100 mL of 0.0018%(w/v) solution of Cl^{-} io | vedclass

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(B) The concentration is given as $0.0018\%(w/v)$,which means $0.0018 \ g$ of $Cl^{-}$ is present in $100 \ mL$ of the solution.Coagulation value is d

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$1$

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(B) The concentration is given as $0.0018\%(w/v)$,which means $0.0018 \ g$ of $Cl^{-}$ is present in $100 \ mL$ of the solution.Coagulation value is defined as the minimum concentration of an electrolyte in millimoles per litre $(mmol/L)$ required to cause coagulation of a colloidal sol.First,calculate the number of moles of $Cl^{-}$:$Moles = \frac{	ext{mass}}{	ext{molar mass}} = \frac{0.0018 \ g}{35.5 \ g/mol} \approx 5.07 	imes 10^{-5} \ mol$.Convert moles to millimoles:$Millimoles = 5.07 	imes 10^{-5} 	imes 1000 = 0.0507 \ mmol$.Since this amount is present in $100 \ mL$ $(0.1 \ L)$,the concentration in $mmol/L$ is:$Concentration = \frac{0.0507 \ mmol}{0.1 \ L} = 0.507 \ mmol/L$.The nearest integer value is $1$.

$100 \ mL$ 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).

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