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Chapter 1: Problem 61

In normal blood, there are about \(5.4 \times 10^{9}\) red blood cells per milliliter. The volume of a red blood cell is about \(90.0 \times 10^{-12} \mathrm{cm}^{3},\) and its density is \(1.096 \mathrm{g} / \mathrm{mL}\) How many liters of whole blood would be needed to collect \(0.5 \mathrm{kg}\) of red blood cells?

Short Answer

Expert verified
You would need approximately \(938.52 \) liters of whole blood to collect \(0.5 \) kg of red blood cells.

Step by step solution

01

Convert weight to grams

Before proceeding, it's important to convert the weight of the red blood cells from kilograms to grams so it matches the unit in the density which is g/mL. So, \(0.5 \, \mathrm{kg} = 500\, \mathrm{g}\).
02

Calculate the volume of 0.5 kg of red blood cells

With the density formula \( \text{Density} = \frac{\text{Mass}}{\text{Volume}}\), the equation can be rearranged to solve for the volume, which yields: \(\text{Volume} = \frac{\text{Mass}}{\text{Density}}\), or \( \text{Volume} = \frac{500\, \mathrm{g}}{1.096\, \mathrm{g/mL}} = 456.204\, \mathrm{mL}\).
03

Calculate the amount of red blood cells needed

Given that the volume of a single red blood cell is \(90.0 \times 10^{-12} \, \mathrm{mL}\), and the total volume of red blood cells needed is \(456.204 \, \mathrm{mL}\), the total quantity \(Q\) of red blood cells can be found using the formula \(Q = \frac{\text{Total volume}}{\text{Volume per cell}}\), which gives \( Q = \frac{456.204\, \mathrm{mL}}{90.0 \times 10^{-12}\, \mathrm{mL/cell}} = 5.07 \times 10^{15}\) cells.
04

Calculate the volume of whole blood needed

The number of cells required is found. Now, the volume of blood needed can be calculated by using the given proportion \(5.4 \times 10^{9} \) cells/mL. Using cross-multiplication, the volume \(V\) of whole blood can be found using the formula \( V = \frac{Q}{5.4 \times 10^{9} \, \text{cells/mL}}\). This yields \( V = \frac{5.07 \times 10^{15} \, \text{cells}}{5.4 \times 10^{9} \, \mathrm{cells/mL}} = 938519\, \mathrm{mL}\) or about \(938.52 \, \mathrm{L}\).

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Most popular questions from this chapter

A family/consumer science class is given an assignment in candy-making that requires a sugar mixture to be brought to a "soft-ball" stage \(\left(234-240^{\circ} \mathrm{F}\right)\). A student borrows a thermometer having a range from \(-10^{\circ} \mathrm{C}\) to \(110^{\circ} \mathrm{C}\) from the chemistry laboratory to do this assignment. Will this thermometer serve the purpose? Explain.

In the third century \(\mathrm{BC}\), the Greek mathematician Archimedes is said to have discovered an important principle that is useful in density determinations. The story told is that King Hiero of Syracuse (in Sicily) asked Archimedes to verify that an ornate crown made for him by a goldsmith consisted of pure gold and not a gold-silver alloy. Archimedes had to do this, of course, without damaging the crown in any way. Describe how Archimedes did this, or if you don't know the rest of the story, rediscover Archimedes's principle and explain how it can be used to settle the question.

The following densities are given at \(20^{\circ} \mathrm{C}\) : water, \(0.998 \mathrm{g} / \mathrm{cm}^{3} ;\) iron, \(7.86 \mathrm{g} / \mathrm{cm}^{3} ;\) aluminum, \(2.70 \mathrm{g} / \mathrm{cm}^{3}\). Arrange the following items in terms of increasing mass. (a) a rectangular bar of iron,$$81.5 \mathrm{cm} \times 2.1 \mathrm{cm} \times 1.6 \mathrm{cm}$$ (b) a sheet of aluminum foil,$$12.12 \mathrm{m} \times 3.62 \mathrm{m} \times 0.003 \mathrm{cm}$$ (c) 4.051 L of water

In the United States, volume of irrigation water is usually expressed in acre- feet. One acre-foot is a volume of water sufficient to cover 1 acre of land to a depth of 1 ft \(\left(640 \text { acres }=1 \mathrm{mi}^{2} ; 1 \mathrm{mi}=5280 \mathrm{ft}\right)\) The principal lake in the California Water Project is Lake Oroville, whose water storage capacity is listed as \(3.54 \times 10^{6}\) acre-feet. Express the volume of Lake Oroville in (a) cubic feet; (b) cubic meters; (c) U.S. gallons.

According to the rules on significant figures, the product of the measured quantities \(99.9 \mathrm{m}\) and \(1.008 \mathrm{m}\) should be expressed to three significant figures-\(101 \mathrm{m}^{2} .\) Yet, in this case, it would be more appropriate to express the result to four significant figures-\(100.7 \mathrm{m}^{2} .\) Explain why.
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