In: Chemistry
How do you prepare 25 ml of 8.5 mM NaCl from a 20% NaCl stock? (Hint: convert 20% to g/ml then to molar)
20% NaCl means , in 100 gm of solution there are 20 gm
NaCl.
water =80 gm
since density of water in 1 gm/ml
volume of solution = 80 mL
molar mass of NaCl = 23+35.5 = 58.5 gm
number of moles of NaCl = mass / molar mass
= 20/58.5
=0.342
volume of solution = 80 mL = 0.08 L
Molarity = mol /volume = 0.342/0.08 = 4.275 M
From this solution we need to prepare 25 mL of 8.5 mM
solution
use dilution formula:
Mi*Vi = Mf*Vf
4.275 * V = 8.5*10^-3 * 25
V = 0.05 mL
So, volume of original solution that needs to be taken is just 0.05
mL
Take 0.05 mL of original solution and add enough water to make it
25 mL and you will get your final required solution