# Chemistry

Calculate normality of normal saline at 0.85%?

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asked by Mya
1. I take it that this is 0.85% w/v. If so, that means 0.85 g NaCl per 100 mL solution.
Convert 0.85 g NaCl to moles. moles = grams/molar mass = 0.85/58.5 (you may want to refine this figures) = 0.0145 moles/100 mL which is 0.145 moles/L soln which is 0.145 M or 0.145 N.

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2. The normality is twice as the molarity because in this case normality is used in reference to tonicity (osmotic pressure) and NaCl splits in two ions doubling the number of osmotically active particles. It would be 2 x 0.145 x 0.93 = 0.270
0.93 is a correction factor for an imperfect solution of NaCl. If you obviate this factor the answer is 0.290 osmols which is normal serum osmolality. Called for this reason "Normal Saline" but it actually is 0.290Normal

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posted by Rafael
3. 0.145

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posted by h man

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