Question

Calculate the amount of benzoic acid $\left(C_6{H}_{5}COOH\right)$ required for preparing $250mL$ of $0.15M$ solution in methanol.

Answer 1

Number of moles of benzoic acid required $=0.15\times \frac{250}{1000}=0.0375$ moles

Molar mass of benzoic acid $=7\left(12\right)+6\left(1\right)+2\left(16\right)=84+6+32=122$ g/mol

Mass of benzoic acid required $=122\times 0.0375=4.575$ g