How many Kilograms of water must be added to 5.0 grams of oxalic acid (H2C2O4) to prepare a 0.40 Molality?
moles H2C2O4 = grams/molar mass.
Solve for mols.
m = moles/kg solvent
Solve for kg solvent.
(5.0 g) / (90.03488 H2C2O4/mol) / 0.40m = 45.01744
You punch a wrong button on your calculator. Try doing it in two steps instead of one.
5/90 = ??
??/0.4 =?? I get something about 0.139 kg.
Ok, I'm sorry. Not taking chemistry, but I'm taking a practice test, and I didn't know what to do. I want to be ready for next year. Thanks!
To determine the number of kilograms of water needed to prepare a 0.40 molality solution of oxalic acid (H2C2O4), you first need to understand the concept of molality.
Molality (m) is defined as the number of moles of solute per kilogram of solvent. In this case, the solute is oxalic acid (H2C2O4), and the solvent is water.
To calculate the molality, you can use the following formula:
m = moles of solute / mass of solvent in kg
In this case, the molality (m) is given as 0.40.
To find the moles of solute, you need to know the molar mass of oxalic acid, which is calculated by adding up the atomic masses of all the atoms in the molecule.
H2C2O4:
2(1.01 g/mol) + 2(12.01 g/mol) + 4(16.00 g/mol) = 90.03 g/mol
The molar mass of oxalic acid is 90.03 g/mol.
Now, let's calculate the moles of solute (oxalic acid) using the given mass of 5.0 grams:
moles of solute = mass of solute / molar mass
moles of solute = 5.0 g / 90.03 g/mol
Next, we need to convert the moles of solute to kilograms of water. Since we know the molality (0.40) and the moles of solute, we can use the following equation:
molality = moles of solute / mass of solvent in kg
Rearranging this equation to solve for the mass of solvent in kg, we get:
mass of solvent in kg = moles of solute / molality
Substituting the given values, we have:
mass of solvent in kg = (5.0 g / 90.03 g/mol) / 0.40
Simplifying the expression, we get:
mass of solvent in kg = 0.125 kg
So, you would need to add 0.125 kilograms (or 125 grams) of water to 5.0 grams of oxalic acid to prepare a 0.40 molality solution.