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Question: A 1.2% solution (w/v) of \(NaCl\) is isotonic with 7.2% solution (w/v) of glucose. Calculate degree ...

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

Explanation

Solution

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 π=icRT\pi = icRT
Where,
π\pi is the osmotic pressure
ii is the Van't Hoff’s index
cc is the molar concentration
cc= given massmolar mass\dfrac{{given{\text{ }}mass}}{{molar{\text{ }}mass}}
RR is the ideal gas constant
TT is the temperature in Kelvin
Degree of ionization α=i−1n′−1\alpha = \dfrac{{i - 1}}{{n' - 1}}
Where,
α\alpha is the degree of ionization
ii is the Van't Hoff Factor

Complete step by step answer:
Molar mass of NaCl=58.5g/molNaCl = 58.5g/mol
Molar mass of glucose = 180g/mol180g/mol
Osmotic pressure expression is given by,
π=icRT\pi = icRT
Since both are isotonic, both of their osmotic pressures are the same. That is, the osmotic pressure of Sodium Chloride π(NaCl)\pi (NaCl) is equal to the osmotic pressure of glucose π(glucos⁡e)\pi (glu\cos e).
Equating both the osmotic pressures since they are equal,
π(NaCl)\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)}}Sinceglucoseisanonelectrolyte,itdoesn’tdissociatewithaqueoussolution.Henceit’sVan′tHoffFactoris1. 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 Taking\dfrac{{1.2}}{{58.5}}ontherighthandsize,theVan′tHofffactorofSodiumChloride,on the right hand size, the Van't Hoff factor of Sodium Chloride, i(NaCl) = 1.95TheVan′tHoffFactorforSodiumChloridehasnowbeenfound.LetusstartfindingthedegreeofionizationforSodiumChloride.Asweknow,theformulaforfindingthedegreeofionization, 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}}alsowehavefoundthevalueofalso we have found the value ofiinthepreviouscalculation,nowwecansubstitutethevalues.Thevalueofin the previous calculation, now we can substitute the values. The value ofn'is2becausetherearetwoionsformedindissolution−theNa+ionsandtheCl−ions.Now,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)anddegreeofionizationand degree of ionization(0.95)arefoundwhichmatchwiththeoptionB−are found which match with the option B-(0.95)and 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.