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Grade:
        
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
one month ago

Answers : (1)

Arun
21962 Points
							

As we learnt in

Relation between concentration terms (Molality & Molarity) -

\dpi{100} m=\frac{1000M}{1000d-Mm_{1}}

- wherein

m=Molality

M=Molarity

m_{1}=Molar\: mass\: of \: solute

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}

one month ago
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