how many grams of NaCl (molecular weight 58.5) should be used to prepare 20 liters of a solution containing 154mEq/L?

Answer 1

For NaCl, 1 equivalent is the same as 1 mole.
154 mEq/L = 0.154 Eq/L = 0.154 mole/L
So you want g = moles x molar mass = 0.154 x 58.5 = grams/L. For 20 L you want 20x that.

If you don't want to go through the conversion each time, just remember that
mEq x milliequivalent weight = grams.
So you want 154 x 0.0585 = ?? grams and that is in 1L, so 20x that is the amount for 20 L.

Answer 2

180.20