What mass of caco3 (in grams) is needed to make 10.5 l of a 1.50 m caco3 solution?

Data:

M = 1.50 (assiming the data is Molarity and not molality, which is written with capital letter)

V= 10.5 l

n = ?

Formulas:

M = n / V => n = M * V

n = mass in grams / molar mass => mass in grams = n * molar mass

Solution

n = 1.50 * 10. 5 l = 15.75 moles

molar mass CaCO3 = 40.08 g/mol + 12.01 g/mol + 3* 16 g/mol = 100.09 g/mol

mass in grams = 15.75 moles * 100.09 g/mol = 1,576.4 g.

1,576 g

Tolerance to high levels of salinity