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Standard XII

Chemistry

Percentage Composition

27.8 g of F...

Question

$27.8$ g of $F e S O_{4} . x H_{2} O$ is treated with excess of $B a C l_{2}$ to give $2.33$ g of white precipitate. Calculate the value of $x$ .
$(B a = 137, S = 32, F e = 56, H = 1, O = 16)$

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Solution

The correct option is B $7$
$F e S O_{4} . x H_{2} O + B a C l_{2} \to B a S O_{4}$
$1 m o l e$ $1 m o l e$
$2.78 g m$ $2.33 g m$
$96 + 56 + 18 x = 137 + 96$
So, $x = 7$ .

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mL of a solution of

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K_{2} S O_{4}

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(Given that the molecular weight of

B a = 137, S = 32, O = 16, K = 39

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Q. The equivalent weight of

F e {S O}_{4} . {({N H}_{4})}_{2} {S O}_{4} .24 H_{2} O

is:
(At. wt of

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(Given that molecular weight of

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Q. 100 mL of 20.8%

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(B a = 137, C l = 35.5, S = 32, H = 1, O = 16)

B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l

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27.8 g of FeSO4.xH2O is treated with excess of BaCl2 to give 2.33 g of white precipitate. Calculate the value of x.(Ba=137,S=32,Fe=56,H=1,O=16)

The correct option is B 7FeSO4.xH2O+BaCl2→BaSO4 1 mole 1 mole 2.78 gm 2.33 gm 96+56+18x=137+96So, x=7.

$27.8$ g of $F e S O_{4} . x H_{2} O$ is treated with excess of $B a C l_{2}$ to give $2.33$ g of white precipitate. Calculate the value of $x$ .
$(B a = 137, S = 32, F e = 56, H = 1, O = 16)$

The correct option is B $7$
$F e S O_{4} . x H_{2} O + B a C l_{2} \to B a S O_{4}$
$1 m o l e$ $1 m o l e$
$2.78 g m$ $2.33 g m$
$96 + 56 + 18 x = 137 + 96$
So, $x = 7$ .