Question

Density Of A 2.05M Solution Of Acetic Acid water is 1.02g/mL. The molality of the solution is (a) 1.14mol kg–1 (b) 3.28mol kg–1 (c) 2.28mol kg–1 (d) 0.44mol kg–1.

Answer 1

Answer:

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}

Correct option is 3.

Option 1)

1.14 mol kg^{-1}

This is an incorrect option.

Option 2)

3.28 mol kg^{-1}

This is an incorrect option.

Option 3)

2.28 mol kg^{-1}

This is the correct option.

Option 4)

0.44 mol kg^{-1}

This is an incorrect option.

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Answer 2

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$density = \frac{mass}{volume}$

mass of acetic acid : 60 g/ mol

$2.05 = \frac{60}{m}$

$m = 29.27$

now formula for molarity

$M = \frac{wb}{mb} \times \frac{1000}{vs}$

$M = \frac{29.27}{60} \times \frac{1000}{1.02}$

$M = 2.28 \: mol\: kg^{ - 1} \:$

option C is correct

hope it help❣