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  4. How much amount of CuSO4 ⋅ 5H2O is required for liberation of 2.54 g of I2 when titrated with KI ?

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Q. How much amount of $CuSO_{4} \cdot 5H_{2}O$ is required for liberation of 2.54 g of $I_{2}$ when titrated with KI ?

AIIMSAIIMS 2011Some Basic Concepts of Chemistry

Solution:

$2 CuSO _{4} \cdot 5 H _{2} O +4 KI \to Cu_2I_2 +2 K_2 SO_4+ \underset{254\,g}{I_{2}}+10H_{2}O$

Molecular weight of $2 CuSO _{4} \cdot 5 H _{2} O$

$[2(63.5+32+64)+10(18)] g=499 \,g$

$254 \,g$ of $I _{2}$ is liberated by $499 \,g \,CuSO _{4} \cdot 5 H _{2} O$

$2.54\, g$ of $I _{2}$ will be liberated by $x \,g\, CuSO _{4} \cdot 5 H _{2} O$

$x=\frac{499}{254} \times 2.54=4.99 \,g$