QUESTION
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The density of 1.25 M Na2SO4
solution at 25°C is 1.25 gram / ml. Calculate -
¡] - mole fraction of Na2SO4
ii] - mass % of Na2SO4
¡¡¡] - the molality of the solution with respect to Na+ion
Note - Don't give wrong answers please I am Class 12 th board student. __/\__
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HARSH PRATAP SINGH
Answers
Answer:
mole fraction = 0.02
% by mass = 14.2 %
molality = 1.17
Explanation:
Given: Density of Na2SO4 = 1.25 g/ml
Molarity of solution = 1.25 M (1.25 moles in 1L)
Moles of Na2SO4 = n = 1.25
Moles of water =
Solution:
density = mass / Volume
Weight of solution = Volume of solution X Density
= 1000ml X 1.25
= 1250 g
Moles = weight of solute / molar mass
Weight of solute = moles X molar mass
= 1.25 X 142
= 177.5
Weight of solution = wgt of solute + wgt of solvent
Weight of solvent = wgt of solution—wgt of solute
= 1250 — 177.5
= 1072.5 g
1. Mole fraction X(Solute) = n/(n+N)
moles of solvent = weight / Molar mass
= 1072.5/18
= 59.58
Mole fraction of solute X = 1.25/(1.25+59.58)
= 1.25/60.83
= 0.02
2. Mass % = (mass of solute/mass of solution)X100
= (177.5/1250)X100
= 14.2 %
3. Molality=(moles ofsolute/mass of solvent)X1000
= (1.25/1072.5)X1000
= 1.17 mì
Answer:
mole fraction = 0.02
% by mass = 14.2 %
molality = 1.17
Given: Density of Na2SO4 = 1.25 g/ml
Molarity of solution = 1.25 M (1.25 moles in 1L)
Moles of Na2SO4 = n = 1.25
Moles of water =
Solution:
density = mass / Volume
Weight of solution = Volume of solution X Density
= 1000ml X 1.25
= 1250 g
Moles = weight of solute / molar mass
Weight of solute = moles X molar mass
= 1.25 X 142
= 177.5
Weight of solution = wgt of solute + wgt of solvent
Weight of solvent = wgt of solution—wgt of solute
= 1250 — 177.5
= 1072.5 g
1. Mole fraction X(Solute) = n/(n+N)
moles of solvent = weight / Molar mass
= 1072.5/18
= 59.58
Mole fraction of solute X = 1.25/(1.25+59.58)
= 1.25/60.83
= 0.02 ans.
2. Mass % = (mass of solute/mass of solution)X100
= (177.5/1250)X100
= 14.2 % ans.
3. Molality=(moles ofsolute/mass of solvent)X1000
= (1.25/1072.5)X1000
= 1.17 m ans.
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