Question

Calculate the mass of Ca(NO3)2 needed to prepare 3.5 L of a 0.8 M solution?

Calculate the mass of `Ca(NO_3)_2` needed to prepare 3.5 L of a 0.8 M solution.

Answer 1

459.45 g

Explanation:

Given:
concentration = 0.8 M = `0.8 (mol)/L`
volume = 3.5 L
mass = ?
molecular weight of `Ca(NO_3)_2` = `164.088 g/(mol)`

Stoichiometry:

`0.8 color(red)(cancel(mol) / cancel(L)) * 3.5 color(red)cancel(L) * 164.088 g/color(red)(cancel(mol)) = 459.45 g`