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:
The correct answer is - 5 carbon compounds due to low to high intermolecular forces between their molecules.
Explanation:
Bottle C has gas in it and we know that alkane has carbon and hydrogen only which means they have a single sigma bond between them and very low intermolecular forces in between molecules and are present mostly at gaseous state. Thus, bottle C has alkane.
Alcohols have -OH group that can form rarely two pi bonds which means they have intermediate intermolecular force whereas acids have -cooH group with a high molecular force so bottle B with liquid is alcohol and A has acid.
4. NaOH
____________________________________
