NEET UG CHEMISTRY MCQ: SURFACE CHEMISTRY

1 Easy

1 of 1

Application +4 | -1

In an experiment,

200 \mathrm{~mL}

0.5 \mathrm{M}

oxalic acid is shaken with

10 \mathrm{~g}

of activated charcoal and filtered. The concentration of the filtrate is reduced to

0.4 \mathrm{M}

. The amount of adsorption

\left(\frac{x}{m}\right)

View answer and explanation

Correct option: C

0.18

Option C (0.18) is correct because it represents the amount of oxalic acid adsorbed per gram of charcoal, calculated from the change in concentration.

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}}

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 :