A 1.577 M solution of AgNO3 has a density of 1.220 gram per centimeter cube. What is the molality of the solution?
1.577 M is 1.577 mols/L solution. With density = 1.220 g/mL, then substitute volume and density to find mass of the solution.
1.577 mols AgNO3 x molar mass AgNO3 = grams AgNO3.
Then mass solution - g AgNO3 = mass solvent. Convert that to kg.
m = 1.577 mols AgNO3/kg solvent
0.01
To find the molality, we need to know the number of moles of solute (AgNO3) per kilogram of solvent.
First, let's calculate the molar mass of AgNO3.
Ag = 107.87 g/mol
N = 14.01 g/mol
O = 16.00 g/mol (x3)
Total molar mass = (107.87 + 14.01 + 16.00 * 3) g/mol = 169.87 g/mol
Next, let's determine the mass of AgNO3 in the solution.
Density = mass/volume
Mass = density × volume
The volume is not given, but we can assume it to be 1 liter since it is a common unit for concentration.
Mass = 1.220 g/cm³ × 1000 cm³ = 1220 g
Since the solution is in 1.577 M, it means there are 1.577 moles of AgNO3 in 1 liter of solution.
So, the mass of AgNO3 in the solution will be:
Mass of AgNO3 = 1.577 moles × 169.87 g/mol = 267.71 g
Now, we can calculate the mass of the solvent (water).
Mass of solvent = Total mass of solution - Mass of solute
Mass of solvent = 1220 g - 267.71 g = 952.29 g
Finally, we can calculate the molality.
Molality = moles of solute / mass of solvent (in kg)
Molality = 1.577 moles / 0.95229 kg = 1.6540 mol/kg
Therefore, the molality of the solution is approximately 1.6540 mol/kg.