JEE MAIN - Chemistry (2020 - 3rd September Evening Slot - No. 4)
Complex A has a composition of H12O6Cl3Cr. If the complex on treatment with conc.H2SO4
loses
13.5% of its original mass, the correct molecular formula of A is :
[Given: atomic mass of Cr = 52 amu and Cl = 35 amu]
[Given: atomic mass of Cr = 52 amu and Cl = 35 amu]
[Cr(H2O)5Cl]Cl2.H2O
[Cr(H2O)4Cl2]Cl.2H2O
[Cr(H2O)3Cl3].3H2O
[Cr(H2O)6]Cl3
Explanation
Let x molecule of water are lost then
13.5 = $$\left[ {{{x \times 18} \over {6 \times 18 + 3 \times 35 + 52}}} \right] \times 100$$
$$ \Rightarrow $$ x = 1.99 $$ \simeq $$ 2
$$ \therefore $$ Around two moles of water are lost during heating.
$$ \therefore $$ Formula of complex could be
[Cr(H2O)4Cl2]Cl.2H2O
13.5 = $$\left[ {{{x \times 18} \over {6 \times 18 + 3 \times 35 + 52}}} \right] \times 100$$
$$ \Rightarrow $$ x = 1.99 $$ \simeq $$ 2
$$ \therefore $$ Around two moles of water are lost during heating.
$$ \therefore $$ Formula of complex could be
[Cr(H2O)4Cl2]Cl.2H2O
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