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NEET UG CHEMISTRY MCQ: SURFACE CHEMISTRY

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A detergent

\left(\mathrm{C}_{12} \mathrm{H}_{25} \mathrm{SO}_4^{-} \mathrm{Na}^{+}\right)

solution becomes a colloidal sol at a concentration of

10^{-3} \mathrm{M}

. On an average

10^{13}

colloidal particles are present in

1 \mathrm{~mm}^3

. What is the average number of ions which are contained by one colloidal particle (micelle)?

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Correct option: A

60

Option A (60) is correct because it involves calculating the number of colloidal particles per liter and then dividing the number of detergent molecules per liter by the number of colloidal particles per liter to find the average number of ions per micelle.

More Questions from SURFACE CHEMISTRY

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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}}

\log {P}

is a straight line inclined at an angle

\theta=45^{\circ}

. When pressure is

0.5

\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}

{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 :