Question

Question: Silver dissolves in molten lead. Compute the osmotic pressure of a $0.010M$solution of silver in lead at $423°C$. Compute the height of a column of molten lead $(p=11.4gc{m}^{-3})$ to which this pressure corresponds.

Short answer

Osmotic pressure is $0.57atm$ and the height of the column is $0.52m$.

Step-by-step solution

Given data

The density of the molten lead is $p=11.4gc{m}^{-3}$.

Concept of Osmotic pressure

Osmotic pressure is the minimum pressure which needs to be applied to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane.

Formula Used:

$\mathrm{\pi }=\mathrm{icRRl}$

Calculation of the osmotic pressure

The osmotic pressure can be calculated as follows:

$\mathrm{\pi }=\mathrm{cRT}$

Since, the absolute temperature is given as:

$$\begin{gathered} T=423+273 \\ T=696K \end{gathered}$$

Substitute the values in the equation of the osmotic pressure.

$$\begin{gathered} \mathrm{\pi }=0.010\frac{\mathrm{mol}}{\mathrm{L}}\times 0.08206{\mathrm{Latmmol}}^{-1}{\mathrm{K}}^{-1}\times 696\mathrm{K} \\ \mathrm{\pi }=0.57\mathrm{atm} \end{gathered}$$

Calculation of the height of the column

The height of a column can be calculated as:

$\mathrm{\pi }=\mathrm{hpg}$

Convert the density in

$$\begin{gathered} \frac{g}{c{m}^3}to\frac{kg}{m^3} \\ 11.4\frac{g}{c{m}^3}=11.4\frac{g}{c{m}^3}\times \frac{kg}{1000g}\times \frac{{10}^{6}c{m}^3}{m^3} \\ 11.4\frac{g}{c{m}^3}=1.14\times {10}^4\frac{kg}{m^3} \end{gathered}$$

Substitute the values in the above equation of the osmotic pressure.

$0.57atm=h\times 1.14\times {10}^4\frac{kg}{m^3}\times 9.80665\frac{m}{s^2}$

Put the numbers on the right-hand side and the height on the left side.

$$\begin{gathered} h=\frac{0.57atm}{1.14\times {10}^4{\displaystyle \frac{km}{m^3}}\times 9.80665{\displaystyle \frac{m}{s^2}}} \\ h=\frac{0.57atm{\displaystyle \frac{1.01325\times {10}^{5}Pa}{1atm}}}{1.14\times {10}^4{\displaystyle \frac{kg}{m^3}}\times 9.80665{\displaystyle \frac{m}{s^2}}} \\ h=0.52Pa\frac{kg}{m{s}^2}m \\ h=0.52m \end{gathered}$$

Thus, osmotic pressure is $0.57atm$ and the height of the column is $0.52m$.