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3851.

On the basis of collision theory, explain the action of a catalyst on the rate of reaction

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Solution :The chemical reactions can be carried out at a FASTER rate by INCREASING the TEMPERATURE , but this is not possible in case of all reaction, Moreover, the reactions can be carried out of a lower temperature by using a catalyst . Since the process takes place at a lower place at a lower temperature, the cost of production gets reduced. A catalyst increases the rate of reaction by providing an alternative path for the reaction. This alternative path is associated with lower activation energy compared to the uncatalysed reaction, which is evident from the following graph.

In most of the reactionscatalysts increases the rate of reactions. These types of catalysts are called positive catalysts.
Example : `2SO_(2) + O_(2) overset(NO)to 2SO_(3)`
There are few reactions in which catalysts slow down the rate of reactions. These type of catalysts are called negative catalysts.
Example: Decomposition of hydrogen peroxide can be slowed down if some substances LIKE glycerine, urea, ACETANILIDE, sodium pyrophosphate etc. are added to `H_(2)O_(2)`.
3852.

On reaharing themobile________ arereversed .

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SOLUTION :CHEMICAL REACTIONS
3853.

On passing exces of CO_(2) gas in an aqueous solution of calcium carbonate milkness of the solution

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Persists
Fades
Deepens
Disappears

Answer :B
3854.

On passing excess carbon dioxide gas through lime water, it first turns milky and then becomes colourless, explain why ? Write all the chemical equation related to it.

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Solution :On passing carbon DIOXIDE through LIME water, initially a precipitate of calcium carbonate is formed and on passing excess of carbon cioxide the calcium carbonate is REDUCED to calcium bicarbonate, which is soluble in aqueous medium , hence the solution becomes colourless.
`Ca(OH)_(2)(aq.)+CO_(2)(g)toCaCO_(3)(s)+H_(2)O(l)`
`CaCO_(3)(s)+CO_(2)(g)+H_(2)O(l) TOCA(HCO_(3))_(2)(aq.)`.
3855.

On passing electricity through an acidified solution of copper (II) sulphate - A. hydrogen ions are reduced at cathode ; B. copper (II) ions are reduced at cathode ; C. copper (II) ions are oxidised at anode ; D. hydorgen gas liberates at cathode .

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hydrogen ions are reduced at cathode .
COPPER (II) ions are reduced at cathode .
copper (II) ions are oxidised at ANODE .
HYDORGEN gas liberates at cathode .

Answer :B
3856.

On passing a mixture of ethane and oxygen through a hot copper tube is _____formed (ethanol/CO_2)

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SOLUTION :ETHANOL
3857.

On oxidation, ethanol gives ethanoic acid, with alkaline :

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`KMnO_(4)`
`K_(2)Cr_(2)O_(7)`
`KNO_(3)`
`NaMnO_(4)`

Answer :A
3858.

On moving down in a group, the atomic radii of elements increases gradually.

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Solution :On moving down in a GROUP, the atomic number of element INCREASES.
A new shell of electrons is added with increase in atomic numbers of ELEMENTS.
Thus, the DISTANCE between the valence shell and nucleus increases and atomic SIZE increases down the group inspite of the increase in nuclear charge.
3859.

On moving down in any group, the metallic character of the elements ............

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SOLUTION :INCREASES
3860.

On increasing temperature, conduction in metallic conductors ______

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INCREASES
decreases
remains constant
none of these

Answer :B
3861.

On immersing the Zn rod in the solution of copper sulphate, you will observe ...

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DEPOSITION of Cu on Zn
deposition of Zn on Cu.
`Cu^(2+)` oxidises.
blue coloured SOLUTION become more dark.

Solution :deposition of Cu on Zn
3862.

On immersing an iron nail in CuSO_(4) solution for few minutes, what will you observe ?

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SOLUTION :Solution TURNS LIGHT GREEN
3863.

When a solid is heated, it turns directly into a gas. This process is called

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sublimation
evaporation
diffusion
condensation

Solution :The conversion of SOLID DIRECTLY into GAS or gas into solid is CALLED sublimation.
3864.

On heating potassiumchlorate ,itdecomposesinto________and_______gas.

