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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

Respuesta :

Answer:

D.) 1.71 M NaCl

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

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