When 100 mL of 0.2 M solution of CoCl_3 cdo | vedclass

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(B) The number of moles of $CoCl_3 \cdot x NH_3$ is calculated as: $n = M 	imes V(L) = 0.2 \ mol/L 	imes 0.1 \ L = 0.02 \ mol$.Given that $3.6 	ime

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$6$

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(B) The number of moles of $CoCl_3 \cdot x NH_3$ is calculated as: $n = M 	imes V(L) = 0.2 \ mol/L 	imes 0.1 \ L = 0.02 \ mol$.Given that $3.6 	imes 10^{22}$ ions are precipitated by $AgNO_3$,the number of moles of $AgCl$ precipitated is $n(AgCl) = \frac{3.6 	imes 10^{22}}{6 	imes 10^{23}} = 0.06 \ mol$.Since $0.02 \ mol$ of the complex produces $0.06 \ mol$ of $AgCl$,$1 \ mol$ of the complex produces $\frac{0.06}{0.02} = 3 \ mol$ of $AgCl$.This implies that there are $3$ chloride ions outside the coordination sphere,meaning the formula is $[Co(NH_3)_x]Cl_3$.For a coordination number of $6$ for $Co^{3+}$,$x$ must be $6$.

When $100 \ mL$ of $0.2 \ M$ solution of $CoCl_3 \cdot x NH_3$ is treated with excess of $AgNO_3$ solution,$3.6 \times 10^{22}$ ions are precipitated. The value of $x$ is $\left(N_A = 6 \times 10^{23} \ mol^{-1}\right)$

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