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The mole fraction of glucose in 0.15 M solution (density= 1.10 g mL-1 ) is

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a
2.40 x 10-3
b
2.0 x 10-3
c
2.82 x 10-3
d
2.40 X 10-2

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detailed solution

Correct option is B

Let we have

Amount of glucose, n2 = 0.15 mol;   Volume of solution, V = 1000 mL

Mass of glucose, m2 = n2M2 = (0.15 mol) (180 g mol -1) = 27.0 g.

Mass of solution, m = Vρ = (1000 mL) (1.10 g mL-1 ) = 1100 g

Mass of water, m1 = m1 - m2 = 1100 g - 27 g = 1073 g

Amount of water, n1=m1M1=10/3g18gmol1=59.61mol

Mole fraction of glucose, x2=n2n1+n2=0.15mol(59.61+0.15)mol=0.002

 

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