The minimum quantity of sodium chloride which | vedclass

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(D) The molar mass of $NaCl$ is $58.5 \ g/mol$. Number of moles $= \frac{	ext{mass}}{	ext{molar mass}} = \frac{0.585 \ g}{58.5 \ g/mol} = 0.01 \ mol

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(D) The molar mass of $NaCl$ is $58.5 \ g/mol$. Number of moles $= \frac{	ext{mass}}{	ext{molar mass}} = \frac{0.585 \ g}{58.5 \ g/mol} = 0.01 \ mol = 10 \ mmol$. The flocculation value is defined as the minimum concentration of an electrolyte in $mmol/L$ required to cause precipitation of a sol. Flocculation value $= \frac{	ext{millimoles of electrolyte}}{	ext{Volume of sol in litres}} = \frac{10 \ mmol}{10 \ L} = 1 \ mmol/L$.

The minimum quantity of sodium chloride which is necessary to precipitate $10 \ L$ of sol in two hours is $0.585 \ g$. The flocculation value of sodium chloride is

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