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
Learn more about molarity: brainly.com/question/10103895
Answer:- 
molecules.
Solution:- The grams of tetrabromomethane are given and it asks to calculate the number of molecules.
It is a two step unit conversion problem. In the first step, grams are converted to moles on dividing the grams by molar mass.
In second step, the moles are converted to molecules on multiplying by Avogadro number.
Molar mass of 
= 12+4(79.9) = 331.6 g per mol
let's make the set up using dimensional analysis:
= molecules
So, there will be molecules in 250 grams of .
1 mol = 6.023x10^23 number of molecules (Avogadro's number)
1 : 6.023x10^23
X : 4.91x10^22
(6.023x10^23)X = 4.91x10^22
X = 4.91x10^22/6.023x10^23
X = 0.082 Moles
Answer:
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