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  • Calculate the amount of benzoic acid (C_6 H_5 COOH) required for preparing 250 mL of 0.15 M solution in methanol.

2.30    Calculate the amount of benzoic acid (C_{6}H_{5}COOH) required for preparing 250\; mL  of  0.15 \; M solution in methanol.

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Molar mass of benzoic acid = 7(12) + 6(1) + 2(16) = 122u.

We are given with the molarity of solution.

                                                  Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution\ in\ litre}

or                                                          0.15 = \frac{Moles\ of\ solute}{\frac{250}{1000}}

or                                      Moles\ of\ solute= \frac{0.15\times 250}{1000} = 0.0375\ mol

So mass of benzoic acid :

                                              = Moles\ of\ benzoic\ acid \times Molar\ mass

                                              =0.0375\times 122 = 4.575\ g

Hence the required amount of benzoic acid is 4.575 g.

Posted by

Devendra Khairwa

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