Question

The amount of benzoic acid, $\left(C_6{H}_{6}COOH\right)$ required for preparing 250 mL of 0.15 M solution in methanol is

• A. $0.4575$ g
• B. $4.575$ g
• C. $45.75$ g
• D. $0.04575$ g

Answer 1

Answer: (B)

$4.575$ g

0.15 M solution means 0.15 mole of benzoic acid is present in 1L or 100 mL. Molar mass of $C_6{H}_{5}COOH=122gmo{l}^{-1}$
0.15 mole of benzoic acid $=015\times 122=18.3g$
1000 mL of solution contains = 18.3g benzoic acid
$\therefore$ 250 mL of solution will contain
$=\frac{250\times 18.3}{1000}=4.575g$ benzoic acid