Excess of silver nitrate solution is added to $100 m L$ of $0.01 M$ pentaaqua chloro chromium (III) chloride solution. The mass of silver chloride obtained in grams is:
[Atomic mass of silver is $108$ ]

287 \times 10^{- 3}

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143.5 \times 10^{- 3}

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143.5 \times 10^{- 2}

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287 \times 10^{- 2}

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Solution

The correct option is B $287 \times 10^{- 3}$
$[C r (H_{2} O_{5}) C l] C l_{2}$ is the reagent
i.e. $2 {C l}^{(-)}$ can be combined with $A g$ to give $A g C l$
mass of $A g C l$ $p p t$ ,
no. of moles of ${C l}^{(-)} = \frac{2 \times 100 \times 0.01}{1000} = 0.002$
$∴$ moles of $A g C l$ $p p t = 0.002$
$∴$ mass of $A g C l = 0.002 \times 143.5$
$= 2.87 \times 10^{- 3} g$

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Excess of silver nitrate solution is added to 100mL of 0.01M pentaaqua chloro chromium (III) chloride solution. The mass of silver chloride obtained in grams is: [Atomic mass of silver is 108]

The correct option is B 287×10−3[Cr(H2O5)Cl]Cl2 is the reagenti.e. 2Cl(−) can be combined with Ag to give AgClmass of AgCl ppt,no. of moles of Cl(−)=2×100×0.011000=0.002∴ moles of AgCl ppt=0.002∴ mass of AgCl=0.002×143.5 =2.87×10−3g

Excess of silver nitrate solution is added to $100 m L$ of $0.01 M$ pentaaqua chloro chromium (III) chloride solution. The mass of silver chloride obtained in grams is:
[Atomic mass of silver is $108$ ]