Chapter 1, Problem 1SP

Suppose you have a culture of bacteria, where the density of each bacterium is $2.0\text{g}/{\text{cm}}^3$.

1. If each bacterium is $5\mu \text{m}\times 5\mu \text{m}\times 20\mu \text{m}$ in size, find the number of bacteria if their total mass is 30 grams. Recall that $1\mu \text{m}={10}^{-6}$ meters.
2. Suppose that you learn that the sizes of bacteria range from $4\mu \text{m}\times 5\mu \text{m}\times 15\mu \text{m}$ to $5\mu \text{m}\times 6\mu \text{m}\times 25\mu \text{m}$ What is the range of the possible number of bacteria making up the total mass of 30 grams?

To determine

To find: The number of bacteria when total mass of culture of bacteria is given.

Answer to Problem 1SP

The number of bacteria in 30 gram is $3\times {10}^{10}$ .

Explanation of Solution

Given:

The size of each bacteria is $\text{5μm×5μm×20μm}$ total mass of bacteria is 30 grams and density of bacterium is 2.0g/cm³_, and $1\mu \text{m=1}{\text{0}}^{\text{-6}}\text{m}$ .

Formula used:

Mass of a bacteria = Size of bacterium $\times$ density of bacteria

Calculation:

The size of bacterium

  $V=5\mu \text{m}\times 5\mu \text{m}\times 20\mu \text{m=}\left(5\times {10}^{-4}\text{cm}\right)\times \left(5\times {10}^{-4}\text{cm}\right)\times \left(20\times {10}^{-4}\text{cm}\right)=500\times {10}^{-12}\text{c}{\text{m}}^{\text{3}}$

  ( $\because 1\mu \text{m}={10}^{-6}\text{m}={10}^{-4}\text{cm}$ )

Mass of bacterium $m_B=V\times$ density of bacterium $=500\times {10}^{-12}\text{c}{\text{m}}^{\text{3}}\times 2.0\frac{g}{\text{c}{\text{m}}^{\text{3}}}=1000\times {10}^{-12}g={10}^{-9}g$ .

Therefore, number of bacteria in 30 g mass $=\frac{30}{m_B}=\frac{30}{{10}^{-9}}=30\times {10}^9=3\times {10}^{10}$.

Conclusion:

The number of bacteria in 30 gram is $3\times {10}^{10}$ .

To determine

To find: The range of possible number of bacteria of different size in 30 gram mass.

Answer to Problem 1SP

The range of number of bacteria of size $5\mu \text{m}\times 6\mu \text{m}\times 25\mu \text{m}$ and $4\mu \text{m}\times 5\mu \text{m}\times 15\mu \text{m}$ in 30 gram mass is $2\times {10}^{10}-5\times {10}^{10}$ .

Explanation of Solution

Given:

The density of bacteria given in part (a)

Formula used:

The same as used in part (a).

Calculation:

The mass of bacteria of size $4\mu \text{m}\times 5\mu \text{m}\times 15\mu \text{m}$

  $m_1{}_B=\left(4\times \text{1}{\text{0}}^{\text{-4}}\text{cm}\right)\times \left(5\times \text{1}{\text{0}}^{\text{-4}}\text{cm}\right)\times \left(15\times \text{1}{\text{0}}^{\text{-4}}\text{cm}\right)\times 2.0\frac{\text{g}}{\text{c}{\text{m}}^{\text{3}}}=300\times {10}^{-12}\times 2=6\times {10}^{-10}g$

Therefore, the number of bacteria of this kind in 30 gram mass $=\frac{30}{m_1{}_B}=\frac{30}{6\times {10}^{-10}}=5\times {10}^{10}$.

The mass of bacteria of size $5\mu \text{m}\times 6\mu \text{m}\times 25\mu \text{m}$

  $m_2{}_B=\left(5\times \text{1}{\text{0}}^{\text{-4}}\text{cm}\right)\times \left(6\times \text{1}{\text{0}}^{\text{-4}}\text{cm}\right)\times \left(25\times \text{1}{\text{0}}^{\text{-4}}\text{cm}\right)\times 2.0\frac{\text{g}}{\text{c}{\text{m}}^{\text{3}}}=750\times {10}^{-12}\times 2=1.5\times {10}^{-9}g$

Therefore, the number of bacteria of this kind in 30 gram mass $=\frac{30}{m_2{}_B}=\frac{30}{1.5\times {10}^{-9}}=20\times {10}^9=2\times {10}^{10}$

Thus, the range of number of bacteria of size $5\mu \text{m}\times 6\mu \text{m}\times 25\mu \text{m}$ and $4\mu \text{m}\times 5\mu \text{m}\times 15\mu \text{m}$ in 30 gram mass is $2\times {10}^{10}-5\times {10}^{10}$ .

Conclusion:

The range of number of bacteria of size $5\mu \text{m}\times 6\mu \text{m}\times 25\mu \text{m}$ and $4\mu \text{m}\times 5\mu \text{m}\times 15\mu \text{m}$ in 30 gram mass is $2\times {10}^{10}-5\times {10}^{10}$ .