1.)An aqueous solution is 34.0% H3PO4 by mass and has a density of 1.207 g/mL. What is the molarity of the solution? AND what is the molality?
density = 1.20 g/mL so 1000 mL has a mass of
1.20 x 1000 = 1200 grams
How much of that is H3PO4? 34.0%.
1200 x 0.34 = 408 g.
How many moles is that?
408/98 = 4 something. You do it exactly. That is 4 something moles in 1 L so Molarity = 4 something.
Molality:
1000 mL has a mass of 1200 grams and 408 of that is H3PO4.
You know the number of moles from above. But the g solvent is 1200-408 = ??. Change that to kg and m = moles/kg.
Close to 5 m. You do it exactly.
thank you
To find the molarity of the solution, we need to convert the mass percentage of H3PO4 to moles and then divide by the volume in liters.
Step 1: Mass of the solution
Assume we have 100g of solution, then the mass of H3PO4 in the solution is 34.0g.
Step 2: Moles of H3PO4
The molar mass of H3PO4 is:
(3 * 1.00794) + (1 * 15.999) + (4 * 1.00794) = 97.9946 g/mol
So, the moles of H3PO4 in the solution can be calculated as:
moles = mass / molar mass = 34.0g / 97.9946 g/mol = 0.3469 mol
Step 3: Volume of the solution
The density of the solution is given as 1.207 g/mL. Since density = mass/volume, we can rearrange the equation to solve for volume:
volume = mass / density = 100g / 1.207 g/mL = 82.817 mL = 0.0828 L
Step 4: Molarity
Molarity is defined as moles of solute divided by volume of solution in liters. Therefore, we can calculate the molarity of the solution as:
Molarity = moles / volume = 0.3469 mol / 0.0828 L ≈ 4.19 M
To find the molality of the solution, we need to convert the mass percentage of H3PO4 to moles and divide by the mass of the solvent (water) in kilograms.
Step 5: Moles of H3PO4
We already found the moles of H3PO4 in step 2, which is 0.3469 mol.
Step 6: Mass of water
Assuming the mass of the solution is 100g and the mass of H3PO4 is 34.0g, the mass of water is:
mass of water = mass of solution - mass of H3PO4 = 100g - 34.0g = 66.0g = 0.066 kg
Step 7: Molality
Molality is defined as moles of solute divided by the mass of the solvent (in kg). Therefore, we can calculate the molality of the solution as:
Molality = moles / mass of solvent = 0.3469 mol / 0.066 kg ≈ 5.26 m
To find the molarity of the solution, we need to determine the concentration of H3PO4 in moles per liter (M). To do this, we'll need the molar mass of H3PO4.
The molar mass of H3PO4 is calculated by adding up the atomic masses of its constituent elements:
H = 1.008 g/mol, P = 30.97 g/mol, O = 16.00 g/mol
Molar mass of H3PO4 = (3 * 1.008) + 30.97 + (4 * 16.00) = 97.99 g/mol
Now, let's calculate the number of moles of H3PO4 in the solution.
We have a 34.0% H3PO4 solution, meaning 34.0 g of H3PO4 per 100 g of solution.
Therefore, the mass of H3PO4 in the solution can be calculated as:
(34.0 g/100 g) * Total mass of the solution
Given that the density of the solution is 1.207 g/mL, we can assume that 1 mL of the solution weighs 1.207 g.
So, the total mass of the solution can be calculated as:
(1.207 g/mL) * Volume of the solution in mL
Since we need the molarity (M), the Volume of the solution (V) should be in liters.
Finally, using the equation Molarity (M) = Moles of solute / Volume of solution (in liters), we can calculate the molarity.
Now, let's find the molality of the solution.
Molality (m) is another way to express the concentration of a solution, using the number of moles of solute per kilogram of solvent.
To calculate the molality, we need to find the mass of the solvent in kilograms. Since the density of the solution is given in g/mL, we can assume that 1 mL of solution is equal to 1 g. With this information, we can calculate the mass of the solvent in kilograms.
Once we have the mass of the solvent in kilograms, we can use the equation Molality (m) = Moles of solute / Mass of solvent (in kg) to find the molality of the solution.
Now you have the step-by-step explanation on how to calculate the molarity and molality of the given solution. Please provide the total mass of the solution and the volume of the solution in order to proceed with the calculation.