Question

Question #03e18

Answer 1

`"8 g CaCl"_2`

Explanation:

For starters, you need to know exactly how many moles of chloride anions are present in your solution.

Since you know that the chloride anions have a concentration of `"0.500 mol L"^(-1)`, you can say that your sample will contain

`300 color(red)(cancel(color(black)("mL solution"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.500 moles Cl"^(-)/(1color(red)(cancel(color(black)("L solution")))) = "0.15 moles Cl"^(-)`

Now, calcium chloride is soluble in water, which means that it dissociates completely in aqueous solution to produce calcium cations and chloride anions.

`"CaCl"_ (2(aq)) -> "Ca"_ ((aq))^(2+) + color(red)(2)"Cl"_ ((aq))^(-)`

Notice that every mole of calcium chloride that dissociates produces `color(red)(2)` moles of chloride anions.

This means that in order for the solution to contain `0.15` moles of chloride anions, it must dissolve

`0.15 color(red)(cancel(color(black)("moles Cl"^(-)))) * "1 mole CaCl"_2/(color(red)(2)color(red)(cancel(color(black)("moles Cl"^(-))))) = "0.075 moles CaCl"_2`

Finally, to convert this to grams, use the molar mass of calcium chloride

`0.075 color(red)(cancel(color(black)("moles CaCl"_2))) * "110.98 g"/(color(red)(cancel(color(black)("mole CaCl"_2)))) = color(darkgreen)(ul(color(black)("8 g")))`

The answer must be rounded to one significant figure, the number of sig figs you have for the volume of the solution.