The molarity of aqueous lithium bromide, LiBr solution is 0.2 M
We'll begin by calculating the number of mole of Pb(NO₃)₂ in the solution.
- Volume = 10 mL = 10 / 1000 = 0.01 L
- Molarity of Pb(NO₃)₂ = 0.250 M
- Mole of Pb(NO₃)₂ =?
Mole = Molarity x Volume
Mole of Pb(NO₃)₂ = 0.25 × 0.01
Mole of Pb(NO₃)₂ = 0.0025 mole
Next, we shall determine the mole of LiBr required to react with 0.0025 mole of Pb(NO₃)₂
Pb(NO₃)₂ + 2LiBr —> PbBr₂ + 2LiNO₃
From the balanced equation above,
1 mole of Pb(NO₃)₂ reacted with 2 mole of LiBr.
Therefore,
0.0025 mole of Pb(NO₃)₂ will react with = 2 × 0.0025 = 0.005 mole of LiBr
Finally, we shall determine the molarity of the LiBr solution
- Mole = 0.005 mole
- Volume = 25 mL = 25 / 1000 = 0.025 L
- Molarity of LiBr =?
Molarity = mole / Volume
Molarity of LiBr = 0.005 / 0.025
Molarity of LiBr = 0.2 M
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Answer:
Option 3. The catalyst does not affect the enthalpy change (
) of a reaction.
Explanation:
As its name suggests, the enthalpy change of a reaction () is the difference between the enthalpy of the products and the reactants.
On the other hand, a catalyst speeds up a reaction because it provides an alternative reaction pathway from the reactants to the products.
In effect, a catalyst reduces the activation energy of the reaction in both directions. The reactants and products of the reaction won't change. As a result, the difference in their enthalpies won't change, either. That's the same as saying that the enthalpy change of the reaction would stay the same.
Refer to an energy profile diagram. Enthalpy change of the reaction measures the difference between the two horizontal sections. Indeed, the catalyst lowered the height of the peak. However, that did not change the height of each horizontal section or the difference between them. Hence, the enthalpy change of the reaction stayed the same.
Answer:
The van't hoff factor of 0.500m K₂SO₄ will be highest.
Explanation:
Van't Hoff factor was introduced for better understanding of colligative property of a solution.
By definition it is the ratio of actual number of particles or ions or associated molecules formed when a solute is dissolved to the number of particles expected from the mass dissolved.
a) For NaCl the van't Hoff factor is 2
b) For K₂SO₄ the van't Hoff factor is 3 [it will dissociate to give three ions one sulfate ion and two potassium ions]
Out of 0.500m and 0.050m K₂SO₄, the van't hoff factor of 0.500m K₂SO₄ will be more.
c) The van't Hoff factor for glucose is one as it is a non electrolyte and will not dissociate.
