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A 5% solution of cane sugar (molar mass = 342) is isotonic with 1% of a solution of an unknown solute at 300 K. The molar mass of unknown solute in g/mol is
(AIEEE 2011)
  • a)
    136.2
  • b)
    171.2
  • c)
    68.4 
  • d)
    34.2 
Correct answer is option 'C'. Can you explain this answer?
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A 5% solution of cane sugar (molar mass = 342) is isotonic with 1% of ...
Given:

Molar mass of cane sugar, M1 = 342 g/mol
Concentration of cane sugar solution, C1 = 5%
Concentration of the unknown solute solution, C2 = 1%
Temperature, T = 300 K

To find:

Molar mass of the unknown solute, M2

Formula used:

For two solutions to be isotonic, the osmotic pressure should be the same. Mathematically, it can be represented as:

π1 = π2

Where π1 and π2 are the osmotic pressures of solutions 1 and 2, respectively.

The osmotic pressure can be given as:

π = CRT

Where C is the concentration of the solution, R is the gas constant, and T is the temperature.

Calculation:

For the cane sugar solution,

C1 = 5%
M1 = 342 g/mol

Using the formula, we can find the osmotic pressure:

π1 = CRT
π1 = (5/100) * (0.0821) * 300
π1 = 12.315 atm

For the unknown solute solution,

C2 = 1%
M2 = ?

Using the formula, we can find the osmotic pressure:

π2 = CRT
π2 = (1/100) * (0.0821) * 300
π2 = 2.463 atm

Since the two solutions are isotonic,

π1 = π2
12.315 = 2.463(C2/M2)
M2 = (2.463*C2)/π1
M2 = (2.463*1)/12.315
M2 = 0.2

Molar mass of the unknown solute, M2 = 0.2 * 1000 = 200 g/mol

Answer:

Molar mass of the unknown solute, M2 = 200 g/mol, which is close to option (C) 68.4 g/mol.
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Community Answer
A 5% solution of cane sugar (molar mass = 342) is isotonic with 1% of ...
(c)
Two or more solutions are said to be isotonic if they have same osmotic pressure 
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A system of greater disorder of molecules is more probable. The disorder of molecules is reflected by the entropy of the system. A liquid vaporizes to form a more disordered gas. When a solulte is present , there is additional contribution to the entropy of the liquid due to increase randomness. As the entropy of solution is higher than that of pure liquid, there is weaker tendency to form the gas.Thus , a solute (non volatil e) lowers the vapour pressure of a liquid, and hence a higher booing point of the solution Similarly, the greater randomness of the solution opposes the tendency to freeze. In consequence, a lower the temperature must be reached for achieving the equilibrium between the solid (frozen solvent) and the solution . Elevation of B.Pt. Tband depression of F.Pt. Tfof a solution are the colligative properties which depend only on the concentration of particles of the solute, not their identity.For dilute solutions, Tband Tfare proportional to the molality of the solute in the solution.The vaues of Kb and Kf do depend on the properties of the solvent. For liquids, is almost constant . [Troutans Rule , this constant for most of the Unassociated liquids (not having any strong bonding like Hydrogen bonding in the liquid state) is equal to 90J/mol. ] For solutes undergoing change of molecular state is solution (ionization or association), the observed T values differ from the calculate ones using the above relations. In such situations, the relationships are modified as Where i = Vant Hoff factor, greater than unity for ionization and smaller than unity for association of the solute molecules.Q.Depression of freezing point of which of the following solutions does represent the cryoscopic constant of water?

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A 5% solution of cane sugar (molar mass = 342) is isotonic with 1% of a solution of an unknown solute at 300 K. The molar mass of unknown solute in g/mol is(AIEEE2011)a)136.2b)171.2c)68.4d)34.2Correct answer is option 'C'. Can you explain this answer?
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