Chapter 5, Problem 5F.15E

(a)

Interpretation Introduction

Interpretation:

The osmotic pressure of $\text{0}\text{.050}\text{M}{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}{}_{\text{(aq)}}$ at ${\text{20}}^{\text{°}}\text{C}$ has to be calculated.

Concept Introduction:

The pressure that is needed to stop the flow of solvent is known as Osmotic pressure.

The osmotic pressure of dilute solution is related to the molar concentration of the solute in the solution.  The osmotic pressure can be calculated by using the following equation:

  $\begin{array}{l}\text{Π}\text{=}\text{i}\text{R}\text{T}\text{c}\\ \text{where,}\\ \text{Π}\text{=}\text{Osmotic}\text{pressure}\\ \text{i}\text{=}\text{i}\text{factor}\\ \text{R}\text{=}\text{Gas}\text{constant}\\ \text{c}\text{=}\text{molar}\text{concentration}\\ \text{T}\text{=}\text{Temperature}\end{array}$

Explanation of Solution

Given, $\text{0}\text{.050}\text{M}{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}{}_{\text{(aq)}}$ at ${\text{20}}^{\text{°}}\text{C}$.

The osmotic pressure can be calculated as follows:

  $\begin{array}{l}\text{Given,}\text{i}\text{=}\text{1,}\text{as}{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\text{does}\text{not}\text{dissociate}\text{into}\text{ions}\\ \text{R}\text{=}\text{0}\text{.0821}\text{L}\text{atm}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \text{T}\text{=}{\text{20}}^{\text{°}}\text{C}\text{+}\text{273}\text{=}\text{293}\text{K}\\ \\ \text{Π}\text{=}\text{i}\text{R}\text{T}\text{c}\\ \text{Π}\text{=}\text{1}\text{×}\text{0}\text{.0821}\text{L}\text{atm}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{×}\text{293}\text{K}\text{×}\text{0}\text{.050}\text{M}\\ \text{=}\text{1}\text{.203}\text{atm}\end{array}$

The osmotic pressure of $\text{0}\text{.050}\text{M}{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}{}_{\text{(aq)}}$ is $\text{1}\text{.203}\text{atm}$.

(b)

Interpretation Introduction

Interpretation:

The osmotic pressure of $\text{0}\text{.0010}\text{M}{\text{NaCl}}_{\text{(aq)}}$ at ${\text{20}}^{\text{°}}\text{C}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given, $\text{0}\text{.0010}\text{M}{\text{NaCl}}_{\text{(aq)}}$ at ${\text{20}}^{\text{°}}\text{C}$.

The osmotic pressure can be calculated as follows:

  $\begin{array}{l}\text{NaCl}\stackrel{}{\to }{\text{Na}}^{\text{+}}\text{+}{\text{Cl}}^{\text{-}}\\ \text{Given,}\text{i}\text{=}\text{2}\\ \text{R}\text{=}\text{0}\text{.0821}\text{L}\text{atm}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \text{T}\text{=}{\text{20}}^{\text{°}}\text{C}\text{+}\text{273}\text{=}\text{293}\text{K}\\ \\ \text{Π}\text{=}\text{i}\text{R}\text{T}\text{c}\\ \text{Π}\text{=}\text{2}\text{×}\text{0}\text{.0821}\text{L}\text{atm}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{×}\text{293}\text{K}\text{×}\text{0}\text{.0010}\text{M}\\ \text{=}\text{0}\text{.0481}\text{atm}\end{array}$

The osmotic pressure of $\text{0}\text{.0010}\text{M}{\text{NaCl}}_{\text{(aq)}}$ is $\text{0}\text{.0481}\text{atm}$.

(c)

Interpretation Introduction

Interpretation:

The osmotic pressure of a saturated aqueous solution of $\text{AgCN}$ of solubility $\text{23}\text{μg/100}\text{g}$ of water at ${\text{20}}^{\text{°}}\text{C}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given, aqueous solution of $\text{AgCN}$ of solubility $\text{23}\text{μg/100}\text{g}$ of water at ${\text{20}}^{\text{°}}\text{C}$.

The molarity of the AgCN has to be calculated to calculate the osmotic pressure of the solution.

  $\begin{array}{l}\text{Molarity}\text{of}\text{AgCN}\text{can}\text{be}\text{calculated}\text{as}\\ \\ \text{Moles}\text{of}\text{AgCN}\text{=}\frac{\text{mass}}{\text{molar}\text{mass}}\\ \text{=}\frac{\text{23}\text{μg}}{\text{133}\text{.89}\text{g/mol}}\text{×}\frac{{\text{10}}^{\text{-6}}\text{g}}{\text{1}\text{μg}}\\ \text{=}\text{1}\text{.72}\text{×}{\text{10}}^{\text{-7}}\text{mol}\\ \text{Volume}\text{of}\text{water}\text{=}\text{100}\text{g}\text{×}\frac{\text{1}\text{kg}}{\text{1000}\text{g}}\text{×}\frac{\text{1}\text{L}}{\text{1}\text{kg}}\\ \text{=}\text{0}\text{.1}\text{L}\\ \\ \text{Molarity}\text{of}\text{AgCN}\text{=}\frac{\text{moles}}{\text{Volume}\left(\text{in}\text{L}\right)}\\ \text{=}\frac{\text{1}\text{.72}\text{×}{\text{10}}^{\text{-7}}\text{mol}}{\text{0}\text{.1}\text{L}}\text{=}\text{1}\text{.72}\text{×}{\text{10}}^{\text{-6}}\text{mol}\end{array}$

The osmotic pressure can be calculated as follows:

  $\begin{array}{l}\text{AgCN}\stackrel{}{\to }{\text{Ag}}^{\text{+}}\text{+}{\text{CN}}^{\text{-}}\\ \text{Given,}\text{i}\text{=}\text{2}\\ \text{R}\text{=}\text{0}\text{.0821}\text{L}\text{atm}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \text{T}\text{=}{\text{20}}^{\text{°}}\text{C}\text{+}\text{273}\text{=}\text{293}\text{K}\\ \\ \text{Π}\text{=}\text{i}\text{R}\text{T}\text{c}\\ \text{Π}\text{=}\text{2}\text{×}\text{0}\text{.0821}\text{L}\text{atm}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{×}\text{293}\text{K}\text{×}\left(\text{1}\text{.72}\text{×}{\text{10}}^{\text{-6}}\text{M}\right)\\ \text{=}\text{8}\text{.27}\text{×}{\text{10}}^{\text{-5}}\text{atm}\end{array}$

The osmotic pressure of aqueous solution of $\text{AgCN}$ of solubility $\text{23}\text{μg/100}\text{g}$ of water is $\text{8}\text{.27}\text{×}{\text{10}}^{\text{-5}}\text{atm}$.