Collection of Solved Problems in Physics

Density of Blood

Task number: 1757

To determine the density of blood, drops of blood were added to a mixture of xylene with a density of \(0{.}867 \mathrm{\frac{g}{cm^3}}\)and bromobenzene with a density of \(1{.}497 \mathrm{\frac{g}{cm^3}}\). The ratio of xylene and bromobenzene was being changed until the drops were floating, i. e. they did not rise nor sink. This happened when the mixture contained 72 % of xylene and 28 % of bromobenzene. What was the density of blood?

• Given values

  \(\rho_{x}={0.867 \mathrm{\frac{g}{cm^3}}}\)	density of xylene
  \(\rho_{b}=1{.}497 \mathrm{\frac{g}{cm^3}}\)	density of bromobenzene
  \(V\)	volume of the mixture
  \(V_{x}=0{.}72 \mathrm{V}\)	volume of xylene
  \(V_{b}=0{.}28 \mathrm{V}\)	volume of bromobenzene
  \(\rho_{blood}= ? \mathrm{\frac{g}{cm^3}}\)	density of blood

• Analysis

  What does it mean that the drops of blood are floating, and what holds for their density in such a situation? After answering these questions, you can determine the density of the mixture.

• Hint 1

  First remind yourself, how to define the density of a substance.

• Solution of Hint1

  The density \(\rho\) of a substance is defined as the quotient of mass \(m\) and volume \(V\) of the substance: \[\rho=\frac{m}{V}.\tag{1}\]

• Hint 2

  You know that the blood drops are floating. What applies to the forces acting on the drops in this situation? What can you say about the density of the blood drops?

• Solution of Hint2

  Since the blood drops do not move up or down, the force of gravity \(F_{\mathrm{g}}\) must be equal to the buoyancy force \(F_{\mathrm{b}}\): \[F_{\mathrm{g}}=F_{\mathrm{b}}.\] Therefore: \[\rho_{\mathrm{blood}}V_{\mathrm{blood}}g=\rho V_{\mathrm{blood}}g,\] where \(V_{\mathrm{blood}}\) is the volume of a blood drop, \(g\) is the acceleration of gravity.

  By dividing both sides of the equation by \(V_{\mathrm{blood}}g\), we obtain: \[\rho_{\mathrm{blood}}=\rho.\]

  The density of the blood is the same as the density of the mixture.

• Hint 3

  Choose a volume of the mixture and determine the volume of xylene and bromobenzene.

• Solution of Hint 3

  For example we choose a volume of \(V=100 \mathrm{cm^3}.\) Then the volume of xylene is \(V_{x}=0.72{\cdot}100 \mathrm{cm^3}=72 \mathrm{cm^3}\) and the volume of bromobenzene is \(V_{b}=0{.}28{\cdot}100 \mathrm{cm^3}=28 \mathrm{cm^3}\).

• Hint 4

  What is the mass of xylene and bromobenzene (corresponding to the selected volume of the mixture)?

• Solution of Hint 4

  From (1) we evaluate a relation we use for calculating the mass: \[m=\rho{\cdot}V.\]

  Thus, the mass of xylene \(m_{\mathrm{{x}}}\) is: \[m_{\mathrm{x}}=\rho_{\mathrm{x}}V_{\mathrm{x}}.\] For given values: \[m_{\mathrm{x}}=0{.}867{\cdot}72 \mathrm{\frac{g}{cm^3}}\cdot\mathrm{cm^3}=62{.}424 \mathrm{g}.\]

  The mass of bromobenzene \(m_{\mathrm{b}}\) is: \[m_{\mathrm{b}}=\rho_{\mathrm{b}}V_{\mathrm{b}}.\] For given values: \[m_{\mathrm{b}}=1{.}497{\cdot}28 \mathrm{\frac{g}{cm^3}}\cdot\mathrm{cm^3}=41{.}916 \mathrm{g}.\]

• Hint 5

  What is the density of the mixture (corresponding to the selected volume of the mixture)?

