100 mL of 0.0018% (w/v) solution of $Cl^-$ ion was the minimum concentration of $Cl^-$ required to p

Question

100 mL of 0.0018% (w/v) solution of $Cl^-$ ion was the minimum concentration of $Cl^-$ required to precipitate a negative sol in one h. The coagulating value of $Cl^-$ ion is (Nearest integer)

Professor PC Thomas Classes · Answer

0.0018 g $Cl^-$ in 100 mL. Moles in 1 L = $(0.018~g/L) / (35.5~g/mol) = 0.000507$ mol/L = 0.507 mmol/L $\approx 1$.