How many grams of NaI would be used to produce a 23.0 M solution with a volume of 1.50 L?
The molar mass of NaI is 22.99 + 126.90 = 149.89 g/mol.
It takes 1.50 * 23.0 * 150 = 5180 g of NaI to produce 1.50 L of its 23.0 M solution.
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