol.
Using Gibbs free energy equation
Δ G = Δ H - T (ΔS)
at equilibrium (when the liquid is boiling), Δ G = 0
so, 0 = ΔH - T (Δ S)
T (Δ S) = Δ H
and ΔS = ΔH / T
Δ S = (31400 J/mol.) / 334.7 K
Δ S = 93.8 J/mol-K
Answer:
p = 260 kilogram/cubic meter
Explanation:
ρ = 
= 
= 0.26 gram/milliliter
= 260 kilogram/cubic meter