Question

The boiling point of water at 735 torr is ${{99}^{\circ }}C$ . The mass of NaCl added in 100gm of water $\left( {{K}_{b}}=0.51\text{ kg }{{\text{m}}^{-3}} \right)$ to make its boiling point at ${{100}^{\circ }}C$ is-
[A] 10.68g
[B] 5.73g
[C] 2.67g
[D] 26.7g

Answer 1

From the above question, the difference of the new and the present boiling points can be calculated. Use it in the relation of elevation of boiling point with the constant K. Find the molality from there and use the definition of molality to calculate the number of moles of the solvent. The total amount of NaCl to be added can be calculated by multiplying molar mass by number of moles.

Complete step by step answer:
Firstly, let us write down the chemical reaction that is taking place. In addition to water, sodium chloride will dissociate into ions as it is an ionic compound. Therefore, we can write the reaction as-
$NaCl+{{H}_{2}}O\to N{{a}^{+}}+C{{l}^{-}}$

Now it is given in the question that the boiling point changes from 99 to 100 degrees. Therefore, the change in boiling point is-

 $$\Delta {{T}_{b}}=100-99={{1}^{\circ }}C$$   

Now, the difference will be the same in kelvin as well as Celsius scale. Therefore, we can write that the elevation in boiling point is of 1K.
Now, we know that the relation between the elevation in boiling point and the constant K is -
$\Delta {{T}_{b}}={{K}_{b}}\times m\times i$
Here, m is the concentration of the solution in molality and i is the Van’t Hoff factor.
For NaCl, i = 2 (as it dissociates into two ions)

We know that molality is the number of moles of solute per kilogram of solvent. If we calculate the molality from here, we can find the grams of NaCl added.
Therefore, $m=\dfrac{\Delta {{T}_{b}}}{{{K}_{b}}\times i}=\dfrac{1}{0.51\text{ kg }{{\text{m}}^{-3}}\times 2}=0.98$
Now, from definition, $molality=\dfrac{number\text{ of moles of solute}}{volume\text{ of solvent (kg)}}$

Here, the solvent is water and we have 100gm of water i.e. 0.1 kg.
Therefore, we can write that - $0.98=\dfrac{number\text{ of moles of solute}}{\text{0}\text{.1 kg}}$
Or, number of moles of solute = 0.098 mol.
Now to find the mass of the solvent we need to determine its molar mass.
Molar mass of NaCl = 58.44 g/mol
We mass is the product of molar mass and number of moles.
Therefore, mass = $58.44g/mol\times 0.98mol$ = 5.73 g.

Therefore, the correct answer is 5.73g
So, the correct answer is “Option B”.

Note: There are four colligative properties. It depends only on the number of solute particles. The dissociation of solute molecules into ions results in an increase in the number of particles and hence affects the colligative properties.
The four colligative properties are elevation in boiling point, depression in freezing point, osmotic pressure and lowering of vapour pressure.