Answer:
15. 2.66 moles .
16. 2.09L.
Explanation:
Molarity of a solution is simply defined as the mole of solute per unit litre of the solvent. Mathematically, it is represented as:
Molarity = mole /Volume.
With the above formula, let us answer the questions given above
15. Data obtained from the question include the following:
Volume of solution = 1.4L
Molarity = 1.9M
Mole of solute =.?
Molarity = mole /Volume
1.9 = mole / 1.4
Cross multiply
Mole = 1.9 x 1.4
Mole = 2.66 moles
Therefore, the mole of the solute present in the solution is 2.66 moles.
16. Data obtained from the question include the following:
Mole of solute = 0.46 mole
Molarity = 0.22M
Volume of solvent (water) =.?
Molarity = mole /Volume
0.22 = 0.46/Volume
Cross multiply
0.22 x Volume = 0.46
Divide both side 0.22
Volume = 0.46/0.22
Volume = 2.09L
Therefore, 2.09L of water is required.
I pick but I'm not sure about it though 1and3
____NaNO3 + ___PbO --> ___Pb(NO3)2 + ___Na[2]O
To balace the eqaution, you need to have the same number of atoms for each element on both the reactant (left) and product (right) side.
To start off, you wanna know the number of atoms in each element on both sides, so take it apart:
[reactants] [product]
Na- 1 Na- 2
N- 1 N- 2(it's 2 because the the subscript [2] is outside of the parenthesis)
O- 4 O- 7 (same reason as above)
Pb- 1 Pb- 1
Na is not balanced out, so add a coefficient to make it the same on both sides.In this case, multiply by 2:
2NaNO3
Now Na is balanced, but the N and O are also effected by this, so they also have to be multiplied by 2 and they become:
Na- 2 Na- 2
N- 2 N- 2 (it balanced out)
O- 7 (coefficient times subscript, plus lone O) O- 7 (balanced out)
Pb was already balanced so no need to mess with it, just put a 1 where needed (it doesn't change anything).
Now to put it back together, it will look like this:
2NaNO3 + 1PbO --> 1Pb(NO3)2 + 1Na[2]O
<span>Answer is: Van't Hoff factor
(i) for this solution is 1.051 .
Change in boiling point from pure solvent to solution: ΔT
=i · Kb · b.
Kb - </span><span>molal boiling point elevation constant</span><span> is 0.512°C/m.
b - molality, moles of solute per kilogram of solvent.
b = 1.26 m.
ΔT = 101.63°C - 100</span>°C = 1.63°C.
i = 1.63°C ÷ (0.512°C/m · 1.26 m).
i = 1.051.
