Weight of LiOH: 18.0 g
Volume of solution: 850.0 mL
To find the percent concentration of the solute (LiOH), we first need to calculate the molarity of the solution.
Molarity (M) = moles of solute / liters of solution
First, let's calculate the moles of LiOH:
Molar mass of LiOH = 6.941 g/mol (Li) + 15.999 g/mol (O) + 1.008 g/mol (H) = 23.948 g/mol
Moles of LiOH = 18.0 g / 23.948 g/mol = 0.751 moles
Now, convert the volume of the solution from mL to liters:
Volume of solution = 850.0 mL / 1000 = 0.850 L
Now calculate the molarity:
Molarity = 0.751 moles / 0.850 L = 0.884 M
To determine the percent concentration, we use the formula:
Percent concentration = (mass of solute / mass of solution) x 100
Mass of solution = 18.0g (LiOH) + 850g (water, assuming density of water is 1 g/mL) = 868.0 g
Percent concentration = (18.0 g / 868.0 g) x 100 = 2.07%
Therefore, the percent concentration of LiOH in the solution is 2.07%.
What is the
weight
volume
significant figures.
percent concentration of the solute given 18.0 g of LiOH in 850. mL of solution? Be sure your answer has the correct number of
%
LiOH
x10
1 answer