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Which choice is closest to the molality of 3.15 M aqueous Ca(NO3)2 (molar mass = 164.1...

Which choice is closest to the molality of 3.15 M aqueous Ca(NO3)2 (molar mass = 164.1 g/mol) with a density of 1.09 g/mL?

A. 2.93 m
B. 6.21 m    
C. 3.78 m    
D. 5.50 m
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Homework Answers

Answer #1

Molarity = moles of Ca(NO3)2/Volume of solution in litres

Thus, if volume of solution is 1 litres (i.e. 1000 ml), then moles of Ca(NO3)2 in 3.15 M solution = 3.15 moles.......(1)

Now, mass of solution = volume of solution*density of solution = 1000*1.09 = 1090 g...........(2)

Mass of Ca(NO3)2 = moles*molar mass = 3.15*164.1 = 516.915 g

Thus, mass of solvent = mass of solution - mass of Ca(NO3)2 = 573.085 g

Now, molality of the solution = moles of Ca(NO3)2/mass of solvent in kg = 3.15/0.573085 = 5.5 m

Hence the correct option is :- D

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