Question #c7336

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Question #c7336

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Answer 1 · 2017-08-13T02:33:42

Taking atomic masses of constituent elements of the complex as

$C r \to 52 g / m o l$ $Crto52g"/"mol$

$C l \to 35.5 g / m o l$ $Cl to35.5g"/"mol$

$O \to 16 g / m o l$ $O to16g"/"mol$

$H \to 1 g / m o l$ $H to1g"/"mol$

We have formula mass of the complex

$H_{12} O_{6} C l_{3} C r \to 12 \times 1 + 6 \times 16 + 3 \times 35.5 + 1 \times 52 = 266.5 g / m o l$ $H_12O_6Cl_3Cr->12xx1+6xx16+3xx35.5+1xx52=266.5g"/"mol$

On heating with conc $H_{2} S O_{4}$ $H_2SO_4$ ,the complex looses 6.75% of its original weight.Actually conc $H_{2} S O_{4}$ $H_2SO_4$ dehydrates it.

So loss of total mass of water from formula mass is

$= 266.5 \times 6.75 % g \approx 18 g$ $=266.5xx6.75%g~~18g$

Molar mass of water $18 g / m o l$ $18g"/"mol$

Hence number of moles of water released $1 m o l$ $1mol$ . So one moecule of water was not in complex sphere.

Hence the formula of the complex as monohydrate should be

$[C r {(H_{2} O)}_{5} C l] C l_{2} \cdot H_{2} O$ $[Cr(H_2O)_5Cl]Cl_2 * H_2O$

And its actual dehydrated formula is

$[C r {(H_{2} O)}_{5} C l] C l_{2}$ $color(magenta)([Cr(H_2O)_5Cl]Cl_2)$

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Question #c7336

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