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SOLUTION :Potassiumchloride, OXYGEN
3865.

On heating in air , iron forms ……………… .

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SOLUTION : `Fe_(3)O_(4)`
3866.

On heating, hydrated crystalline salts lose their ....................

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SOLUTION :WATER of CRYSTALLIZATION
3867.

On heating ferrous sulphate srystals, we obtain

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a mixture of FERRIC oxide and sulphur dioxide.
a mixture of ferrous oxide and sulphur trioxide.
a mixture of ferric oxide, sulphur dioxide and sulphur trioxide.
a mixture of ferrous oxide, sulphur dioxide and sulphur trioxide.

SOLUTION :The reaction takes PLACE as under
`2FeSO_(4)(s)OVERSET("Heat")(to)underset("Ferric oxide")(Fe_(2)O_(3)(s))+SO_(2)(g)+SO_(3)(g)`
3868.

On heating copper sulphate pentahydrate crystals .....................

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BLUE COLOUR changes into GREEN
green colour changes into blue
blue colour changes into colourless
no colour changes

Solution :blue colour changes into colourless
3869.

On heating blue coloured powder of copper (II) nitrate in a boiling tube, copper oxide (black), oxygen gas and a brown gas X is formed. What could be the pH range of aqueous solution of the gas X ?

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SOLUTION :`NO_(2)` is an OXIDE of non-metal. THEREFORE on dissolving in water, it will produce an acidic solution.
The pH of the solution will be LESS than 7.
3870.

On heating blue coloured powder of copper(II) nitrate in a boiling tube, copper oxide (black), oxygen gas an a brown gas 'X' is formed. Identify the brown gas 'X'.

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Solution :`UNDERSET("Copper(II) nitrate")(2CU(NO_(3))_(2)(s))OVERSET("Heat")rarrunderset("Copper oxide")(2CuO(s))+O_(2)(g)+4NO_(2)(g)`
.X. is nitrogen dioxide gas.
3871.

On heating blue coloured powder of copper (II) nitrate in a boiling tube, copper oxide (black), oxygen gas and a brown gas X is formed. Identify the type of reaction.

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SOLUTION :It is a DECOMPOSITION REACTION.
3872.

On heating blue coloured powder of copper (II) nitrate in a boiling tube, copper oxide (black), oxygen gas and a brown gas X is formed. Write a balanced chemical equation of the reaction.

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SOLUTION :`2CU(NO_(3))_(2)overset("Heat")(to)2CuO+4NO_(2)+O_(2)`
3873.

On heating blue coloured powder of copper (II) nitrate in a boiling tube, copper oxide (black), oxygen gas and a brown gas X is formed. Identify the brown gas X evolved.

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Solution :The BROWN gas in NITROGEN DIOXIDE `(NO_(2))`.
3874.

On dropping a small piece of sodium in a test-tube containing carbon compound 'X' with molecular formula C_(2)H_(6)O, a brisk effervescence is observed and a gas Y is produced. On bringing a burning splinter at the mouth of the test-tube the gas evolved burns with a pop sound. Identify 'X' and 'Y'. Also write the chemical equation for the reaction. Write the name and structure of the product formed, when you heat 'X' with excess conc. sulphuric acid.

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SOLUTION :`X-C_(2)H_(5)OH,Y-H_(2)` GAS
`2C_(2)H_(5)OH+2Na rightarrow 2C_(2)H_(5)ONa+H_(2)uparrow`
Ethene `C_(2)H_(4).`
3875.

On dilutingsodium meta aluminatewith water , a precipitateof ________ si formed .

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SOLUTION :ALUMINIUM HYDROXIDE
3876.

On diluting a solution of pH of 4, its pH :

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REMAINS same
DECREASES
INCREASES
Becomes 14

Solution : Increases
3877.

On cooling,pure ethanol is frozen to form ice like flakes.

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Solution :On cooling, PURE ETHANOIC ACID is frozen to FORM ice like FLAKES.
3878.

On combustion, one mole of a hydrocarbon gives 2 moles of carbon dioxide and two moles of water and a large quantity of heat is produced. The hydrocarbon is

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methane
ETHANE
ethene
ethyne

ANSWER :C
3879.

