Answer:
Option a.
0.01 mol of CaCl₂ will have the greatest effect on the colligative properties, because it has the biggest i
Explanation:
To determine which of the solute is going to have a greatest effect on colligative properties we have to consider the Van't Hoff factor (i)
These are the colligative properties:
ΔP = P° . Xm . i → Lowering vapor pressure
ΔT = Kb . m . i → Boiling point elevation
ΔT = Kf . m . i → Freezing point depression
π = M . R . T → Osmotic pressure
Van't Hoff factor are the numbers of ions dissolved in the solution. For nonelectrolytes, the i values 1.
CaCl₂ and KNO₃ are two ionic solutes. They dissociate as this:
CaCl₂ → Ca²⁺ + 2Cl⁻
We have 1 mol of Ca²⁺ and 2 chlorides, so 3 moles of ions → i = 3
KNO₃ → K⁺ + NO₃⁻
We have 1 mol of K⁺ and 1 mol of nitrate, so 2 moles of ions → i = 2
Option a, is the best.
Answer:
A. 15859.2 L or 15900 L
B. 0.629 mol
Explanation:
At STP, one mole is equal to approximately 22.4 L
L or mL is volume, so you are attempting to solve for L or mL.
A.
708 mol x (22.4 L/1 mol) = 15859.2 L (w/ significant figures included - 15900 L)
B.
(14.1 L) x (1 mole/ 22.4 L) = 0.629 mol.
Answer: d) 16.34 grams
Explanation:
To calculate the moles :
According to stoichiometry :
As 2 moles of 
are produced by= 4 moles of 
Thus 0.292 moles of are produced by= =
of
Mass of 
Thus 16.34 g of will be needed
