The minimum quantity of sodium chloride which | vedclass

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

Question

(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

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(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

Similar Questions

Which among the following is an example of a one-dimensional nanostructure?

Which of the following is not a colloidal solution?

How many of the following are true solutions?
$I :$ Urea solution,
$II :$ Gelatin,
$III :$ Glucose solution,
$IV :$ $NaCl$ solution,
$V :$ Butter,
$VI :$ Blood

If $Fe(OH)_3$ sol is filled in container $B$,then water level in which container will increase?

Bredig's arc method cannot be used to prepare a colloidal solution of which of the following?

Vedclass Test Series

Exam Paper Generator

Online Exam Module