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
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0
Answer:
2.5 is the amount of CuSO4.5H2O for liberication of 2.54gl, when titrated with Kl
Answered by
0
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
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