Frances wanted to prepare a 0.0285M standard solution of LiCl. She weighed out 27.5g of solid LiCl. She had plenty of water available. What volume of the solution was she able to prepare?
how many moles in 27.5g?
We only need .0285 moles/liter
so divide the # moles of salt by the molarity to get the volume.
Thank you so much for your help. Once I got the formula to convert grams to moles, this was a lot of help. But could you check my work for me? I converted the grams to moles and got 0.648737909 mol LiCl. Then I divided that by 0.0285M and got 22.76273365 L which I rounded to be 22.8 liters of LiCl.
So is 22.8 the correct answer?
I'm almost 100% sure it is.
To find the volume of the solution Frances was able to prepare, we need to use the formula:
Molarity (M) = moles of solute (mol) / volume of solution (L)
First, let's calculate the number of moles of LiCl in the 27.5g of solid using its molar mass. The molar mass of LiCl is 6.94g/mol (for Li) + 35.45g/mol (for Cl) = 42.39g/mol.
Moles of LiCl = mass of LiCl (g) / molar mass of LiCl (g/mol)
= 27.5g / 42.39g/mol
≈ 0.65 mol (rounded to two decimal places)
Now, we have the moles of solute. To calculate the volume of the solution, we rearrange the formula:
Volume of solution (L) = moles of solute (mol) / Molarity (M)
Plugging in the values:
Volume of solution (L) = 0.65 mol / 0.0285 M
≈ 22.81 L (rounded to two decimal places)
Therefore, Frances was able to prepare approximately 22.81 liters of the 0.0285M LiCl solution.