Question

How many grams of LiBr are needed to make 1.5 L of a 3.0 M solution?

Answer 1

Answer: 391 g

Step-by-step explanation:

For this problem, we need to know that molarity is. Molarity is moles of solute/liters of solution. it is also denoted as M=n/V, which is also mol/L. We are given that the molarity is 3.0 M and the liter is 1.5 L. All we have to do is plug in 3.0 and 1.5 into our formula and solve for moles.

`3.0M=(n)/(1.5L)`

`n=4.5 mols`

Now that we have moles, we can convert moles to grms by using the molar mass of LiBr.

`4.5 mols*(86.844 g)/(1 mol) =391g`