Question #198297

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?

Expert's answer

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"

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