Answer:
The volume of a 0.0015 [tex]\frac{moles}{liters}[/tex] calcium sulfate solution that contains 25.0 g of calcium sulfate CaSO₄ is 122.53 liters.
Explanation:
Molarity (M) is the number of moles of solute that are dissolved in a given volume. Molarity is expressed by:
[tex]Molarity=\frac{number of moles of solute}{Dissolution volume}[/tex]
Molarity is expressed in units [tex]\frac{moles}{liter}[/tex]
In this case you have 25.0g of calcium sulfate CaSO₄. First of all you need to know the amount of moles that mass represents. For that you must first know the atomic masses of each element:
Then the molar mass of the compound calcium sulfate is:
molar mass= 40 g/mol + 32 g/mol + 4*16 g/mol= 136 g/mol
It is then possible to apply a rule of three as follows: if 136 g represents 1 mol of the compound calcium sulfate, 25 g how many moles are they?
[tex]moles=\frac{25 g*1 mole}{136 g}[/tex]
moles≅0.1838
Now you can apply a rule of three knowing the molarity of 0.0015 [tex]\frac{moles}{liters}[/tex]: if 0.0015 moles represents 1 liter of solution, 0.1838 moles how many liters are they?
[tex]volume=\frac{0.1838moles*1 liter}{0.0015moles}[/tex]
volume=122.53 liters
The volume of a 0.0015 [tex]\frac{moles}{liters}[/tex] calcium sulfate solution that contains 25.0 g of calcium sulfate CaSO₄ is 122.53 liters.
