What is the existing mass (in grams) of sodium ions in 200 mL of liquid whose density is 80 g/L?
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200mL of NaOH contains 200/1000 * 80 = 16g of NaOH. In 1 mole of NaOH there are 23+16+1 = 40 g of NaOH Therefore, in 200mL of NaOH, the mass of Na is 16/40 = 0.4 moles. So there are 0.4 * 23 = 9.2 grams of Na