Question

How many grams of KBr are dissolved in 92.7 mL of a 0.955 M solution?

Answer 1

To answer first we need to calculate the molar weight of this molecule (KBr):

For this we go to the periodic table and check the molar weight of potassium and bromine:

K: 39.098 g/mol

Br: 79.904 g/mol

So the molar weight of KBr is:

`M_(KBr)=39.098(g)/(mol)+79.904(g)/(mol)=119(g)/(mol)`

Now, we know that the solution is 0.955 M, this means that in 1000 ml there are 0.955 moles of KBr. So we calculate the number of moles in 92.7ml:

`moles_(KBr)=\frac{92.7ml\text{ }x\text{ }0.955(mol)/(L)}{1000(ml)/(L)}=0.0885\text{ mol}`

Now we use the molar weught to calculate the grams in the sample:

`m_(KBr)=0.0855mol\text{ x }119(g)/(mol)=10.17g`

So the answer is 10.17g.