A 42.12 % (w/v) solution of NaCl causes preci | vedclass

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Question

(A) The coagulating value is defined as the minimum concentration of an electrolyte in $m\,mol \, L^{-1}$ required to cause coagulation of a sol in $2

Vedclass · Accepted Answer

$36 \, m\,mol \, L^{-1}$

Vedclass · Answer

(A) The coagulating value is defined as the minimum concentration of an electrolyte in $m\,mol \, L^{-1}$ required to cause coagulation of a sol in $2 \, hours$.$1$. Calculate the molar mass of $NaCl$: $M = 23.0 + 35.5 = 58.5 \, g \, mol^{-1}$.$2$. Calculate the molarity of the given $42.12 \% (w/v)$ solution:$42.12 \% (w/v)$ means $42.12 \, g$ of $NaCl$ in $100 \, mL$ of solution.Concentration in $g \, L^{-1} = 42.12 	imes 10 = 421.2 \, g \, L^{-1}$.Concentration in $mol \, L^{-1} = \frac{421.2}{58.5} = 7.2 \, mol \, L^{-1} = 7200 \, m\,mol \, L^{-1}$.$3$. The question asks for the coagulating value,which is the concentration required for coagulation. Given the data provided is insufficient to relate the time $(10 \, hours)$ to the standard definition $(2 \, hours)$ without a specific rate law or empirical relationship,the provided options suggest a calculation based on the concentration itself. However,based on standard textbook problems of this type,the answer is derived as $7200 \, m\,mol \, L^{-1}$. Since this is not an option,and the provided solution was 'Data insufficient',we maintain this assessment.

$A$ $42.12 \% (w/v)$ solution of $NaCl$ causes precipitation of a certain sol in $10 \, hours$. The coagulating value of $NaCl$ for the sol is $.....$
[Given : Molar mass : $Na = 23.0 \, g \, mol^{-1} ; Cl = 35.5 \, g \, mol^{-1}]$

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$A$ $42.12 \% (w/v)$ solution of $NaCl$ causes precipitation of a certain sol in $10 \, hours$. The coagulating value of $NaCl$ for the sol is $.....$[Given : Molar mass : $Na = 23.0 \, g \, mol^{-1} ; Cl = 35.5 \, g \, mol^{-1}]$