On being heated , baking soda undergoes _____ to give sodium carbonate, water and carbon dioxide.

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ANSWER :DECOMPOSITION
3880.

Onanalysingan impuresampleof sodiumchloride, thepercentageof chloridewasfoundto be45.5whatis thepercentageofpuresodiumchloridein thesample?

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Solution :Molecular(orFormula )
MASSOF pureNaCl= Atomicmassof `Na+`
Atomicmassof CL
`=23+35.5=58.5 u`
Percentageof CHLORINEIN pureNaCl
`=(35.5 )/(58.5 )xx100 =60.6`
Now, ifChlorineis60.6 parts , NACL= 100 Parts
Ifchlorineis 45.5parts,NaCl `=(45.5 )/(60.6 ) xx 100 =75 `
Thus , PERCENTAGE of pure`NaCl =75 %`
3881.

On adding dilute hydrochloric acid to copper oxide powder, the solution formed is blue-green. Predict the new compound formed which imparts a blue - green colour to the solution.

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Solution :The new compound formed is CUPROUS CHLORIDE. It IMPARTS a BLUE - green COLOUR to the solution.
3882.

On adding dilute HCl to copper oxide powder, the solution formed is blue- green. Predict the new compound formed which imparts a blue-green colour to the solution.

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SOLUTION :The new compound formed is copper(II) chloride `(CuCl_(2))`, which IMPARTS blue-green colour to the solution.
`UNDERSET("Copper oxide")(CuO)+2HClrarrunderset(("Blue-green"))underset("Copper(II) chloride")(CuCl_(2))+H_(2)O`
3883.

On adding acetic acidto NaHCO_(3) in a test tube , a student observes

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No reaction
A COLOURLESS GAS has with pungent SMELL
BUBBLES of a colourless and odourless gas
A strong smell of vinegar.

Solution :Bubbles of a colourless and odourless gas
3884.

Olfactory indicators are

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Clove
Turmeric
Soap
rose petals

Answer :A
3885.

Oils on treating with hydrogen in the presence of palladium or nickel catalyst form fats. This is an example of …

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ADDITION REACTION.
substitution reaction.
REARRANGEMENT reaction.
OXIDATION reaction.

Answer :A
3886.

Oil used ibn Froth floatation method is …………….

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PINE OIL
NATURAL oil
CRUDE oil
Synthetic oil

Answer :A
3887.

Oil and fat containing food items are flushed with nitrogen. Why?

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SOLUTION :In the presence of oxygen in the air, the fats PRESENT in the fatty FOOD are oxidised to compounds which have a bad smell, i.e., the food BECOMES RANCID. Flushing with nitrogen cuts off the contact of food with oxygen and protects the food from rancidity.
3888.

Oil and fat containing food items are flushed with nitrogen. Why ?

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Solution :Nitrogen being an inert gas, does not REACT easily with the oil and FAT present in the food substances. But oxygen REACTS easily with the food substances and . makes them RANCID. Hence food items containing oil and fat are FLUSHED with nitrogen gas before their packing to remove oxygen present in the pack.
3889.

Of what must substances must the anode be made ?

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SOLUTION :SILVER METAL
3890.

Of the three metals X,Y and Z, X reacts with cold water. Y with hot water and Z with steam only. Identify X,Y and Z and also arrange them in order of increasing reactivity.

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Solution :X is alkali metal Na or K.
Y is an ALKALINE earth metal, MG or CA
Z is Fe
INCREASING reactivity series: `Na gt Mg gt Fe`
3891.

Of the following what is the pH of a solution which turns red litmus blue?2 468

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2
4
6
8

Answer :D
3892.

Of the following pairs, the one containing example of metalloids in the periodic table is

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SODIUM and potassium
fluorine and chlorine
calcium and magnesium
boron and silicon

Answer :D
3893.

Octet rule is not violated in case of

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`CH_(4)`
`BF_(3)`
`SF_(6)`
`AlCl_(3)`

ANSWER :a
3894.

