An aqueous solution is 22.0% by mass silver nitrate, AgNO3, and has a density of 1.22 g/mL. what is the molality of silver nitrate in the solution?
1.58
An aqueous solution of potassium sulfide has a concentration of 0.405 molal. The percent by mass of potassium sulfide in the solution is?
To find the molality of silver nitrate in the solution, we need to determine the number of moles of solute (AgNO3) per kilogram of solvent (water).
Here are the steps to calculate the molality:
Step 1: Calculate the mass of the solution.
Given that the density of the solution is 1.22 g/mL, we can assume that 1 mL of the solution weighs 1.22 grams.
To find the mass of the solution, we multiply the volume by the density. Since we don't know the volume, we can assume 100 grams of the solution (as mass and volume are numerically equivalent for water-based solutions).
Mass of solution = 100 g
Step 2: Calculate the mass of AgNO3 in the solution.
Given that the solution is 22.0% by mass AgNO3, we can calculate the mass of AgNO3 in 100 grams of the solution.
Mass of AgNO3 = 22.0% of 100 g
Mass of AgNO3 = (22.0/100) * 100 g
Mass of AgNO3 = 22 g
Step 3: Calculate the number of moles of AgNO3.
To calculate the number of moles of AgNO3, we need to use its molar mass, which is the sum of the atomic masses of silver (Ag), nitrogen (N), and oxygen (O).
Molar mass of AgNO3 = (1 * atomic mass of Ag) + (1 * atomic mass of N) + (3 * atomic mass of O)
Molar mass of AgNO3 = (1 * 107.87 g/mol) + (1 * 14.01 g/mol) + (3 * 16.00 g/mol)
Molar mass of AgNO3 = 169.87 g/mol
Number of moles of AgNO3 = Mass of AgNO3 / Molar mass of AgNO3
Number of moles of AgNO3 = 22 g / 169.87 g/mol
Step 4: Convert the mass of water to kilograms.
Since molality is defined as moles of solute per kilogram of solvent, we need to convert the mass of water to kilograms.
Mass of water = Mass of solution - Mass of AgNO3
Mass of water = 100 g - 22 g
Mass of water = 78 g
Mass of water in kilograms = 78 g / 1000
Mass of water in kilograms = 0.078 kg
Step 5: Calculate the molality.
Molality = Number of moles of AgNO3 / Mass of water in kilograms
Molality = (22 g / 169.87 g/mol) / 0.078 kg
Now you can calculate the molality using the above equation.
22.0% AgNO3 by mass means 22.0 grams AgNO3 per 100 g solution. That is equivalent to 220 g/1000g solution.
The 1000 g solution is made of 220 g AgNO3 + 780 g water.
How many moles AgNO3 in 220 g AgNO3. That will be 220/molar mass AgNO3 (approximately 1.3 moles.
molality = moles/kg solvent.
m = 1.3 moles/0.780 kg solvent = about 1.7 m.
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