Density of $2.03M$ . Aqueous solution of acetic acid is $1.017gm{L^{ - 1}}$ . Molecular mass of acetic acid is $60$ . Calculate the molality of solution.
(a) $ 2.27$
(b) $1.27$
(c) $3.27$
(d) $4.27$

Hint: Molality is one of the important properties of solutions. It is used to express the concentration of a solute in a solution and mostly depends on the mass of the solvent. Molality is also sometimes referred to as molal concentration.

Complete step by step answer:
Given,
Density $ = 1.017gm{L^{ - 1}}$
Molecular Mass $ = 60$
Molarity $ = 2.03M$
Molarity means moles in solution.
$n = $ Molar Mass given Mass​
Given Mass​,
$2.03 \times 60 = 121.8g$
$121.8g$ of acetic acid in water.
Density $ = 1.07gm{L^{ - 1}} - 11mL$ of solution $ \to 1.017$ of solution .
$1000mL$ of solution $ \to 1017g$ of solution.
Mass of solution = Mass of solvent $ + $ Mass of solute
$1017g = $ mass of solvent $ + 121.8g$.
$1017g - 121.8g = $ Mass of solvents
$895.2g = $ Mass of solvent

Molality $ = \dfrac{{no.\,of\,mole\,of\,acetic\,acid}}{{mass\,of\,solvent\,in\,liqiud}} = \dfrac{{2.03}}{{\left( {\dfrac{{895.5}}{{1000}}} \right)}} = 2.267m$ .
Molality is a measure of the number of moles of solute present in $1kg$ of solvent. This contrasts with the definition of molarity which is based on a specified volume of solution. A commonly used unit for molality in chemistry is $molk{g^{ - 1}}$ . A solution of concentration $1molalk{g^{ - 1}}$ is also sometimes denoted as $1molal$ .

So the correct option is option A

Note: Molality is a property of a solution and is defined as the number of moles of solute per kilogram of solvent. The SI unit for molality is $molk{g^{ - 1}}$ . A solution with a molality of $3molk{g^{ - 1}}$ is often described as or . However, following the SI system of units, $molk{g^{ - 1}}$ or a related SI unit is now preferred.