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500 mL should contain

0.500 (L) * 0.100 (mol / L) * 58.5 (g / mol) = 2.925 g (=grams)

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6y ago
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6y ago

2,922 g dried sodium chloride are necessary.

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6y ago

0.500 L x 0.100 mole/L x 58.5 g/mole = 2.92 g (to 3 significant figures)

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Q: How many grams of sodium choride are required to prepare 500.0mL of a 0.100 M solution?
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