we want to prepare stock solution containing 5 micromole/ml of maltose. for this we have to add 180mg per 100 ml of distilled water. but according to our calculations 0.18 mg of maltose should be added to make 100 ml of distill water
Rabia Shafiq
Thanks Dr.Sarawak but according to net calculations it should b 180/100 MLK but how?
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Roopasingam Kugapreethan
Hi Rabia
you are confusing with millimole and micromole. 180 mg for 5 mM stock and 0.18 mg for 5 uM of stock solution.
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Arnaud John Kombe Kombe
Hi Rabia.
Hope you solved and understood the solution stock preparation issue since then.
Otherwise, just note that:
MW (maltose) = 342.3 g/mol
Volume = 100 mL (0.1 L)
Conc. = 5 mM (0.05 M)
The required mass = 342.3 × 0.1 × 0.05 = 0.17115 g (around 171 mg or 180 mg if you want)
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