Calculation of Molar Concentration of Blood

The osmotic pressure of blood is given as 7.65 atm at 310K. To calculate the molar concentration of blood, we can use the relation:

π = MRT

where π is the osmotic pressure, M is the molar concentration, R is the gas constant, and T is the temperature in Kelvin.

On substituting the given values, we get:

7.65 atm = M x 0.0821 L atm mol^-1 K^-1 x 310 K

M = 0.295 M

This means that the molar concentration of blood is 0.295 M.

Calculation of Molar Concentration of Glucose Solution

To find the molar concentration of an aqueous solution of glucose that is isotonic with blood, we need to first understand what isotonicity means. Isotonicity refers to the property of a solution where the concentration of solutes is the same as that of another solution. In this case, we want to find a glucose solution that has the same osmotic pressure as blood, i.e., is isotonic with blood.

The osmotic pressure of a solution is directly proportional to the molar concentration of its solutes. Therefore, if we can find the molar concentration of a glucose solution that has the same osmotic pressure as blood, then that solution will be isotonic with blood.

We know that the molar concentration of blood is 0.295 M. So, we can find the molar concentration of a glucose solution that is isotonic with blood by equating the osmotic pressure of the two solutions. Let the molar concentration of glucose solution be x M.

Using the same relation as before, we get:

7.65 atm = x M x 0.0821 L atm mol^-1 K^-1 x 310 K

x M = 0.795 M

This means that the molar concentration of an aqueous solution of glucose that is isotonic with blood is 0.795 M.

Calculation of Weight-Volume Percentage of Glucose Solution

To find the weight-volume percentage of glucose solution that is isotonic with blood, we need to know the molecular weight of glucose and the density of the solution. The molecular weight of glucose is 180 g/mol, and the density of the solution is assumed to be 1 g/mL.

We can use the relation:

% w/v = (mass of solute/volume of solution) x 100

The mass of solute required to prepare 100 mL of a 0.795 M glucose solution can be calculated as:

mass of glucose = molar concentration x molecular weight x volume

mass of glucose = 0.795 M x 180 g/mol x 0.1 L

mass of glucose = 14.31 g

Therefore, to prepare 100 mL of an aqueous glucose solution that is isotonic with blood, we need to dissolve 14.31 g of glucose in water and make up the volume to 100 mL.

Using the above relation, we get:

% w/v = (14.31 g/100 mL) x 100

% w/v = 5.41%

Therefore, the correct answer is option A, i.e., 5.41%.