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:
[tex]M_{KBr}=39.098\frac{g}{mol}+79.904\frac{g}{mol}=119\frac{g}{mol}[/tex]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:
[tex]moles_{KBr}=\frac{92.7ml\text{ }x\text{ }0.955\frac{mol}{L}}{1000\frac{ml}{L}}=0.0885\text{ mol}[/tex]Now we use the molar weught to calculate the grams in the sample:
[tex]m_{KBr}=0.0855mol\text{ x }119\frac{g}{mol}=10.17g[/tex]So the answer is 10.17g.
