Question

The temperature at which molarity of pure water is equal to its molality is
(A) 273K
(B) 298K
(C) 277K
(D) None of these.

Answer 1

Both molarity and molality are related to the concentration of the solution. Pure water shows abnormal behaviour at certain temperatures which can be related to concentration.

Complete step by step solution:
Molarity and molality are concentration terms which mean that we can measure the concentration of a solution.
Molarity of a solution is equal to the total number of moles of solute per litre of a solution.
$M = \dfrac{{{n_{solute}}}}{{Volume{\text{ }}of{\text{ }}solution{\text{ }}(in{\text{ }}mL)}} \times 1000$
From the formula of molarity, we can say that molarity depends on the volume of the solution. The volume of any solution is dependent on the temperature. Therefore, the value of molarity is altered by the change in temperature. Molarity is expressed as M. The unit of molarity is mol/litre.
The molality of a solution is equal to the total number of moles of solute per kilogram of the solvent.
$m = \dfrac{{{n_{solute}}}}{{mass{\text{ }}of{\text{ }}solvent{\text{ }}(in{\text{ }}g)}} \times 1000$
From the formula of molality, we can say that molality depends on the mass of the solvent. Mass of the solvent is independent of the temperature. Therefore, the value of molality is not altered by the change in temperature. Molality is expressed as m. The unit of molality is mol/kg.
Molarity and molality
$m = \dfrac{{M \times 1000}}{{(d \times 1000) - M(M')}}$
Here,
m= molality of the solution
M = molarity of the solution
d = density of the solution
M’ = molar mass of the solvent
We can also solve in a different way if the density is equal to unity.
In case of pure water there is no different solution added to it. The density of pure water is 1g/litre at $4{}^\circ C$. So as the density is equal to one and we know that, density is the ratio of mass to volume, therefore in this case mass is equal to volume. So by using the formulas of both molarity and molality we get,
$\dfrac{{{n_{solute}}}}{{Volume{\text{ }}of{\text{ }}solution{\text{ }}(in{\text{ }}mL)}} \times 1000 = \dfrac{{{n_{solute}}}}{{mass{\text{ }}of{\text{ }}solvent{\text{ }}(in{\text{ }}g)}} \times 1000$
$M = m$
Because of this the molarity and molality will be equal. This happens at $4{}^\circ C$.
${\text{T(K) = 4 + 273 = 277K}}{\text{.}}$
So at ${\text{277K}}$ the molarity and molality of pure water are equal.

Hence option (C) is the correct one.

Note: At $4{}^\circ C$, water shows abnormal behaviour. At this temperature, water has the highest density. If you heat it or cool it, in both the cases it will expand.