Question

How much amount of CuSO4.5H2O required for
liberation of 2.54 g I, when titrated with Kl
(1) 2.5 gm
(2) 4.99 gm
(3) 2.4 gm
(4) 1.2 gm​

Answer 1

Answer:

2.5 is the amount of CuSO4.5H2O for liberication of 2.54gl, when titrated with Kl

Answer 2

Answer:

Explanation:

EQUIVALENT OF CUSO4.5H2O = EQUIVALENT OF 2.54 G I2

MOLES*F=MOLES*F

X/249.5*1 = 2.54/254*2

X/249.5 = 0.02

X=4.99 GM