Question

What is the molarity of a solution in which 87 g of NaCl are dissolved in 1.0 L of solution

Answer 1

$Cm = \frac{n}{V}=\frac{1,4886}{1,0}=1.4886 g/l$

$n=\frac{m}{M}=\frac{87}{58.443}=1.4886$