Question

A 1.2% solution (w/v) of $NaCl$ is isotonic with 7.2% solution (w/v) of glucose. Calculate degree of ionization and Van't Hoff factor of $NaCl.$
A: 0.55, 0.45
B: 0.95, 1.95
C: 0.98, 2.05
D: 0.95, 2.76

Answer 1

Hint: Since both of them are isotonic, they will be having equal osmotic pressure (the measure of the tendency of a solution to take in pure solvent by osmosis). Substituting the values of molar concentrations of Sodium Chloride and glucose, you will be able to solve the answer by finding the Van't Hoff Factor.
Formula used:
Osmotic pressure $\pi = icRT$
Where,
$\pi $is the osmotic pressure
$i$ is the Van't Hoff’s index
$c$ is the molar concentration
$c$= $\dfrac{{given{\text{ }}mass}}{{molar{\text{ }}mass}}$
$R$ is the ideal gas constant
$T$ is the temperature in Kelvin
Degree of ionization $\alpha = \dfrac{{i - 1}}{{n' - 1}}$
Where,
$\alpha $ is the degree of ionization
$i$ is the Van't Hoff Factor


Complete step by step answer:
Molar mass of $NaCl = 58.5g/mol$
Molar mass of glucose = $180g/mol$
Osmotic pressure expression is given by,
$\pi = icRT$
Since both are isotonic, both of their osmotic pressures are the same. That is, the osmotic pressure of Sodium Chloride $\pi (NaCl)$ is equal to the osmotic pressure of glucose \[\pi (glu\cos e)\].
Equating both the osmotic pressures since they are equal,
$\pi (NaCl)$$ = $\[\pi (glu\cos e)\]
$ = > $$i(NaCl)c(NaCl)RT = i(gl)c(gl)RT$
$ = > $$i(NaCl)c(NaCl) = i(gl)c(gl){\text{ (Since R and T are same on both sides and get cancelled)}}$
Since glucose is a nonelectrolyte, it doesn’t dissociate with aqueous solution. Hence it’s Van't Hoff Factor is 1.
$ = > i(NaCl) \times \dfrac{{1.2}}{{58.5}} = 1 \times \dfrac{{7.2}}{{180}}{\text{ (From the formula of molar concentration)}}$
Taking $\dfrac{{1.2}}{{58.5}}$ on the right hand size, the Van't Hoff factor of Sodium Chloride,
$i(NaCl) = 1.95$
The Van't Hoff Factor for Sodium Chloride has now been found. Let us start finding the degree of ionization for Sodium Chloride. As we know, the formula for finding the degree of ionization, $\alpha = \dfrac{{i - 1}}{{n' - 1}}$ also we have found the value of $i$ in the previous calculation, now we can substitute the values. The value of $n'$ is 2 because there are two ions formed in dissolution- the Na+ ions and the Cl- ions. Now,
$\alpha = \dfrac{{1.95 - 1}}{{2 - 1}}$
\[\alpha = \dfrac{{0.95}}{1} = 0.95\]
Hence, both the Van't Hoff factor $(1.95)$ and degree of ionization $(0.95)$ are found which match with the option B- $(0.95)$ and$(1.95)$.
Hence option B is the right answer.

Note:
For a non-electrolyte, the Van't Hoff Factor is always 1, since no dissociation or association takes place .So if you see a non-electrolytic compound, substitute 1 in the place of its Van't Hoff Factor.