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Standard IX

Chemistry

Percentage Composition

Complex A has...

Question

Complex A has a composition of $H_{12} O_{6} {Cl}_{3} Cr$ . If the complex treatment with conc. $H_{2} {SO}_{4}$ loses $13.5 %$ of its original mass, the correct molecular formula of A is: [Given: the atomic mass of $Cr = 52 amu and Cl = 35 amu$ ].

$[Cr {(H_{2} O)}_{5} Cl] {Cl}_{2} . H_{2} O$

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$[Cr {(H_{2} O)}_{4} {Cl}_{2}] Cl . 2 H_{2} O$

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$[Cr {(H_{2} O)}_{3} {Cl}_{3}] . 3 H_{2} O$

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$[Cr {(H_{2} O)}_{6}] {Cl}_{3}$

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Solution

The correct option is B
$[Cr {(H_{2} O)}_{4} {Cl}_{2}] Cl . 2 H_{2} O$

B is the correct option.
Explanation for the correct option:
Let $x$ molecule of water be lost then
$% mass of water = \frac{x \times 18}{(12 + 6 \times 16 + 35 \times 3 + 52)} \times 100 = 1.99 ≃ 2$
Hence, B $[Cr {(H_{2} O)}_{4} {Cl}_{2}] Cl . 2 H_{2} O$ is the correct molecular formula of A.
Explanation for the incorrect option:
Since, the $%$ mass of water is found to be $2$ .
Therefore, A, C and D are incorrect options.

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Similar questions

An acid of phosphorus has the following percentage composition phosphorus = 38.27% hydrogen = 2.47% oxygen = 59.26 . Find the empirical formula of the acid and its molecular formula, given that its relative molecular mass is 162.

A, B and C are three complexes of chromium (III) with the empirical formula

H_{12} O_{6} C l_{3} C r

. All the three complexes have water and chloride ions as ligands. Complex A does not react with concentrated

H_{2} S O_{4}

, whereas complexes B and C lose 6.75% and 13.5% of their original weight respectively, on treatment with concentrated

H_{2} S O_{4}

Thus, select the correct statement :

10.66 g

of a chromium (III) complex having molecular formula :

C r C l_{3} (H_{2} O)_{6}

when treated with excess

A g N O_{3}

solution,

11.48 g

of white precepitate is obtained. How much water is lost by

133.25 g

of the given complex (in g) when it is treated with conc.

H_{2} S O_{4}

? (Assume the complex is octahedral) Atomic mass of

C r = 52 u

and

A g = 108 u

)

A, B

and

C

are three complexes of chromium(III) with the empirical formula:

H_{12} O_{6} C l_{3} C r

. All the three complexes have water and chloride ions as ligands. Complex

A

does not react with concentrated

H_{2} S O_{4}

, whereas complexes

B

and

C

lose

6.75 %

and

13.5 %

of their original weight, respectively, on treatment with concentrated

H_{2} S O_{4}

. If

C

[C r (H_{2} O)_{x} C l_{y}] C l \cdot z H_{2} O,

find the x, y and z.

A, B and C are three complexes of chromium (III) with the empirical formula

H_{12} O_{6} C l_{3} C r

. All the three complexes have water and chloride ions as ligands. Complex A does not react with concentrated

H_{2} S O_{4}

, whereas complexes B and C lose 6.75% and 13.5% of their original weight respectively, on treatment with concentrated

H_{2} S O_{4}

Thus, loss in weight on treatment with concentrated

H_{2} S O_{4}

is due to :

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Complex A has a composition of H12O6Cl3Cr. If the complex treatment with conc. H2SO4loses 13.5%of its original mass, the correct molecular formula of A is: [Given: the atomic mass of Cr=52amuandCl=35amu].

Complex A has a composition of $H_{12} O_{6} {Cl}_{3} Cr$ . If the complex treatment with conc. $H_{2} {SO}_{4}$ loses $13.5 %$ of its original mass, the correct molecular formula of A is: [Given: the atomic mass of $Cr = 52 amu and Cl = 35 amu$ ].