Answer:
Part A → 7.82 atm
Part B → The unknown solution had the higher concentration
Part C → 0.83 mol/L
Explanation:
Part A
Osmotic pressure (π) = M . R. T . i
NaCl → Na⁺ + Cl⁻ (i =2)
0.923 g of NaCl must be dissolved in 100 mL of solution.
0.923 g / 58.45 g/m = 0.016 moles
Molarity is mol/L → 0.016 m / 0.1L = 0.16M
π = 0.16M . 0.08206 L.atm/molK . 298K . 2 ⇒ 7.82atm
Part. B
The solvent moves toward the solution of higher concentration (to dilute it) until the two solutions have the same concentration, or until gravity overtakes the osmotic pressure, Π. If the level of the unknown solution drops when it was connected to solution in part A, we can be sure that had a higher concentration.
Part. C
π = M . R . T
20.1 atm = M . 0.08206 L.atm/mol.K . 294K
20.1 atm / (0.08206 L.atm/mol.K . 294K) = 0.83 mol/L
Answer:
![r = k . [CO] .[Cl_{2}]](https://tex.z-dn.net/?f=r%20%3D%20k%20.%20%5BCO%5D%20.%5BCl_%7B2%7D%5D)
Explanation:
Let´s consider the following reaction:
CO + Cl₂ ⇒ COCl₂
The general rate law is:
![r = k . [CO]^{m}. [Cl_{2}]^{n}](https://tex.z-dn.net/?f=r%20%3D%20k%20.%20%5BCO%5D%5E%7Bm%7D.%20%5BCl_%7B2%7D%5D%5E%7Bn%7D)
where,
r is the rate of the reaction
k is the rate constant
[CO] and [Cl₂] are the molar concentrations of each reactant
m and n are the reaction orders for each reactant
Since the reaction is first order in CO, m = 1. The overall order is the sum of all the individual orders. In this case, the overall order m + n = 2. Then,
m + n = 2
n = 2 - m = 2 - 1 = 1
The reaction is first order in Cl₂.
The rate law is:
![r = k . [CO]. [Cl_{2}]](https://tex.z-dn.net/?f=r%20%3D%20k%20.%20%5BCO%5D.%20%5BCl_%7B2%7D%5D)
<span> It requires much more energy to vaporize the substance than to melt it</span>
