Question

the number of gram-equivalent of h3po4 present in its 0.1 molar,250 ml solution is

Answer 1

H3PO4

3+31 +64 = 98 , molecular mass.

1 litre of 0.1 molar solution contains 98 x 0.1 = 9.8 g

250 ml contains 9.8/4 = 2.45 g

eq. mass of H3PO4 = 98/3 = 32.67 g

there fore no.of equivalents 250 ml of 0.1 molar = 2.45/32.67 g

= 0.075

Answer 2

Answer:NORMALITY = BASICITY OF ACID * MOLARITY

BASICITY OF H3PO4 = 3

N= 3*0.1

=0.3N

NORMALITY = NO. OF GRAM EQUIVALENTS/ VOLUME

WE KNOW NORMALITY = 0.3

VOLUME = 250 ML = 0.25L

NO OF GRAM EQUIVALENTS = 0.3 * 0.25

= 0.075

Explanation: