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;

$M =\frac{ x}{molecular mass of given subs\tan ce}\times \frac{1000}{required volume}$

here x= amount of substance required 

$0.15M =\frac{ x}{122}\times \frac{1000}{250}$

Amount of benzoic acid required

$$\begin{gathered} =\frac{122}{1000}\times 250\times 0.15 \\ = \frac{4575}{1000} = 4.575 \mathrm{g}. \end{gathered}$$