If the molar concentration of sodium sulfate (Na2SO4) is 0.10 M, what is the concentration of sodium ion
2 answers:
Answer : The concentration of sodium ions in a solution is 0.20 M
Explanation :
The dissociation reaction of in solution will be:
From the reaction we conclude that, 1 mole of dissociate in solution to give 2 moles of
ion (sodium ion) and 1 mole of
ion (sulfate ion).
As per question,
0.10 M of dissociate in solution to give
of
ion (sodium ion) and 0.10 M of
ion (sulfate ion).
Hence, the concentration of sodium ions in a solution is 0.20 M
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The volume of N₂ at STP=56 L
<h3>Further explanation</h3>Given
2.5 moles of N₂
Required
The volume of the gas
Solution
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, the volume per mole of gas or the molar volume-Vm is 22.4 liters/mol.
So for 2.5 moles gas :
The question is incomplete, the table of the question is given below
Answer:
I) xA= 0.34, yA= 0.55
ii) 76.2 mole % vapor
iii) Percentage of vapor volume = 98%
Explanation:
i) xA= 0.34, yA= 0.55
xA= 0.34, yA= 0.55
ii) 0.50 = 0.55 nv + 0.34 nL
Therefore, nV = 0.762 mol vapor and nL = 0.238 mol liquid
This shows 76.2 mole % vapor
iii) ρA= 0.791 g/cm3 and, ρE = 0.789 g/cm3
Therefore, ρ = 0.790 g/cm3
Now, we have:
MA = 58.08 g/mol and ME= 46.07 g/mol
So Ml = (0.34 x 58.08)+[(1 -0.34) x 46.07] = 50.15 g/mol
1 mol liquid = (0.762 mol vapor/0.238 mol liquid) = 3.2 mol vapor
Liquid volume = Vl= [1 mol x (50.15 g/mol)] / (0.790 g/cm3) = 63.48 cm3
Vapour volume = Vv = 3.2 mol x(22400 cm3/mol) x [(65+273)/273] = 88747 cm3
Therefore, percentage of vapour volume = 88747 / (88747+63.48) = 99.9 %
