Question

You need to make an aqueous solution of 0.495 M lithium chloride for an experiment in the lab, using a 125 mL volumetric flask. How much solid lithium chloride should you add?

a) 5.475 g

b) 0.0619 g

c) 3.275 g

d) 2.756 g

Answer 1

To prepare a 0.495 M lithium chloride solution in a 125 mL volumetric flask, you need to add approximately 2.619 grams of solid lithium chloride.Hence the closest option is option D.

Step-by-step explanation:

To prepare the 0.495 M lithium chloride solution using a 125 mL volumetric flask, you need to calculate the amount of solid lithium chloride to add.

Molarity (M) is defined as moles of solute divided by liters of solution. To calculate the moles of lithium chloride, you can use the formula:

Moles = Molarity x Volume (in liters)

Given that the molarity is 0.495 M and the volume is 0.125 L (converted from mL), you can calculate:

Moles = 0.495 M x 0.125 L = 0.061875 moles

The molar mass of lithium chloride is 42.39 g/mol. To find the mass of lithium chloride, you can use the formula:

Mass = Moles x Molar mass

Mass = 0.061875 moles x 42.39 g/mol = 2.619 g