Question

A beaker contains 11.30 M osimium (lll) fluoride in 0.0673 liters of solution. How many grams of osimium (lll) fluoride are dissolved in this solution?

Answer 1

It is given that:

Molarity (M) = 11.30 M

Volume (V) = 0.0673 liters

Now :

$Molarity(M)=\dfrac{moles}{Volume(liters)}$

$11.30=\dfrac{moles}{0.0673}$

So Moles of osimium(III) fluoride $=11.30\times0.0673=0.76moles$

Now mass of osimium(III) fluoride $=moles\times molar mass$

$=0.76\times304.2=231.192 gm$