
What is the molality, molarity, and mole fraction of $FeC{{l}_{3}}$ in a 22.2 mass % aqueous solution?
(d = 1.280 g/ mL)
Answer
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Hint: 22.2 mass % aqueous solution of ferric chloride means there 100 gram of solution in which 22.2 grams of ferric chloride is present and 77.8 grams of water is present. Calculate the moles of ferric chloride by dividing the given mass of ferric chloride by 162.2 grams.
Complete answer:
First, we have to find the molality of the solution and the molality of the solution is the ratio of moles of solute (ferric chloride) and the mass of the solvent (water) in kilograms.
22.2 mass % aqueous solution of ferric chloride means there 100 grams of solution in which 22.2 grams of ferric chloride is present and 77.8 grams of water is present.
So, the moles of ferric chloride will be:
$Moles=\dfrac{22.2}{162.2}=0.137$
The mass of the solvent is 77.8 grams, in kilograms, it will be 0.0778 Kg.
Now, we can calculate the molality as:
$Molality=\dfrac{0.137}{0.0778}=1.76\text{ }m$
So, the molality is 1.76 m.
Now, we have to find the molarity of the solution and the molarity of the solution is the ratio of moles of solute (ferric chloride) and the volume of the solution in liters.
We are given the density of the solution as 1.280 g/ mL so, we can calculate the volume as:
$Volume=\dfrac{100}{1.280\text{ x 1000}}=0.0783\text{ }L$
Now, we can calculate the molarity as:
$Molarity=\dfrac{0.137}{0.0781}=1.75\text{ }M$
So, the molarity of the solution is 1.75 M
Now, we have to calculate the mole fraction of solution, and mole fraction is the ratio of moles of solute (ferric chloride) and the total moles of the solution (ferric chloride + water).
The molecular mass of water is 18.015grams/ mol and the given mass is 77.8 grams. The moles of water will be:
${{H}_{2}}O\text{ }moles=\dfrac{77.8}{18.015}=4.32$
Now, we can calculate the mole fraction as:
$Mole\text{ }fraction=\dfrac{0.137}{0.137\text{ }+\text{ }4.32}=0.0307$
So, the mole fraction of the solution is 0.0307.
Note:
The volume of the solution is calculated by using the formula: volume is the ratio of mass and the density of the solution. Since the density was given in mL, it was multiplied by 1000 to convert into liters.
Complete answer:
First, we have to find the molality of the solution and the molality of the solution is the ratio of moles of solute (ferric chloride) and the mass of the solvent (water) in kilograms.
22.2 mass % aqueous solution of ferric chloride means there 100 grams of solution in which 22.2 grams of ferric chloride is present and 77.8 grams of water is present.
So, the moles of ferric chloride will be:
$Moles=\dfrac{22.2}{162.2}=0.137$
The mass of the solvent is 77.8 grams, in kilograms, it will be 0.0778 Kg.
Now, we can calculate the molality as:
$Molality=\dfrac{0.137}{0.0778}=1.76\text{ }m$
So, the molality is 1.76 m.
Now, we have to find the molarity of the solution and the molarity of the solution is the ratio of moles of solute (ferric chloride) and the volume of the solution in liters.
We are given the density of the solution as 1.280 g/ mL so, we can calculate the volume as:
$Volume=\dfrac{100}{1.280\text{ x 1000}}=0.0783\text{ }L$
Now, we can calculate the molarity as:
$Molarity=\dfrac{0.137}{0.0781}=1.75\text{ }M$
So, the molarity of the solution is 1.75 M
Now, we have to calculate the mole fraction of solution, and mole fraction is the ratio of moles of solute (ferric chloride) and the total moles of the solution (ferric chloride + water).
The molecular mass of water is 18.015grams/ mol and the given mass is 77.8 grams. The moles of water will be:
${{H}_{2}}O\text{ }moles=\dfrac{77.8}{18.015}=4.32$
Now, we can calculate the mole fraction as:
$Mole\text{ }fraction=\dfrac{0.137}{0.137\text{ }+\text{ }4.32}=0.0307$
So, the mole fraction of the solution is 0.0307.
Note:
The volume of the solution is calculated by using the formula: volume is the ratio of mass and the density of the solution. Since the density was given in mL, it was multiplied by 1000 to convert into liters.
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