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How many mL of 0.650 M LiF solution will contain 30.0g LiF?​

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Answer:

First, we need to calculate the molar mass of LiF:

LiF: Li = 6.941 g/mol, F = 18.998 g/mol

1 Li + 1 F = 6.941 g/mol + 18.998 g/mol = 25.939 g/mol

So, 1 mole of LiF weighs 25.939 g.

Now, we can calculate how many moles of LiF are in 30.0 g:

moles = mass ÷ molar mass

moles = 30.0 g ÷ 25.939 g/mol

moles = 1.157 mol

Finally, we can use the formula for molarity to find the volume of the solution:

Molarity = moles ÷ volume (in liters)

We want to solve for volume in milliliters, so we can rearrange the formula:

Volume (in liters) = moles ÷ molarity

Volume (in mL) = (moles ÷ molarity) × 1000

Plugging in the values, we get:

Volume (in mL) = (1.157 mol ÷ 0.650 mol/L) × 1000 = 1778.5 mL

Rounding to three significant figures, the answer is:

The solution contains 1780 mL of 0.650 M LiF.

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User Frenchy
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