Chapter: Pharmaceutical Calculations
Lecture-3
21. A solution (100 mL) that contains 409.5 mg of NaCl/100 mL has how many mEq of Na+
and Cl-?
❑Solution
We know, mEq = mmol/valence
According to question, amount of NaCl in the solution= 409.5 mg= 0.4095 g
and MW of NaCl is (23+35.5) = 58.5 g/mole
therefore, no. of moles of NaCl in solution= mass/M.W= 0.4095g÷58.5 g/moles= 7x10-3 moles
We also know, 1 mole= 1000 milli moles (mmol)
So, 7x10-3 moles= 7 mmol
Now, as per the equation of mEq, we get, mEq= mmol/valence
Therefore, mEq of Na+ = 7÷1=7 mg equivalent (Ans.)
Similarly, mEq of Cl- = 7÷1=7 mg equivalent (Ans.)
22. What is the Equivalent wt of ferrous ion (Fe2+) which has a gram atomic
weight of 55.85?
❑Solution:
We know, Equivalent weight=molar mass ÷ total charge on cation or anion
1 mole Fe2+ = 55.85 g= molar mass of Fe2+
So, Equivalent weight = 55.85/2
= 27.925 gram equivalent (Ans.)
23. How many mEq of Na are in 60 ml of an 5% solution saccharine (g mol wt: 241 g,
valence:1)?
❑Solution:
Amount of Na is 5% in 60 ml
❑100 ml solution contains = 5 gm Na
❑60 ml solution contains = 5X60/100 gm Na = 3 gm
Moles of Na present in saccharine= 3 g ÷ 241 g/moles= 0.0124 moles
= 12.4 mmol
Now we know, mEq = mmol/valence
