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Question: The osmotic pressure of human blood is \(7.85atm\) at \({30^ \circ }C\). Then the freezing point of ...

The osmotic pressure of human blood is 7.85atm7.85atm at 30C{30^ \circ }C. Then the freezing point of blood will be.
[Given: Kf=1.86 K Kg moll1{K_f} = 1.86{\text{ K Kg mol}}{{\text{l}}^{ - 1}}]
A.+0.587C + {0.587^ \circ }C
B.0.587C - {0.587^ \circ }C
C.+1.174C + {1.174^ \circ }C
D.1.174C - {1.174^ \circ }C

Explanation

Solution

Depression in freezing point: It is defined as the decrease in freezing point of a solvent in addition to non-volatile solute. It is directly proportional to the molality of the solution.
Osmotic pressure: It is defined as the minimum pressure which needs to be applied to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane.

Complete step by step answer:
In this question we have to work on osmotic pressure. Osmotic pressure is represented by π\pi and it is calculated as π=CRT\pi = CRT, where CC is the concentration of the solution, RR is gas constant, TT is temperature and π\pi is osmotic pressure. Here in the question we are first given with the osmotic pressure and then we have to find the depression in freezing point. So first we will find the molarity i.e. concentration of the solution by using the osmotic pressure as C=πRTC = \dfrac{\pi }{{RT}}. Here in the question π=7.85atm\pi = 7.85atm, R=0.0821R = 0.0821 and T=30C=303KT = {30^ \circ }C = 303K.
So, concentration will be as C=7.850.0821×303=0.315C = \dfrac{{7.85}}{{0.0821 \times 303}} = 0.315.
Here in the question we are not given the density of the solution so we assume that molarity i.e. concentration of the solution is the same as molality of the solution.
Now depression in freezing point is as ΔTf=Kf×m\Delta {T_f} = {K_f} \times m where ΔTf\Delta {T_f} is the depression in freezing point, Kf{K_f} is molal depression constant and mm is molality of the solution.
Here Kf=1.86 K Kg moll1{K_f} = 1.86{\text{ K Kg mol}}{{\text{l}}^{ - 1}} and m=0.315m = 0.315 so the value of ΔTf\Delta {T_f} will be 1.86×0.315=0.587C1.86 \times 0.315 = {0.587^ \circ }C. And hence the freezing point of the blood will be 00.587C=0.587C0 - {0.587^ \circ }C = - {0.587^ \circ }C.

Hence option B is correct.

Note:
Van’t Hoff factor: It is defined as the ratio of actual number of particles in the solution after dissociation or association to the number of particles for which no ionization takes place. It is represented by ii.