I am following a paper in that it written 0.7M Zinc acetate and 0.7 M monoethanolamine in 2 methoxy ethanol of 5 ml. What amount is zinc acetate and monoethanolamine amount in gram.
Aniruddh Bahadur Yadav I understand that you need to know what is the mass of zinc acetate and monoethanolamine to prepare 5 mL of solution in 2-methoxyethanol? If it is, you have to make the next calculations:
m (Zn acetate) = c (Zn acetate) * M (Zn acetate) * V (Zn acetate)
m (Zn acetate) = 0.7 mol/L * 183.48 g/mol * 0.005 L
m (Zn acetate) = 0.6422 g
In the case of monoethanolamine, it is a liquid reagent, so you have to consider the density of the substance:
m (monoethanolamine) = c (monoethanolamine) * M (monoethanolamine) * V
m (monoethanolamine) = 0.7 mol/L * 61.08 g/mol * 0.005 L
m (monoethanolamine) = 0.2103 g
Now, you have to use the density to know the volume of the reagent containing 0.2103 g of monoethanolamine:
d(monoethanolamine) = 1.01 g/mL
d = m/V
V = m/d
V = 0.2103 g/1.01 g/mL
V = 0.2082 mL
Liquid ragents have a purity grade, so, the final volume will be:
Vf = V/P
where P is the purity of the ragent/100 (if the purity is 98%, P = 98/100=0.98)
Vf= 0.2082 mL/0.98 = 0.2124 mL
I wrote 98% as an example, you should use the % that is reported in your reagent flask.
I hope it be useful for you. Best regards.
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