Question

Sea water contains roughly 28.0 g of nacl per liter

Answer 1

Answer:

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Explanation:

58.5 g/molMolecular weight of NaCl28 gWeight of NaCl1 literVolume of the solution

M=\frac{wt*1000}{mw*volume}M=mw∗volumewt∗1000

M=\frac{28*1000}{58.5*1000}M=58.5∗100028∗1000

M=0.479\,mol/LM=0.479mol/L

Molarity formula