JEE MAIN - Chemistry (2020 - 8th January Morning Slot - No. 10)
The volume (in mL) of 0.125 M AgNO3 required to quantitatively precipitate chloride ions in 0.3 g of
[Co(NH3)6]Cl3 is ________.
M[Co(NH3)6Cl3] = 267.46 g/mol
MAgNO3 = 169.87 g/mol
M[Co(NH3)6Cl3] = 267.46 g/mol
MAgNO3 = 169.87 g/mol
Answer
26.80to27.00
Explanation
[Co(NH3)6]Cl3 + 3AgNO3 $$ \to $$ 3AgCl
$$ \Rightarrow $$ $${{0.3} \over {267.46}} = {{0.125 \times v \times {{10}^{ - 3}}} \over 3}$$
$$ \Rightarrow $$ v = 26.92 ml
Mole of [Co(NH3)6]Cl3
1
=
Mole of AgNO3
3
$$ \Rightarrow $$ $${{0.3} \over {267.46}} = {{0.125 \times v \times {{10}^{ - 3}}} \over 3}$$
$$ \Rightarrow $$ v = 26.92 ml
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