Question

calculate the amount of benzoic acid (C6H5COOH) required for preparing 250ml of 0.15M solution in methanol.​

Answer 1

Amount of benzoic required = 4.575 g

\fbox{\texttt{\pink{Given\::}}}

Given:

Amount of solution = 250 mL of 0.15 M (in methanol)

\fbox{\texttt{\orange{To\:find\::}}}

Tofind:

Amount of benzoic acid required to prepare the solution.

\fbox{\texttt{\red{How\:to\:Find\::}}}

HowtoFind:

\bold{Molarity=\frac{Moles\:of\:solute}{Vol.\:of\:solution\:(in\:mL)}}Molarity=

Vol.ofsolution(inmL)

Molesofsolute

$$\implies\bold{0.15=\frac{Moles\:of\:benzoic\:acid}{250}\times{1000}}$$ $$\begin{lgathered}\\\\\end{lgathered}$$

$$\implies\bold{Moles\:of\:benzoic\:acid=\frac{0.15\times{250}}{1000}=0.0375}$$ $$\begin{lgathered}\\\\\end{lgathered}$$

$$\bold{Molecular\:mass\:of\:benzoic\:acid(C_6H_5COOH)}$$

$$\bold{=7\times{12}+6\times{1}+2\times{16}}$$

$$\begin{lgathered}\bold{=122}\\\\\end{lgathered}$$

$$\bold{\therefore Amount\:of\:benzoic\:acid=0.0375\times{122