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An 11 M solution containing 449.4 g of ammonium sulfate would contain how many liters?

User Shabinjo
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1 Answer

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Answer : The volume of solution will be, 0.309 L

Explanation : Given,

Mass of
(NH_4)_2SO_4 = 449.4 g

Molar mass of
(NH_4)_2SO_4 = 132 g/mole

Concentration = 11 M

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :


\text{Molarity}=\frac{\text{Mass of }(NH_4)_2SO_4}{\text{Molar mass of }(NH_4)_2SO_4* \text{Volume of solution (in L)}}

Now put all the given values in this formula, we get:


114M=\frac{449.4g}{132g/mole* \text{Volume of solution (in L)}}


\text{Volume of solution (in L)}=0.309L

Therefore, the volume of solution will be, 0.309 L

User Ben Jacobs
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