What is the molarity of the solution prepared by concentrating, by evaporation of solvent, 755 mL...

Question

Question

What is the molarity of the solution prepared by concentrating, by evaporation of solvent, 755 mL of 0.200% (m/v) NaNO3 to a

Answer 1

Answer #1

Mass of NaNO3 in the initial solution= 0.200 x 755/100 = 1.51 g [ as it is 0.200% mass/volume]

The new volume = 573mL = 0.573L

mass of NaNO3 does not change during evaporation, only solvent goes.

molar mass of NaNO3 = 85g/mol

thus molarity = moles/V(L)

= [ mass/ molar mass] / V(L)

= [1.51g/85g/mol] /0.573L

=0.031 mol/L or 0.031M

Answer 2

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