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Question

0.1M of urea is isotonic to

a)0.1M NaCl solution

b) 0.1M glucose solution

C)0.02M KCl solution

d) 0.1 BaCl2 solution

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Solution

Answer: Option (b)
Isotonic solutions are those which have same osmotic pressure and hence same molar concentration. We know that osmotic pressure,
straight pi =i CRT
where i = total no. of ions in the electrolyte solution.


Option (a) 0.1 M Urea.
Urea does not ionize.
So, straight pi = CRT = 0.1 RT

0.1 M NaCl.
NaCl rightwards harpoon over leftwards harpoon stack Na to the power of plus with 1 space ion below plus stack Cl to the power of minus with 1 space ion below
i = 1 + 1 = 2
therefore straight pi = i CRT = 2 × 0.1 RT = 0.2 RT
Since, 0.1 RT and 0.2 RT differ, these are not isotonic solutions.

Option (b) 0.1 M Glucose
Glucose does not ionize. So, straight pi = CRT = 0.1 RT

Urea (0.1 M) does not ionize.
So, straight pi = CRT = 0.1 RT
Hence, 0.1 RT and 0.1 RT are same, these are isotonic solutions.

Option (c) 0.02 M KCl
KCl rightwards harpoon over leftwards harpoon stack straight K to the power of plus with 1 space ion below plus stack Cl to the power of minus with 1 space ion below
therefore i = 1+ 1 = 2
thereforestraight pi = 2 × 0.02 RT = 0.04 RT

Urea (0.1 M) does not ionize.
So, straight pi = CRT = 0.1 RT
Since 0.04 RT and 0.1 RT differ, these are not isotonic solutions.


Option (d) 0.1 M BaCl subscript 2
BaCl subscript 2 rightwards harpoon over leftwards harpoon stack Ba to the power of plus 2 end exponent with 1 space ion below plus stack 2 Cl to the power of minus with 2 space ion below
therefore i = 1+ 2 = 3
thereforestraight pi = 3 × 0.1 RT = 0.3 RT

Urea (0.1 M) does not ionize.
So, straight pi = CRT = 0.1 RT
Since 0.3 RT and 0.1 RT differ, these are not isotonic solutions.

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