25.0g of potassium chloride were dissolved in 80g of distilled water at 300 0 C. Calculate the solubility of the solute in mol dm3 3 . [K =39, Cl = 35.5]

25.0g of potassium chloride were dissolved in 80g of distilled water at 300 ${}^0$C. Calculate the solubility of the solute in mol dm3 ${}^3$. [K =39, Cl = 35.5]

Answer Details

To calculate the solubility of potassium chloride (KCl) in mol dm³, we need to follow these steps:

1. First, determine the molar mass of potassium chloride (KCl).
• Molar mass of K (Potassium) = 39 g/mol
• Molar mass of Cl (Chlorine) = 35.5 g/mol
2. Add these to get the molar mass of KCl:

Molar mass of KCl = 39 + 35.5 = 74.5 g/mol



3. Calculate moles of KCl:
• We have 25.0 g of KCl dissolved.

Moles of KCl = Mass of KCl / Molar mass of KCl = 25.0 g / 74.5 g/mol = 0.3356 mol



4. Next, convert the mass of water to liters.
• 80 g of distilled water corresponds approximately to 80 ml, because the density of water is roughly 1 g/ml.

Convert ml to liters: 80 ml = 0.080 L



5. Calculate the concentration (solubility) in mol dm³ (which is equivalent to mol/L):

Concentration = Moles of solute / Volume of solvent in liters = 0.3356 mol / 0.080 L = 4.195 mol/dm³

The solubility of potassium chloride at 30°C in mol/dm³ is therefore approximately 4.2 mol/dm³.