The volume (in mL) of 0.125 M AgNO3 required to quantitatively precipitate chloride ions in 0.3 g of [Co(NH3)]Cl3 is. M[Co(NH3)6]Cl3 = 267.46 g/mol MAgNO3 = 169.87 g/mol Given -

Question

The volume (in mL) of 0.125 M AgNO3 required to quantitatively precipitate chloride ions in 0.3 g of [Co(NH3)]Cl3 is. M[Co(NH3)6]Cl3 = 267.46 g/mol MAgNO3 = 169.87 g/mol Given - (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Number of moles of Cl- precipitated in [Co(NH3)6]Cl3 is equal to number of moles of AgNO3 used. (0.3/267.46)×3=(0.125× V/1000) where V is volume of AgNO3 (in mL) V = 26.92 mL

Q. The volume (in mL) of 0.125 M $A g N O_{3}$ required to quantitatively precipitate chloride ions in 0.3 g of $[C o (N H_{3})] C l_{3}$ is__________.
$^{M} [C o (N H_{3})_{6}] C l_{3} = 267.46$ g/mol $M_{A g N O_{3}} = 169.87$ g/mol
Given -

Number of moles of $C l^{-}$ precipitated in $[C o (N H_{3})_{6}] C l_{3}$ is equal to number of moles of $A g N O_{3}$ used.
$\frac{0.3}{267.46} \times 3 = \frac{0.125 \times V}{1000}$
where $V$ is volume of $A g N O_{3}$ (in $m L$ )
$V = 26.92 m L$

Q. The volume (in mL) of 0.125 M AgNO3​ required to quantitatively precipitate chloride ions in 0.3 g of [Co(NH3​)]Cl3​ is__________. M[Co(NH3​)6​]Cl3​=267.46 g/mol MAgNO3​​=169.87 g/mol Given -

Number of moles of Cl− precipitated in [Co(NH3​)6​]Cl3​ is equal to number of moles of AgNO3​ used. 267.460.3​×3=10000.125×V​ where V is volume of AgNO3​ (in mL) V=26.92mL

Q. The volume (in mL) of 0.125 M $A g N O_{3}$ required to quantitatively precipitate chloride ions in 0.3 g of $[C o (N H_{3})] C l_{3}$ is__________.
$^{M} [C o (N H_{3})_{6}] C l_{3} = 267.46$ g/mol $M_{A g N O_{3}} = 169.87$ g/mol
Given -

Number of moles of $C l^{-}$ precipitated in $[C o (N H_{3})_{6}] C l_{3}$ is equal to number of moles of $A g N O_{3}$ used.
$\frac{0.3}{267.46} \times 3 = \frac{0.125 \times V}{1000}$
where $V$ is volume of $A g N O_{3}$ (in $m L$ )
$V = 26.92 m L$