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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 $10\text{mg}=\text{0.01g}$.

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

Osmotic pressure of solution is,

$\begin{array}{rcl}\mathrm{Π}& =& \text{0.340torr}\\ & =& \text{ }0.340\text{torr}×\frac{\text{1atm}}{\text{760torr}}\\ & =& 4.47×{10}^{−4}\text{atm}\end{array}$

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

$M=\frac{\mathrm{Π}}{RT}$ ….. (1)

Here, $M$is molarity, $\mathrm{Î }$is osmotic pressure, $R$is gas constant and $T$is temperature.

The gas constant,$R=0.082\raisebox{1ex}{${\displaystyle \text{atm}â‹\dots \text{L}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}â‹\dots \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).

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

$M=\frac{\text{³¾´Ç±ô±ð²õ o´Ú s²¹³¾±è±ô±ð}}{\text{±¹´Ç±ô³Ü³¾±ð o´Ú s´Ç±ô³Ü³Ù¾±´Ç²Ô}\left(\text{L}\right)}$ ….. (2)

Rearrange the above equation for the moles of sample.

$\begin{array}{rcl}\text{³¾´Ç±ô±ð²õ o´Ú s²¹³¾±è±ô±ð}& =& M×\text{±¹´Ç±ô³Ü³¾±ð o´Ú s´Ç±ô³Ü³Ù¾±´Ç²Ô}\left(\text{L}\right)\\ & =& 1.83×{10}^{−5}\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.×0.03\text{L}\\ & =& 5.49×{10}^{−7}\text{mol}\end{array}$

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

$\begin{array}{rcl}\text{³¾´Ç±ô²¹°ù​â¶Ä³¾²¹²õ²õ o´Ú s²¹³¾±è±ô±ð}& =& \frac{\text{³¾²¹²õ²õ​â¶Ä‰o´Ú s²¹³¾±è±ô±ð}}{\text{³¾´Ç±ô±ð²õ o´Ú s²¹³¾±è±ô±ð}}\\ & =& \frac{0.01\text{g}}{5.49×{10}^{−7}\text{mol}}\\ & =& 1.82\text{ }×{10}^{−4}\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\end{array}$

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

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

$\mathrm{Δ}{T}_b=K_b×m$ ….. (3)

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

Determine the mass of solvent as below.

$\begin{array}{rcl}\text{³¾²¹²õ²õ o´Ú s´Ç±ô±¹±ð²Ô³Ù}& =& \text{density}×\text{±¹´Ç±ô³Ü³¾±ð o´Ú s´Ç±ô±¹±ð²Ô³Ù}\\ & =& 0.997\text{g}/\text{mL}×30\text{L}\\ & =& 29.91\text{g}\end{array}$

Define the molality as below.

$\begin{array}{rcl}m& =& \frac{\text{³¾´Ç±ô±ð²õ â¶Ä‹o´Ú s²¹³¾±è±ô±ð}}{\text{³¾²¹²õ²õ â¶Ä‹o´Ú w²¹³Ù±ð°ù}}\\ & =& \frac{5.49×{10}^{−7}\text{mol}}{29.91\text{ }×{10}^{−3}\text{kg}}\\ & =& 1.83×{10}^{−5}\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.\end{array}$

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

$\begin{array}{rcl}\mathrm{Δ}{T}_b& =& 1.86\raisebox{1ex}{${\displaystyle \text{°°ä}}$}\!\left/ \!\raisebox{-1ex}{$\text{m}$}\right.×1.83×{10}^{−5}\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.\\ & =& (3.40×{10}^{−5})\text{°°ä}\end{array}$

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