Question 1.36 - Chemistry: Molecular Nature Of Matter And Change

Question

The volume of a certain bacterial cell is 2.56 mm³ . (a) What is its volume in cubic millimeters ( mm³ )? (b) What is the volume of 10⁵ cells in liters (L)?

Step-by-Step Solution

Answer

Answer

 

(a) The volume of a certain bacterial cell in the units of cubic millimeters is $2.56\times {10}^{-9}{\mathrm{mm}}^3$  .

(b) The volume of 10⁵ cells in the units of liters is 10^-10 L.

Convert the volume of the cell from cubic micrometers to cubic millimeters.

The unit conversion equation that shows the relation between micrometers and meters can be shown as follows:

$$\begin{gathered} \mathbf{1}\mathbf{\mu m}\mathbf{ }\mathbf{=}\mathbf{ }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{m} \\ \mathrm{The} \mathrm{conversion} \mathrm{factor} \mathrm{used} \mathrm{to} \mathrm{convert} \mathrm{from} \mathrm{micrometers} \mathrm{to} \mathrm{meters} \mathrm{is} \mathrm{as} \mathrm{follows}: \\ \frac{{10}^{-6}\mathrm{m}}{1\mathrm{\mu m}} \\ \mathrm{The} \mathrm{unit} \mathrm{conversion} \mathrm{equation} \mathrm{for} \mathrm{the} \mathrm{conversion} \mathrm{between} \mathrm{meters} \mathrm{and} \mathrm{millimeters} \mathrm{is} \mathrm{shown} \mathrm{below}: \\ 1\mathrm{mm} = {10}^{-3}\mathrm{m} \\ \mathrm{The} \mathrm{unit} \mathrm{conversion} \mathrm{ratio} \mathrm{for} \mathrm{converting} \mathrm{meters} \mathrm{into} \mathrm{millimeters} \mathrm{is} \mathrm{written} \mathrm{below}: \\ \frac{1\mathrm{mm}}{{10}^{-3}\mathrm{m}} \\ \mathrm{The} \mathrm{unit} \mathrm{analysis} \mathrm{setup} \mathrm{for} \mathrm{this} \mathrm{conversion} \mathrm{can} \mathrm{be} \mathrm{written} \mathrm{as} \mathrm{follows}: \\ 2.56\mu m^3 \times {\left(\frac{{10}^{-6}\mathrm{m}}{1\mathrm{\mu m}}\right)}^3 \times {\left(\frac{1\mathrm{mm}}{{10}^{-3}\mathrm{m}}\right)}^3 = {\text{2.56×10}}^{-9}m{m}^3 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{volume} \mathrm{of} \mathrm{a} \mathrm{certain} \mathrm{bacterial} \mathrm{cell} \mathrm{in} \mathrm{the} \mathrm{units} \mathrm{of} \mathrm{cubic} \mathrm{millimeters} \mathrm{is} 2.56\times {10}^{-9}{\mathrm{mm}}^3 \end{gathered}$$     

Calculate the volume of 10 5 cells in the units of liters .

The unit conversion equation that describes the relation between meters and liters is shown as follows:

$$\begin{gathered} \mathbf{1}\mathbf{ }\mathbf{L}\mathbf{ }\mathbf{=}\mathbf{ }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{m}}^{\mathbf{3}} \\ \mathrm{The} \mathrm{conversion} \mathrm{factor} \mathrm{for} \mathrm{converting} \mathrm{from} \mathrm{meters} \mathrm{to} \mathrm{liters} \mathrm{by} \mathrm{using} \mathrm{the} \mathrm{above} \mathrm{equation} \mathrm{is} \mathrm{shown} \mathrm{below}: \\ \frac{1 \mathrm{L}}{{10}^{-3}{\mathrm{m}}^3} \\ \mathrm{Use} \mathrm{the} \mathrm{above} \mathrm{necessary} \mathrm{conversion} \mathrm{factors} \mathrm{to} \mathrm{calculate} \mathrm{the} \mathrm{volume} \mathrm{of} 105 \mathrm{cells} \mathrm{in} \mathrm{liters} \mathrm{is} \mathrm{written} \mathrm{as} \mathrm{follows}: \\ {10}^5 cells \times \frac{2.56\mu m}{1 cell} \times {\left(\frac{{10}^{-6}m}{1\mu m}\right)}^3 \times \frac{1 L}{{10}^{-3}{m}^3} = 1.72 \times {10}^{-10 }L = {10}^{-10} L \\ \mathrm{Hence}, \mathrm{the} \mathrm{volume} \mathrm{of} 105 \mathrm{cells} \mathrm{in} \mathrm{the} \mathrm{units} \mathrm{of} \mathrm{liters} \mathrm{is} 1.72 \times {10}^{-10 }\mathrm{L} = {10}^{-10} \mathrm{L} . \end{gathered}$$