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        Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is:a.3.28 mol/Kg       b.0.44 mol/Kg
4 months ago

Arun
22956 Points
							As we learnt inRelation between concentration terms (Molality & Molarity) -$\dpi{100} m=\frac{1000M}{1000d-Mm_{1}}$- wherein$\dpi{100} m=Molality$$\dpi{100} M=Molarity$$\dpi{100} m_{1}=Molar\: mass\: of \: solute$$\dpi{100} d=density\: of \: solution \: g/mol$   $Molality(m)= \frac{M}{1000d-MM_{1}}$Where M = Molarity,  d = Density, M1= Molecular mass of solute$m= \frac{2.05}{1000\times 1.02-2.05\times 60}= \frac{2.05}{897}$$= 2.28\times 10^{-3}mol g^{-1}=\: 2.28\: mol \: kg^{-1}$

4 months ago
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