The volume (in mL ) of $0.125 \mathrm{M} \mathrm{AgNO}_{3}$ required to quantitatively precipitate chloride ions in 0.3 g of $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ is ____ $$ \begin{aligned} & { }^{\mathrm{M}}\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{Cl}_{3}\right]=267.46 \mathrm{~g} / \mathrm{mol} \\ & { }^{\mathrm{M}} \mathrm{AgNO}_{3}=169.87 \mathrm{~g} / \mathrm{mol} \end{aligned} $$

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Coordination Compounds

The volume (in mL ) of

0.125 M AgNO_{3}

required to quantitatively precipitate chloride ions in 0.3 g of

[Co (NH_{3})_{6}] Cl_{3}

is ____

^{M} [Co (NH_{3})_{6} Cl_{3}] = 267.46 g / mol^{M} AgNO_{3} = 169.87 g / mol

Answer: 26.92

Solution

All three chloride ions in

[Co (N H_{3})_{6}] C l_{3}

are ionic and precipitate as AgCl, following

[Co (N H_{3})_{6}] C l_{3} + 3 AgN O_{3} \to 3 AgCl + [Co (N H_{3})_{6}] (N O_{3})_{3}

. The moles of complex are

\frac{0.3}{267.46} = 1.122 \times 1 0^{- 3}

, requiring

3 \times 1.122 \times 1 0^{- 3} = 3.366 \times 1 0^{- 3}

mol of

AgN O_{3}

. At

0.125

M the volume needed is

\frac{3.366 \times 1 0 ^{- 3}}{0.125} = 0.02693

L, that is about

26.92

mL.

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Numerical answer

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