Question

11. Mass of benzoic acid required to prepare
500 mL of 0.3 M solution in ethanol is
[NCERT Pg. 53]
(2) 15.2 g
(1) 25.2 g
13718.39
(4) 21.5 g​

Answer 1

Answer:18.3 g

Explanation:

Moles of benzoic acid= MxVolume(in L)

= 0.3x(500/1000) = 3/20 moles

Molar mass of benzoic acid (C6H5COOH) = 6(12)+5(1)+12+2(16)+1 = 122g/mol

Mass of benzoic acid required = moles x molar mass

=3/20 x 122 = 18.3 g