A variation of the indicator-dilution method (see preceding problem) is used to measure total blood volume. A known amount of a tracer is injected into the bloodstream and disperses uniformly throughout the circulatory system. A blood sample is then withdrawn, the tracer concentration in the sample is measured, and the measured concentration [which equals (tracer injected)/(total blood volume) if no tracer is lost through blood vessel walls] is used to determine the total blood volume. In one such experiment, 0.60 \mathrm{cm}^{3}0.60cm 3 of a solution containing 5.00 mg/L of a dye is injected into an artery of a grown man. About 10 minutes later, after the tracer has had time to distribute itself uniformly throughout the bloodstream, a blood sample is withdrawn and placed in the sample chamber of a spectrophotometer. A beam of light passes through the chamber, and the spectrophotometer measure the intensity of the transmitted beam and displays the value of the solution absorbance (a quantity that increases with the amount of light absorbed by the sample). The value displayed is 0.18. A calibration curve of absorbance A versus tracer concentration C (micrograms dye/liter blood) is a straight line through the origin and the point (A=0.9, C=3 \mu \mathrm{g} / \mathrm{L})(A=0.9,C=3μg/L). Estimate the patient's total blood volume from these data.

Solution :

Given :

The spectrophotometer reading is (A) = 0.18

Absorbdance Vs Concentration follows a straight line relation passing throght the origin.

Let the relation be of the form A = mC  -------(1)

where, A = absorbdance

            C = concentration (μg/L)

For a given point, A = 0.9

                             C = 3 μg/L

From equation (1), we have

0.9 = m x 3

m = 0.3

Therefore the complete equation of the spectrophometer calibration curve becomes A = 0.3C

Now we have the relation relating the concenrtation of the traces and the absorbdance. Let us now calculate the concentration of the traces in sample for which A = 0.18.

   = 0.6

Traces concentration = 0.6 μg/L

Now in order to calculate the volume of the blood in a grown man we must know how much dye(mass) was injected into his blood.

For that it is given that the trace solution of 0.6 and 5 μg/L was injected.

∴ Mass of the dye in solution injected

                                                              = 3 μg

This much of mass is diluted throughout the blood volume.

Thus if V(L) is total blood volume, then 0.6 μg/L x V = 3 μg

This means that the total mass of dye in the body is what we injected initially i.e. 3 μg and must be holding by V(L) of blood.

∴

       = 5 L

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acid : acids are those which accepts an electron-pairs.

base : bases are those which donates ab electron-pairs.

alkali : it is a base that dissolves in water. it is basically a base and the ph greater than 7.

hydroxide ion : it is a poly-atomic ion which is negatively charged.