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Grade upto college level Physical Chemistry

8.0575 * 10-2 kg of Glauber’s salt is dissolved in water to obtain 1 dm3 of a solution of density 1077.2 kg m-3. Calculate the molarity, molality and mole fraction of Na2SO4 in the solution.

Profile image of Amit Saxena
12 Years agoGrade upto college level
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2 Answers

Profile image of Navjyot Kalra
12 Years ago
(i) Density = Mass/Volume
(ii) Molarity = Moles of solute/Volume of solution in L
(iii) Molality = Moles of solute/Mass of solvent in kg
(iv) Mole fraction of solute = Moles of solute/Total moles
The formula of Glauber’s salt is Na2SO4. 10H2O
Molecular mass of Na2SO4. 10H2O
= [2 * 23 + 32.1 + 4 * 16] + 10 (1.01 * 2 + 16) = 322.3 g mol-1
Weight of the Glauber’s salt taken = 80.575 gm
Out of 80.575 g of salt, weight of anhydrous Na2SO4
= 142.1/322.3 * 80.575 = 35.525 g
Number of moles of Na2SO4 per dm3 of the solution
= 35.525/142.1 = 0.25
Molarity of the solution = 0.25 M
Density of solution = 1077.2 kgm-3
= 1077.2 *103/106gm cm-3 = 1.0772 g cm-3
Total weight of sol = V * d = 1 dm3 * d
= 1000 cm3 * 1.0772 gcm-3 = 1077.2 g
Weight of water = 1077.2 – 35.252 = 1041.67 g
Molality of sol. 0.25/1041.67 g * 1000 g = 0.2399 = 0.24 m
Number of moles of water in the solution = 1041.67/18 = 57.87
Mole fraction of Na2SO4
= No. of moles of Na2SO4/Total number of moles = 0.25/0.25 + 57.87
= 0.0043 = 4.3 * 10-3
Profile image of ankit singh
5 Years ago

The formula for Glauber's salt is Na2​SO4​⋅10H2​O.
Given mass =8.0575×10−2kg=80.575kg
Molecular weight =322gmol−1
Molecular weight of Na2​SO4​=142gmol−1
Mass of Na2​SO4​ in 80.575 g sampe =322142​×80.575 =35.53g
Molarity (M) =14235.52​×1=0.2502M
Density=1077.2kgm−3=1077.2gL−1
Density=VolumeMass​
Mass of solution =density×volume =1077.2×1=1077.2g
Mass of water =1077.2−35.53=1041.67g
Molality(m)=14235.53​×1041.671​×100=0.2402

Moles of Na2​SO4​=14235.53​=0.2502

Moles of H2​O=181041.67​=57.87
Moles fraction (Na2​SO4​)=Total MolesMoles ofNa2​SO4​​ =0.2502+57.870.2502​ =4.304×10−3

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