# How many liters of 0.88 M solution can be made with 25.5 grams of lithium fluoride?

The concentration of a solution can be expressed in many ways. One of them is as the molarity of the solution. A solution with molarity equal to one has one mole of the solute dissolved in every liter of the solutions.

A solution of lithium fluoride has to be prepared...

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The concentration of a solution can be expressed in many ways. One of them is as the molarity of the solution. A solution with molarity equal to one has one mole of the solute dissolved in every liter of the solutions.

A solution of lithium fluoride has to be prepared here and the number of grams of lithium fluoride available for the preparation of the solution is 25.5 g.

As the molecular mass of lithium fluoride is 25.939 g/mole, 25.5 grams constitute 0.983 moles.

To make one liter of 0.88 M solution we need 0.88 moles of lithium fluoride. We have 0.983 moles. This can be used to make 0.983/0.88 = 1.117 L of the solution.

With 25.5 g of lithium fluoride 1.117 L of 0.88 M solution of lithium fluoride can be made.

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