NEET UG CHEMISTRY MCQ: SURFACE CHEMISTRY

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The density of gold is

19 \mathrm{~g} / \mathrm{cm}^3, 1.9 \times 10^{-3} \mathrm{~g}

of gold is dispersed in one litre of water to give a sol having spherical gold particles of radius

10.5 \mathrm{~nm}

.
The gold particles are occupying

50 \%

of the available space. The number of gold particles per

\mathrm{mm}^3

is about

10^n

. Find the value of

n

View answer and explanation

Correct option: A

Option A (7) is correct because it involves calculating the volume of a single gold particle, the total number of particles, and then the number of particles per unit volume, considering the space occupied.

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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}}

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 :