0.9% NaCl w/v means 0.9g NaCl/100 mL and that is the same as 9g NaCl/L.
The equivalent weight of NaCl is the same as the formula weight of NaCl; therefore, 9/58.4 = moles NaCl/L = equivalents/L = 0.154 Eq/L or 154 mEq/L. That may not be the right number of significant figures; I don't know if that 0.9% is 0.9 or 0.900 or what. You take care of that end of it.
The normal concentration of sodium chloride in blood is 0.9% (w/v). Therefore, assuming FW for Nacl of 58.4 g/mole, an isotonic 0.9% (w/v) saline would have how many milli-equivalents per liter (mEq/L) of Na ions?
Can you show me how you came up with this answer?
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