The ratio of water of crystallization present in Mohr’s salt to that of Potash alum is $\dots \dots . . x 10^{- 1}$

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Solution

The explanation for the blank:
Potash alum = $K_{2} ({SO}_{4}) . {Al}_{2} {({SO}_{4})}_{3} . 24 H_{2} O$
Mohr’s salt = ${(NH 4)}_{2} Fe {({SO}_{4})}_{2} . 6 H_{2} O$
$\begin{array}{rcl} The ratio of the water molecule & = & \frac{6 (From Mohr' s Salt)}{24 (From Potash Alum)} \\ = & 0.25 \\ = & 2.5 \times 10^{- 1} \end{array}$
Therefore, the ratio of water of crystallization present in Mohr’s salt to that of Potash alum is $2.5 \times 10^{- 1}$ .

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Similar questions

Potash alum is a double salt having the chemical formula $K_{2} {SO}_{4} . {Al}_{2} {({SO}_{4})}_{3} . 24 H_{2} O$ It dissolves in water to form a clear solution.

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[ Potash Alum

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Q. What is equivalent weight of potash alum , Mohr's salt with explanation?

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The ratio of water of crystallization present in Mohr’s salt to that of Potash alum is……..x10-1

The explanation for the blank:Potash alum = K2(SO4).Al2(SO4)3.24H2OMohr’s salt = (NH4)2Fe(SO4)2.6H2OTheratioofthewatermolecule=6(FromMohr'sSalt)24(FromPotashAlum)=0.25=2.5×10-1Therefore, the ratio of water of crystallization present in Mohr’s salt to that of Potash alum is 2.5×10-1.

The ratio of water of crystallization present in Mohr’s salt to that of Potash alum is $\dots \dots . . x 10^{- 1}$