Answer:
The actual Van't Hoff factor for AlCl3 is 3.20
Explanation:
Step 1: Data given
Molarity of AlCl3 = 0.050 M
osmotic pressure = 3.85 atm
Temperature = 20 °C
Step 2: Calculate the Van't Hoff factor
AlCl3(aq) → Al^3+(aq) + 3Cl^-(aq)
The theoretical value is 4 ( because 1 Al^3+ ion + 3 Cl- ions) BUT due to the interionic atractions the actual value will be less
Osmotic pressure depends on the molar concentration of the solute but not on its identity., and is calculated by:
π = i.M.R.T
⇒ with π = the osmotic pressure = 3.85 atm
⇒ with i = the van't Hoff factor
⇒ with M = the molar concentration of the solution = 0.050 M
⇒ with R = the gas constant = 0.08206 L*atm/K*mol
⇒ with T = the temperature = 20 °C = 293.15 Kelvin
i = π /(M*R*T
)
i = (3.85) / (0.050*0.08206*293.15)
i = 3.20
The actual Van't Hoff factor is 3.20
Answer:
Pb(NO3)2
Cd(NO3)2
Na2SO4
Explanation:
In the first part, addition of HCl leads to the formation of PbCl2 which is poorly soluble in water. This is the first precipitate that is filtered off.
When the pH is adjusted to 1 and H2S is bubbled in, CdS is formed. This is the second precipitate that is filtered off.
After this precipitate has been filtered off and the pH is adjusted to 8, addition of H2S and (NH4)2HPO4 does not lead to the formation of any other precipitate.
The yellow flame colour indicates the presence of Na^+ which must come from the presence of Na2SO4.
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
In this kind of exercises, you should use the "ideal gas" rules: PV = nRT
P should be in Pascal:
445mmHg = 59328Pa
1225mmHg = 163319Pa
V should be in cubic meter:
16L = 0.016 m3
R =
= constant
=
==> P1 * V1 = P2 * V2
V2 =
=
V2 = 0.00581 m3 = 5.81 L
