-DOLCOLODCalculate the amount of benzoic acid (C,H,COOH) required for

1.	-DOLCOLODCalculate the amount of benzoic acid (C,H,COOH) required for preparing 250mL of 0.15 M solution in methanol.
Answer» 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 So250 mL of the solution has mass of benzoic acid = 18.3 × 250/1000 = 4.58 g.

-DOLCOLODCalculate the amount of benzoic acid (C,H,COOH) required for preparing 250mL of 0.15 M solution in methanol.

Answer»

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

So250 mL of the solution has mass of benzoic acid = 18.3 × 250/1000

= 4.58 g.