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
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
To calculate the solubility of potassium chloride (KCl) at X°C in moldm-3, we need to determine the number of moles of KCl that dissolved in the solution.
First, we need to find the number of moles of silver trioxonitrate (V) (AgNO3) used in the experiment.
Given:
Volume of AgNO3 solution = 22.5 cm3
Concentration of AgNO3 solution = 34 g/100 cm3
Converting cm3 to dm3:
Volume of AgNO3 solution = 22.5 cm3 × (1 dm3/1000 cm3) = 0.0225 dm3
Converting g to moles for AgNO3:
Molar mass of AgNO3 = 170 g/mol
Number of moles of AgNO3 = (34 g/100 cm3) × (0.0225 dm3) × (1 mol/170 g) = 0.0009118 mol
Since silver trioxonitrate (V) reacts with potassium chloride according to the equation:
3AgNO3 + KCl → AgCl + 3KNO3
From the equation, we can see that 3 moles of AgNO3 react with 1 mole of KCl to form 1 mole of AgCl.
Therefore, the number of moles of KCl dissolved in the solution is equal to the number of moles of AgNO3 used in the experiment. So, the number of moles of KCl = 0.0009118 mol.
Now, we can calculate the solubility of KCl in moldm-3.
Given:
Volume of solution = 30 m3 = 30 dm3
Molar mass of KCl = 74.5 g/mol
Solubility of KCl in moldm-3 = (Number of moles of KCl/Volume of solution in dm3) = (0.0009118 mol/30 dm3) = 0.00003039 moldm-3
Rounding off to the appropriate number of significant figures, the solubility of KCl at X°C is approximately 0.0000304 moldm-3.
Therefore, the correct answer is A. 1.5 moldm-3 (rounded to the nearest significant figure).