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.
D= M/V so, D=300/40 wich is 7.5 so D= 7.5g/m3
Answer is: concentration of products increases (ammonia nad water).
Chemical reaction: heat + NH₄⁺ + OH⁻ ⇄ NH₃ + H₂<span>O.
</span>According to Le
Chatelier's Principle, the position of equilibrium moves to counteract the
change, because heat is increased, system consume that heat, so equilibrium is shifted to right, by decreasing concentration of reaactants and increasing concentration of product.
We can rearrange the ideal gas equation:
PV = nRT, where n is the number of moles equivalent to:
n = mass / Mr
PV = mRT/Mr
m/V = PMr/RT
density = PMr / RT; where Mr and R are constant.
