Question
Derive van’t Hoff general solution equation.
✓
Answer
$1.$ According to van't Hoff$-$Boyle's law, osmotic pressure of a dilute solution is inversely proportional to the volume containing $1 \ mole$ of solute at constant temperature and according to van't Hoff$-$Charles' law, osmotic pressure of a dilute solution is directly proportional to the absolute temperature, at constant concentration.
$2.$ If $\pi $is the osmotic pressure, $V$ is the volume of the solution and $T$ is the absolute temperature, then
$\pi \propto \frac{1}{V} \ldots(1) \ldots[$ van't Hoff$-$Boyle's law at constant temperature$]$
$\therefore \pi V=$ constant
$\pi \propto T \quad.....(2) ... [$van't Hoff$-$Charles' law at constant concentration$]$
$\therefore \frac{\pi}{T}=$ constant
$3.$ Combining $(1)$ and $(2)$ we get,
$\pi \propto \frac{ T }{ V }$
$\therefore \pi=$ Constant $\times \frac{T}{V}$
$\therefore \pi V=R \ ^{\prime} T$, where $R \ ^{\prime}$ is a constant.
$4.$ This equation is parallel to the ideal gas equation $PV = RT ( n = 1 )$
Since, the calculated value of $R \ '$ is almost same as $R,$ the equation can be written as $\pi V = RT ($ for $1 mole$ of solute $)$
$5.$ This equation was derived for $1 \ mole$ of solute dissolved in $V dm^3.$ If n moles of solute are dissolved in $V \ dm^3$ of solution, the equation becomes
$\pi V=n R T$
$\therefore \pi=\frac{ nRT }{ V }$
$6.$ $c=\frac{n}{V}$
$\therefore \pi= CRT$
where,
$\pi = $osmotic pressure,
$C =$ concentration of solution in moles/litre
$R =$ gas constant $= 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$ or $8.314 \ J \ mol^{-1} \ K^{-1}$
$T =$ absolute temperature
$n = $number of moles of solute,
$V =$ volume of the solution.
$2.$ If $\pi $is the osmotic pressure, $V$ is the volume of the solution and $T$ is the absolute temperature, then
$\pi \propto \frac{1}{V} \ldots(1) \ldots[$ van't Hoff$-$Boyle's law at constant temperature$]$
$\therefore \pi V=$ constant
$\pi \propto T \quad.....(2) ... [$van't Hoff$-$Charles' law at constant concentration$]$
$\therefore \frac{\pi}{T}=$ constant
$3.$ Combining $(1)$ and $(2)$ we get,
$\pi \propto \frac{ T }{ V }$
$\therefore \pi=$ Constant $\times \frac{T}{V}$
$\therefore \pi V=R \ ^{\prime} T$, where $R \ ^{\prime}$ is a constant.
$4.$ This equation is parallel to the ideal gas equation $PV = RT ( n = 1 )$
Since, the calculated value of $R \ '$ is almost same as $R,$ the equation can be written as $\pi V = RT ($ for $1 mole$ of solute $)$
$5.$ This equation was derived for $1 \ mole$ of solute dissolved in $V dm^3.$ If n moles of solute are dissolved in $V \ dm^3$ of solution, the equation becomes
$\pi V=n R T$
$\therefore \pi=\frac{ nRT }{ V }$
$6.$ $c=\frac{n}{V}$
$\therefore \pi= CRT$
where,
$\pi = $osmotic pressure,
$C =$ concentration of solution in moles/litre
$R =$ gas constant $= 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$ or $8.314 \ J \ mol^{-1} \ K^{-1}$
$T =$ absolute temperature
$n = $number of moles of solute,
$V =$ volume of the solution.
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