Question

The average O.P. of human blood is $7.7atm$ at ${40^ \circ }C$. Calculate
(a) the concentration of blood in molarity
(b) the freezing point of blood. Assume molarity and molality to be the same. $\left( {{K_f}} \right) = 1.86$.
A. (a) $0.3M$, (b) $ - {0.558^ \circ }C$
B. (a) $0.3M$, (b) $ + {0.558^ \circ }C$
C. (a) $0.5M$, (b) $ - {0.72^ \circ }C$
D. (a) $0.5M$, (b) $ + {0.72^ \circ }C$

Answer 1

Hint:Osmotic pressure is defined as the minimum pressure that is required by a solution to halt the flow of solvent molecules through the semipermeable membrane. It depends on the concentration of the solute particles that are present in the solution. Also, depression in the freezing point is observed when a solute is added to the solvent.

Complete step by step answer:
Osmotic pressure is defined as the minimum pressure that is required by a solution to halt the flow of solvent molecules through the semipermeable membrane.
It is given by the formula: $\pi = CRT$…..1
Where $\pi = $ osmotic pressure
$C = $ molar concentration
$R = $ Gas constant
$T = $ Temperature
This relationship was first put forward by the Dutch Chemist: Jacobus Van’t Hoff
Freezing point depression: it is defined as the lowering of freezing points of the solvent on the addition of solutes.
It is given by the formula: $\Delta {T_f} = {K_f}m$ …….2
$\Delta {T_f} = $ freezing point depression
${K_f} = $ cryoscopic constant
$m = $ molality
Molarity can be defined as the number of moles of solute present in per liter of solution.
It is given by the formula: $M = \dfrac{n}{V}$
Where,
$M = $ Molarity
$n = $ moles of solute
$V = $ liters of solution
Molality is defined as the number of moles of solute present in per kilogram of solvent.
It is given by the formula: $m = \dfrac{{mol}}{{kg}}$
Where,
$m = $ molality
Mol=moles of solute
Kg= kilogram of solvent.
Given data:
Osmotic pressure=$7.7$ atm
Gas constant : $R = 0.08206$ Latm/mol.K
Temperature=${40^ \circ }C$
$T = 273 + 40$
$T = 313$K

b)
We will use the formula of osmotic pressure given in equation 1 to find the concentration:
$\pi = CRT$
substituting the value from the given data we get,
$7.7 = C \times 0.08206 \times 313$
$C = \dfrac{{7.7}}{{0.08206 \times 313}}$
On further solving ,
$C = \dfrac{{7.7}}{{25.68478}}$
$C = 0.299 \approx 0.3$M
b) Now we will use the freezing point depression formula given in equation 2 to find the freezing point of blood.
$\left( {{K_f}} \right) = 1.86$,$C = m = 0.3$ M
$\Delta {T_f} = {K_f}m$
substituting the value from the given data we get,
$\Delta {T_f} = 1.86 \times 0.3$
On further solving ,
$\Delta {T_f} = {0.558^ \circ }C$
But as we know that the freezing point of water is ${0^ \circ }C$
So the freezing point of blood will be: $0 - 0.558$
Freezing point of blood= $ - {0.558^ \circ }C$

So, the correct answer will be option A :
a) $C = 0.3M$
b) $\Delta {T_f} = {0.558^ \circ }C$.


Note:
In osmotic pressure, we find concentration in terms of molarity whereas to find the freezing point depression we need concentration in molality. Both are colligative properties, that is why they depend on the number of particles of solute dissolved in a definite amount of solvent and are independent of the nature of the solute.