To calculate the solubility of potassium chloride at X°C, we first need to find the number of moles of silver trioxonitrate (V) present in the 22.5cm3 solution.
Number of moles of AgNO3 = Mass / Molar mass
Number of moles of AgNO3 = 34g / 170 g/mol
Number of moles of AgNO3 = 0.2 mol
According to the balanced chemical equation, 1 mole of AgNO3 reacts with 1 mole of KCl to form 1 mole of AgCl. Therefore, the number of moles of AgCl that can be formed in the reaction is also 0.2 mol.
Using the volume of the potassium chloride solution (30m3) and the number of moles of AgCl (0.2 mol), we can calculate the solubility of potassium chloride:
Solubility of KCl = Number of moles of KCl / Volume
Solubility of KCl = 0.2 mol / 30m3
Converting m3 to dm3:
Solubility of KCl = 0.2 mol / (30 * 1000 dm3)
Solubility of KCl = 0.2 mol / 30000 dm3
Solubility of KCl = 1/150 mol/dm3
Therefore, the solubility of potassium chloride at X°C is 1/150 moldm-3.
The correct answer is C. 0.15 moldm-3
A saturated solution of
potassium chloride was prepared
at X˚C. If 30m3
of this solution
required 22.5cm3
of silver
trioxonitrate (V) solution,
containing 34g in 100cm3
solution for complete precipitation.
Calculate the solubility of
potassium chloride at X°C in
moldm-1 (take KCl=74.5,
AgNO3=170)
A. 1.5 moldm-3
B. 2.0 moldm-3
C. 0.15 moldm-3
D. 0.20 moldm-3
1 answer