Question

Find the amount of Benzoic acid ( C6H5COOH) for 250 ml of 0.15 M solution ​

Answer 1

Answer:

Explanation:

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

Use 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.

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Answer 2

${\bold{\underline{Answer}}}$ → 4.575 g

${\bold{\underline{Step\:by\:Step\:Explanation}}}$ →

• Our first step is to calculate the molecular mass of the given compound. The molecular mass is obtained by adding all atomic masses present in a compound . So we have ,

Molecular mass of C6H5COOH →

=> 12 × 6 + 5 × 1 + 12 + 16 + 16 + 1

[ Atomic mass of C = 12 u , O = 16 u and H = 1u ]

=> 122 u

→ Let us consider that the amount of Benzoic Acid is x gms.

$= > { \bold{Number \: of \: moles }}= \frac{given \: mass \: }{molecular \: mass}$

•°• n = $\frac{x}{122}$

Now , let's apply the formula of Molarity :

${ \bold{ \underline{Molarity }}} = \: { \sf{\frac{n}{Volume} \times 1000}}$

Plug the values :

$= > 0.15 = \frac{x}{122 \times 250} \times 1000 \\ \\ = > 0.15 \times 122 = 4x \\ \\ = > 4x = 18.3 \\ \\ = > x = \frac{18.3}{4}$

${ \boxed{ \bold{x = 4.575g}}}$

${\bold{Extra \:Info}}$ :

• Molarity is the number of moles of solute dissolved per litre of solution.
• It is measured in Moles per litre and represented by M.
• While calculating Molarity , we multiply the values by 1000 only if volume is given in smaller units . ( like here it was in ml)