Question

What is the molarity of a solution prepared by dissolving 30.0g NaCl in enough water to make a 0.300 L solution?

A. 1.82 M NaCl

B. 0.154 M NaCl

C. 0.833 M NaCl

D. 1.71 M NaCl

Answer 1

D.) 1.71 M NaCl

Step-by-step explanation:

Molarity equation: M= n/v

n= moles of solute

v=liters of solution

NaCl= 58.443 g/mol

30g NaCl / 58.443g/mol = 0.5133(this is n)

0.5133 mols/0.300 L=1.71115674 M