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Question

# 161×10−2 kg of Glauber's salt is dissolved in water to obtain 1 dm3 of a solution of density 1520 kg m−3. 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 =0.1The 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 mol−1 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 m−3 =1.520×103106 g cm−3 = 1.520 g cm−3 Total weight of solution =V×d=1 dm3×d =1000 cm3×1.520 g cm−3=1520 g Weight of water =1520−710=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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