Question

calculate the amount of benzoic acid required for preparing 250 ml of 0.15 M soln in methnol.

Answer 1

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

Solution

In this problem molarity = 0.15M is given 



Let take volume of solution = 1 liter  = 1000 mL

Us the above formula we get number of moles of solute = 0.15 moles

 Molar mass of benzoic acid (C6H5COOH) = 12 × 6 + 5 × 1 + 12 + 16 + 16 +1

       = 72 + 5 + 12 + 32 + 1

                                                                = 122 g mol-1

Mass of 0.15 mole of benzoic acid = number of moles x molar mass

     =0.15 × 122 g

     = 18.3 g

Thus, 1000 mL of the solution has mass of benzoic acid = 18.3 g

So        250 mL of the solution has mass of benzoic acid = 18.3 × 250/1000

  = 4.58 g.

Answer 2

$\fbox{\texttt{\green{Answer\::}}}$

Amount of benzoic required = 4.575 g

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

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

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

Amount of benzoic acid required to prepare the solution.

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

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

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

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

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

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

$\bold{=122}$

$\bold{\therefore Amount\:of\:benzoic\:acid=0.0375\times{122}=4.575\:g}$