Answer:
0.425 moles of P₂O₅.
Explanation:
As per given balanced equation, four moles of phosphorus reacts with five moles of oxygen to give two moles of P₂O₅.
As given that the initial moles of phosphorus taken = 0.97 moles
moles of phosphorus left after reaction = 0.12 moles
moles of phosphorus reacted = 0.97-0.12 = 0.85 moles
When four moles of P reacts they give two moles of P₂O₅.
when one mole of P will react to give =
0.85 moles of P will react to give = P₂O₅
Answer: True
Explanation:
Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.
is positive when randomness increases and
is negative when randomness decreases.
When the solid changes to liquid to gas ,the randomness of particles increases and hence is positive and thus the entropy increases.
Answer : The correct option is, (A)
Explanation :
Van't Hoff factor : It is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property.
We can determine the van't Hoff factor by the association and dissociation of the compound.
(A)
It is an electrolyte that dissociates to give aluminum ion and chloride ion.
The dissociation of will be,
So, Van't Hoff factor = Number of solute particles = = 1 + 3 = 4
(B)
It is an electrolyte that dissociates to give potassium ion and iodide ion.
The dissociation of will be,
So, Van't Hoff factor = Number of solute particles = = 1 + 1 = 2
(C)
It is an electrolyte that dissociates to give calcium ion and chloride ion.
The dissociation of will be,
So, Van't Hoff factor = Number of solute particles = = 1 + 2 = 3
(D)
It is an electrolyte that dissociates to give magnesium ion and sulfate ion.
The dissociation of will be,
So, Van't Hoff factor = Number of solute particles = = 1 + 1 = 2
(E) Non-electrolyte
The dissociation non-electrolyte is not possible. So, the Van't Hoff factor will always be, 1.
Hence, the highest van't Hoff factor of solute is,