$50 \mathrm{ml}$ of $\mathrm{Fe}(\mathrm{OH})_3$ sol is completty coagulated by $50 \mathrm{ml}$ of $0.2 \mathrm{M} \mathrm{KBr}$. Coagulation value is

NEET UG CHEMISTRY MCQ: SURFACE CHEMISTRY

1 Medium

1 of 1

Application +4 | -1

50 \mathrm{ml}

\mathrm{Fe}(\mathrm{OH})_3

sol is completty coagulated by

50 \mathrm{ml}

0.2 \mathrm{M} \mathrm{KBr}

. Coagulation value is

View answer and explanation

Correct option: A

200

Option A (200) is correct because the coagulation value is defined as the number of milligrams of electrolyte required to coagulate 1 liter of the colloidal solution. We convert the molarity of KBr to mg/L and then adjust for the volume of the sol.

More Questions from SURFACE CHEMISTRY

Practice All Chapter MCQs →

A cationic colloidal electrolyte forms micelle at

10^{-3} \mathrm{M}

concentration in water. Its

1 \mathrm{~mm}^3

solution contains

2 \times 10^{13}

micelle structures. What is the number of cations contained in one micelle? Take

\mathrm{N}_{\mathrm{A}}=6 \times 10^{23}

. Graph between

\log \frac{{x}}{{m}}

and

\log {P}

is a straight line inclined at an angle

\theta=45^{\circ}

. When pressure is

0.5

atm and

\log {k}= 0.699,

the amount of solute adsorbed per

\mathrm{g}

of adsorbent will be: For adsorption of gas on solid surface, the plots of

\log \;x/m\mathrm{ } \;vs.

\log \;P

is linear with a slope equal to : Most suitable salt which can be used for efficient clotting of blood will be The density of gold is

19\;\mathrm{g/c}{\mathrm{m}^3}

. If

{1.9 \times {{10}^{ - 4}}\;\mathrm{g}}

of gold is dispersed in 1 L of water to give a sol having spherical gold particles of radius

10\;\mathrm{ }nm

, then the number of gold particles per

m{m^3}

of the sol will be The magnitude of colligative properties in all colloidal dispersions is …..than solution :