Question

If one litre of calcium carbonate solution contains 5.34 x 10-3 mol, what is the corresponding mass of calcium carbonate in a litre, and therefore the concentration in mg/L (not g/L)?

how do u solve this i am confused

Answer 1

The concentration of `"CaCO"_3"` in mg/L is `"534 mg/L"`.

Explanation:

Determine the mass in grams using the given moles and molar mass of `"CaCO"_3"` `("100.086 g/mol")`. Then convert the mass in grams to mass in mg.

Mass of `"CaCO"_3"` in grams

Multiply moles `"CaCO"_3"` by its molar mass.

`5.34xx10^(-3)color(red)cancel(color(black)("mol CaCO"_3))xx(100.086"g CaCO"_3)/(1color(red)cancel(color(black)("mol CaCO"_3)))="0.534 g CaCO"_3"`

Convert grams to milligrams.

`"1 g"` `=` `"1000 mg"`

`0.534color(red)cancel(color(black)("g"))xx("1000 mg")/(1color(red)cancel(color(black)("g")))="534 mg"`

The concentration of `"CaCO"_3"` in mg/L is `"534 mg/L"`.