Question

What is the solubility of KCl at $25{}^\circ C$?

Answer 1

Hint: Solubility of any substance is the ability to get dissolved up to a certain concentration in any solution. The solubility can be measured in terms of molarity, molality and concentrations. Here we have to assume a certain amount of mass of KCl to detect its solubility at $25{}^\circ C$.

Complete answer:
We have been given to determine the solubility of KCl at room temperature which is $25{}^\circ C$. We have not been given the amount to detect the solubility, so this amount or mass of KCl has to be assumed. We are assuming the mass of KCl is taken as 360 g in 1 kg of water at $25{}^\circ C$.
So, we have the mass of solute and solvent, then from the formula of molality we will take out the solubility as molality. Which is, $molality=\dfrac{no.of\,moles\,of\,solute}{Kg\,of\,solvent}$ , so number of moles, will be calculated by the formula, $moles=\dfrac{given\,mass}{molar\,mass}$ , therefore,
Molality, $m=\dfrac{\dfrac{360\,g}{74.55\,g\,mo{{l}^{-1}}}}{1\,Kg\,water}$
Molality, m = 4.829 mol / kg or molal
Also, the solubility can be expressed in percentage, as mass percent,
$\dfrac{360\,g\,KCl}{1000\,g\,water}\times 100$
So, mass % = $0.36\times 100$
Mass % = 36 %
Hence, the solubility of 360 g of KCl at $25{}^\circ C$is 4.829 mol/kg in terms of molality and 36 % in terms of mass percent.

Note:
The conversion for 1 kg is 1 kg = 1000g. Also the solubility can be calculated in terms of molarity as, $molarity=\dfrac{no.of\,moles\,of\,solute}{L\,of\,solvent}$, so, molarity M =$\dfrac{\dfrac{360\,g}{74.55\,g\,mo{{l}^{-1}}}}{1\,Kg\,water\times \dfrac{1000g}{1Kg}\times \dfrac{1L}{997.0749\,g}}$= 4.815 moles/ L. The conversion of gram into liter is 1 L = 997.0749 g. Another way to calculate solubility is through ppm, which is parts per million.