Calculate the amount of benzoic acid (C6H5COOH) required for prep

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144. Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol. 


Answer:

Mol. mass of benzoic acid, C6H5COOH
= 6 x 12 + 5 x 1 + 12 + 16 + 16 + 1
= 72 + 5 + 12 + 16+16 + 1
= 122 g mol–1

by using formula;

M = xmolecular mass of given substance×1000required volume

here x= amount of substance required 

0.15M = x122×1000250

Amount of benzoic acid required

=1221000×250×0.15= 45751000 = 4.575 g.

 

 

 

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