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Question

161×102 kg of Glauber's salt is dissolved in water to obtain 1 dm3 of a solution of density 1520 kg m3. Which of the following is true?

A
Molarity of the solution =5 M
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B
Molality of the solution =6.17 m
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C
Mole fraction of the solute =0.1
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D
Mole fraction of the solute =0.0043
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Solution

The correct options are
A Molarity of the solution =5 M
B Molality of the solution =6.17 m
C Mole fraction of the solute =0.1
The formula of Glauber's salt is Na2SO4.10H2O
Molecular mass of Na2SO4.10H2O
=[2×23+32+4×16]+10(1.0×2+16)=322 g mol1
Weight of the Glauber's salt taken =1610 g
Out of 1610 g of salt, weight of anhydrous Na2SO4
=142322×1610=710 g
Number of moles of Na2SO4 per dm3 of the solution
=710142=5
Molarity of the solution =5 M
Density of solution =1520 kg m3
=1.520×103106 g cm3 = 1.520 g cm3
Total weight of solution =V×d=1 dm3×d
=1000 cm3×1.520 g cm3=1520 g
Weight of water =1520710=810 g
Molality of solution. =5810 g×1000 g=6.17 m
Number of moles of water in the solution =81018=45
Mole fraction of Na2SO4
=No. of moles of Na2SO4Total number of moles=55+45
=0.1

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