A solution is 36% silver nitrate (AgNO3) by mass. The density of this solution is 1.44
g/mL. The formula weight of AgNO3 is 170
g/mol. Calculate the molality of AgNO3 in
this solution.
36% AgNO3 means 36 g AgNO3/100 g solution.
36 g = 36/170 moles = ??
100 g solution = volume x density. Calculate volume. Then M = moles/L soln.
To calculate the molality of AgNO3 in the solution, we need to understand what molality is and how to calculate it.
Molality (m) is a unit of concentration that represents the number of moles of solute per kilogram of solvent. It differs from molarity (M) which represents the number of moles of solute per liter of solution.
To calculate molality (m), we can use the following formula:
m = moles of solute / mass of solvent (in kg)
In this case, the solute is silver nitrate (AgNO3) and the solvent is the solution. The question provides the concentration of the silver nitrate solution (% by mass) and the density (g/mL), which can help us determine the mass of the solvent.
Let's break down the steps to find the molality of AgNO3 in the solution:
Step 1: Calculate the mass of the solution.
We are given the density of the solution (1.44 g/mL). To find the mass of the solution, we need to use the formula: mass = density * volume. The volume of the solution is not provided, so we cannot calculate the exact mass of the solution. The best we can do is assume a certain volume (e.g. 100 mL) and proceed with the calculation. Keep in mind that the volume you assume will affect the accuracy of the final result.
Step 2: Calculate the mass of AgNO3 in the solution.
Since we know that the solution is 36% AgNO3 by mass, we can calculate the mass of AgNO3 in the solution. Assuming a total mass of 100 g (if you assumed 100 mL as the volume of the solution), the mass of AgNO3 would be 36 g (since it is 36% of 100 g).
Step 3: Calculate the moles of AgNO3.
To calculate the moles of AgNO3, we need to divide the mass of AgNO3 by its molar mass (formula weight). In this case, the formula weight of AgNO3 is 170 g/mol. Therefore, 36 g of AgNO3 would be equal to 36 g / 170 g/mol = 0.212 mol.
Step 4: Calculate the mass of the solvent.
To find the mass of the solvent (water), we subtract the mass of AgNO3 (36 g) from the assumed total mass of the solution (100 g). Using our assumed 100 mL volume, the mass of the solvent would be 100 g - 36 g = 64 g.
Step 5: Calculate the molality of AgNO3.
Finally, we can calculate the molality of AgNO3 using the formula mentioned earlier: molality (m) = moles of solute / mass of solvent (in kg). In this case, the moles of AgNO3 is 0.212 mol, and the mass of the solvent is 64 g (0.064 kg). Plugging these values into the formula gives us: molality (m) = 0.212 mol / 0.064 kg = 3.312 mol/kg.
Therefore, the molality of AgNO3 in the solution is approximately 3.312 mol/kg. Remember that the assumption of the volume of the solution will affect the final result, so it is always best to use the actual volume if available.