How many grams are needed to make 2 L of solution of 4 M LiNO3?

552 g of LiNO₃

Explanation:

From the question given above, the following data were obtained:

Volume of solution = 2 L

Molarity of LiNO₃ = 4 M

Mass of LiNO₃ =?

Next, we shall determine the number of mole of LiNO₃ in the solution. This can be obtained as follow:

Volume of solution = 2 L

Molarity of LiNO₃ = 4 M

Mole of LiNO₃ =?

Molarity = mole /Volume

4 = mole of LiNO₃ / 2

Cross multiply

Mole of LiNO₃ = 4 × 2

Mole of LiNO₃ = 8 moles

Finally, we shall determine the mass of of LiNO₃ needed to prepare the solution. This is can be obtained as follow:

Mole of LiNO₃ = 8 moles

Molar mass of LiNO₃ = 7 + 14 + (16×3)

= 7 + 14 + 48

= 69 g/mol

Mass of LiNO₃ =?

Mole = mass /Molar mass

8 = Molar mass of LiNO₃ /69

Cross multiply

Molar mass of LiNO₃ = 8 × 69

Molar mass of LiNO₃ = 552 g

Thus, 552 g of LiNO₃ is needed to prepare the solution.

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