Question

What weight of monosodium glutamate (MW 187.1) is required to make a 50 ml of 50 mM solution?

Answer 1

Since the number of miligrams per milimole of a given substance is equivalent to that substances molar mass, the number of milimoles required to prepare 1000 ml of monosodium glutamate = 187.1 mgms.

& for preparing 50 mM solution the weight of monosodium glutamate required = 187.1 x 50 mgms per 1000 ml of solution.

But we need to prepare 50 ml solution , so the weight of monosodium glutamate required

............................................................................................................................= 187.1 x 50 x 50 / 1000

........................................................................................................................... = 467.75 mgms

..................................................................................................................or,---- = 4.675 x 10^-1 gms.