If 0.001435 g of AgCl is present in 100 mL | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The solubility $s$ is the concentration in $mol/L$. Given mass of $AgCl = 0.001435 \ g$ in $100 \ mL$. Mass in $1000 \ mL (1 \ L) = 0.001435 	ime

Vedclass · Accepted Answer

$10^{-8}$

Vedclass · Answer

(B) The solubility $s$ is the concentration in $mol/L$. Given mass of $AgCl = 0.001435 \ g$ in $100 \ mL$. Mass in $1000 \ mL (1 \ L) = 0.001435 	imes 10 = 0.01435 \ g/L$. Molar mass of $AgCl = 107.87 + 35.45 \approx 143.5 \ g/mol$. Solubility $s = \frac{0.01435 \ g/L}{143.5 \ g/mol} = 10^{-4} \ mol/L$. For $AgCl ightleftharpoons Ag^+ + Cl^-$,$K_{sp} = [Ag^+][Cl^-] = s^2$. $K_{sp} = (10^{-4})^2 = 10^{-8}$.

If $0.001435 \ g$ of $AgCl$ is present in $100 \ mL$ of water,calculate the $K_{sp}$.

Similar Questions

Which of the following is most soluble in water?

If the solubility of $MX_2$ in water is $0.0002 \ mol/L$,then its $K_{sp}$ is:

Assign $A, B, C, D$ from the given type of reaction.
$ZnS(s) + 2HCl(aq) \longrightarrow ZnCl_2(aq) + H_2S(g)$

What would be the concentration of $H_2SO_4$ necessary to precipitate $BaSO_4$ from a solution of $0.01 \ M \ Ba^{2+}$ ions $(K_{sp} = 1.0 \times 10^{-9})$?

Which of the following has the maximum solubility? (Given $K_{sp}$ values)

Vedclass Test Series

Exam Paper Generator

Online Exam Module