• Solution of Hint 5

  We need to know the total mass of the mixture \(m\), for which the following applies \[m=m_{\mathrm{x}}+m_{\mathrm{b}}.\] For given values: \[m=62{.}424 \mathrm{g}+41{.}916 \mathrm{g}=104{.}34 \mathrm{g}.\]

  We have chosen the volume of the mixture to be: \(V=100 \mathrm{cm^3}.\)

  By substituting the mass and volume of the mixture into the equation (1) we obtain the density of the mixture \(\rho\): \[\rho=\frac{104{.}34}{100} \mathrm{\frac{g}{cm^3}}=1{.}0434 \mathrm{\frac{g}{cm^3}} \dot= 1{.}04 \mathrm{\frac{g}{cm^3}}.\]

  The density of blood is the same as the density of the mixture, that is: \(\rho_{\mathrm{blood}} \dot= 1{.}04 \mathrm{\frac{g}{cm^3}}.\)

• Overall solution

  The definition of density

  The density of a substance \(\rho\) is defined as mass \(m\) over volume \(V\): \[\rho=\frac{m}{V}.\tag{1}\]

   

  The density of blood

  Since the blood drops do not sink nor rise in the mixture, the force of gravity \(F_{\mathrm{g}}\) must be equal to the buoyancy force \(F_{\mathrm{b}}\): \[F_{\mathrm{g}}=F_{\mathrm{b}}.\] Thus: \[\rho_{\mathrm{blood}}V_{\mathrm{blood}}g=\rho V_{\mathrm{blood}}g,\] where \(V_{\mathrm{blood}}\) is the volume of the blood drops, and g is the acceleration of gravity.

  By dividing both sides of the equation by \(V_{\mathrm{blood}}g\) we obtain: \[\rho_{\mathrm{blood}}=\rho.\]

  The density of blood is the same as the density of the mixture.

  The volume of xylene and bromobenzene

  We choose the volume of the mixture to be \(V=100 \mathrm{cm^3}.\) Then the volume of xylene is \(V_{\mathrm{x}}=0{.}72{\cdot}100 \mathrm{cm^3}=72 \mathrm{cm^3}\) and the volume of bromobenzene is \(V_{\mathrm{b}}=0{.}28{\cdot}100 \mathrm{cm^3}=28 \mathrm{cm^3}\).

   

  The mass of xylene and bromobenzene

  From (1) we express a relation for calculating the mass: \[m=\rho{\cdot}V.\]

  Thus, the mass of xylene \(m_{\mathrm{x}}\) is: \[m_{\mathrm{x}}=\rho_{\mathrm{x}}V_{\mathrm{x}}.\] For given values: \[m_{\mathrm{x}}=0{.}867{\cdot}72 \mathrm{\frac{g}{cm^3}}\cdot\mathrm{cm^3}=62{.}424 \mathrm{g}.\]

  The mass of bromobenzene \(m_{\mathrm{b}}\) is: \[m_{\mathrm{b}}=\rho_{\mathrm{b}}V_{\mathrm{b}}.\] For given values: \[m_{\mathrm{b}}=1{.}497{\cdot}28 \mathrm{\frac{g}{cm^3}}\cdot\mathrm{cm^3}=41{.}916 \mathrm{g}.\]

   

  The density of the mixture

  We need to know the total mass of the mixture \(m\), for which the following applies: \[m=m_{x}+m_{b}.\] For given values: \[m=62{.}424 \mathrm{g}+41{.}916 \mathrm{g}=104{.}34 \mathrm{g}.\]

  We have chosen the volume of the mixture to be: \(V=100 \mathrm{cm^3}.\)

  By substituting the mass and volume of the mixture into (1),we obtain its density \(\rho\): \[\rho=\frac{104{.}34}{100} \mathrm{\frac{g}{cm^3}}=1{.}0434  \mathrm{\frac{g}{cm^3}} \dot= 1{,}04 \mathrm{\frac{g}{cm^3}}.\]

  The density of blood is the same as the density of the mixture: \(\rho_{\mathrm{blood}} \dot= 1{.}04 \mathrm{\frac{g}{cm^3}}.\)

• Answer

  The density of blood is the same as the density of the mixture: \(\rho_{\mathrm{blood}} \dot= 1{.}04 \mathrm{\frac{g}{cm^3}}.\)