Q135CP - Chemistry: Molecular Nature Of Matter And Change

Question

: A biochemical engineer isolates a bacterial gene fragment and dissolves a $\mathbf{\text{10.0 mg}}$ sample in enough water to make $\mathbf{\text{30.0 mL}}$ of solution. The osmotic pressure of the solution is $\mathbf{\text{0.340 torr}}$ at $\mathbf{\text{25°C}}$.

(a) What is the molar mass of the gene fragment?

(b) If the solution density is $\mathbf{\text{0.997 }}\raisebox{1ex}{${\displaystyle \mathbf{\text{g}}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{mL}}$}\right.$, how large is the freezing point depression for this solution (${\mathbf{\text{K}}}_{\mathbf{f}}$ of water is  ${\mathbf{\text{1.86 }}}^{\mathbf{\text{o}}}\raisebox{1ex}{${\displaystyle \mathbf{\text{C}}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{m}}$}\right.$)?

Step-by-Step Solution

Answer

(a) Molar mass of gene fragment is $1.82\times {10}^{-4}\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$.

(b) Freezing point depression for this solution is $(3.40\times {10}^{-5})°\text{C}$.

A concept:

Freezing point depression is the decrease of the freezing point of a solvent due to the addition of a solute into the solvent.

Molality can be calculated by dividing moles of solute by mass of solvent.

Mass of a substance can be calculated by dividing mole of substance by mass of substance.

Consider the given data as below.

The mass of sample is $10\text{ mg}=\text{0.01 g}$.

Volume of water is $\text{30 mL}=\text{0.03 L}$.

Osmotic pressure of solution is,

$\begin{array}{rcl}\Pi & =& \text{0.340 torr}\\ & =& 0.340\text{ torr}\times \frac{\text{1 atm}}{\text{760 torr}}\\ & =& 4.47\times {10}^{-4}\text{ atm}\end{array}$  

 

(a) Molar mass of gene fragment:

Now, it can calculate the molarity of solution by using following formula.

$M=\frac{\Pi }{RT}$                                                                                                              ….. (1)

Here, $M$ is molarity, $\Pi$ is osmotic pressure,  $R$ is gas constant and $T$ is temperature.

 

The gas constant, $R=0.082\raisebox{1ex}{${\displaystyle \text{atm}\cdot \text{L}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}\cdot \text{K}$}\right.$

Temperature is,

$\begin{array}{rcl}T& =& {25}^{\text{o}}\text{C}\\ & =& (273+25)\text{ K}\\ & =& 298\text{ K}\end{array}$

Substitute the above values into equation (1).

$\begin{array}{rcl}M& =& \frac{4.47\times {10}^{-4}\text{ atm}}{0.082{\displaystyle \raisebox{1ex}{$\text{atm}\cdot \text{L}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}\cdot \text{K}$}\right.}\times 298\text{ K}}\\ & =& 1.83\times {10}^{-5}\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.\end{array}$ 

 

Now, it can calculate moles of sample or bacterial gene fragment by using the following formula of molarity.

 $M=\frac{\text{moles\hspace{0.17em}of\hspace{0.17em}sample}}{\text{volume\hspace{0.17em}of\hspace{0.17em}solution }\left(\text{L}\right)}$                                                                                 ….. (2)

 

Rearrange the above equation for the moles of sample.

$\begin{array}{rcl}\text{moles\hspace{0.17em}of\hspace{0.17em}sample}& =& M\times \text{volume\hspace{0.17em}of\hspace{0.17em}solution }\left(\text{L}\right)\\ & =& 1.83\times {10}^{-5}\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.\times 0.03\text{ L}\\ & =& 5.49\times {10}^{-7}\text{ mol}\end{array}$ 

 

Now, it has the moles of sample and mass of sample, so you can calculate the molar mass of sample.

$\begin{array}{rcl}\text{molar\hspace{0.17em}mass\hspace{0.17em}of\hspace{0.17em}sample}& =& \frac{\text{mass\hspace{0.17em}of\hspace{0.17em}sample}}{\text{moles\hspace{0.17em}of\hspace{0.17em}sample}}\\ & =& \frac{0.01\text{ g}}{5.49\times {10}^{-7}\text{ mol}}\\ & =& 1.82\times {10}^{-4}\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\end{array}$ 

 

Hence, the molar mass of gene fragment is $1.82\times {10}^{-4}\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$.

 

(b) Freezing point depression:

The following formula can be used to determine the freezing point depression;

$\Delta T_b=K_b\times m$                                                                                                           ….. (3)

Where,  $m$ is molality and $K_b$ is the constant having a value $1.83\times {10}^{-5}\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.$.

 

Determine the mass of solvent as below.

$\begin{array}{rcl}\text{mass\hspace{0.17em}of\hspace{0.17em}solvent}& =& \text{density}\times \text{volume\hspace{0.17em}of\hspace{0.17em}solvent}\\ & =& 0.997\text{ g}/\text{mL}\times 30\text{ L}\\ & =& 29.91\text{ g}\end{array}$

Define the molality as below.

$\begin{array}{rcl}m& =& \frac{\text{moles\hspace{0.17em}of\hspace{0.17em}sample}}{\text{mass\hspace{0.17em}of\hspace{0.17em}water}}\\ & =& \frac{5.49\times {10}^{-7}\text{ mol}}{29.91\times {10}^{-3}\text{ kg}}\\ & =& 1.83\times {10}^{-5}\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.\end{array}$ 

 

Therefore, freezing point depression is defined by substituting known values into equation (3).

$\begin{array}{rcl}\Delta T_b& =& 1.86\raisebox{1ex}{${\displaystyle \text{°C}}$}\!\left/ \!\raisebox{-1ex}{$\text{m}$}\right.\times 1.83\times {10}^{-5}\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.\\ & =& (3.40\times {10}^{-5})\text{°C}\end{array}$ 

 

Hence, the freezing point depression for this solution is $(3.40\times {10}^{-5})°\text{C}$.