Octer rule is not violated in case of

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methane
boron trifluoride
sulphur hexafluoride
aluminnium chloride

Solution :In `CH_(4)` methane C is bonded to four HYDROGENS and has 8 ELECTRONS in its valence shell, hence, it has a complete octet.
In `BF_(3)` and `AlCl_(3)` the central atoms has six electrons while in `SF_(6)`, S has 12 electrons in its valance shell.
3895.

Obtain the IUPAC name of the following compounds systematically. (i) CH_(3)-CH_(2)-CH_(2)-CH_(2)-CH_(3) (ii) CH_(3)-overset(CH_(3))overset(|)(CH)-CH_(2)-CH_(2)-CH_(3)

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Solution :(i) `CH_(3)-CH_(2)-CH_(2)-CH_(2)-CH_(3)`
Step 1: It is a five- CARBON chain and hence the root word is .PENT.. (RULE 1)
Step 2: All the bonds between carbon atoms are single bonds, and thus the suffix is .ane..
So, its name is Pent + ane = Pentane
(II) `CH_(3)-overset(CH_(3))overset(|)(CH)-CH_(2)-CH_(2)-CH_(3)`
Step 1: The longest chain contains five carbon atoms and hence the root word is .Pent..
Step 2: There is a SUBSTITUENT. So, the carbon chain is numbered from the left end, which is closest to the substituent. (Rule 2)
`underset(to)(overset(1)(C)H_(3)-overset(CH_(3))overset(|2)(CH)-overset(3)(C)H_(2)-overset(4)(C)H_(2)-overset(5)(C)H_(3))`
Step 3: All are single bonds between the carbon atoms and thus the suffix is .ane..
Step 4: The substituent is a methyl group and it is located at second carbon atom. So, its locant number is 2. Thus the prefix is .2-Methyl.. (Rule 6).
The name of the compound is 2-Methyl + pent +ane = 2-Methylpentane
3896.

Obtain the IUPAC name of the following compound systematically. CH_(3)CHO (ii) CH_(3)CH_(2)COCH_(3) (iii) ClCH_(2)-CH_(2)-CH_(2)-CH_(3)

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Solution :(i) `CH_(3)CHO`
Step 1 : The parent chain consists of 2 carbon ATOMS.
The root word is ..Eth...
Step 2 : All are single bonds between carbon atom of the chain. So the primary suffix is ..ane...
Step 3 : Since the compound contains the - CHO GROUP, it is a aldehyde. The secondary suffix is ..al...
`therefore` The NAME of the compound is Eth + ane + al = Ethanal
(ii) `CH_(3)CH_(2)COCH_(3)`
Step 1 : The parent chain consists of 4 carbon atoms.
The root word is ..But...
Step 2 : All are single bonds between carbon atom of the chain. So the primary suffix is ..ane...
Step 3 : Since the compound contains the - CO - group, it is a ketone group. The secondary suffix is ..one...
`therefore` The name of the compound is But + ane + one = Butanone
(iii) `ClCH_(2)-CH_(2)-CH_(2)-CH_(3)`
Step 1 : The parentchain consists of 4 carbon atoms.
The root word is But...
Step 2 : All the single bonds between carbon atom of the chain. So the suffix is ..ane...
Step 3 : Since the compound contains the - Cl substituent. The prefix is Chloro.
Step 4 : The locant number of - Cl is 1 and thus prefix is 1 -Chloro.
`therefore` The name of the compound is 1 -Chloro + But - ane = 1 -Chlorobutane.
3897.

Observe the given table and question: answer the following {:("Elements", A, B, C, D, E),("Atomic number", 11,4,2,7,19):} Identify the two elements that belong to the same period and the two elements that belong to the same group. Give reason for your conclusion.

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Solution :The electronic CONFIGURATIONS of the GIVEN elements are: A (11) = (2, 8, 1), B (4) =(2, 2), C (2) = (2), D (7) = (2,5), E (19) = (2, 8, 8, 1). B and D BELONG to same period since they have two shells each. A and E belong to same group since they have same number of valence electrons, i.e., I.
3898.

Observe the given figure. Name the eye defect indicated in the figure and also mention the lens used to correct this defect.

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SOLUTION :MYOPIA
CONCAVE LENS
3899.

Observe the given table and answer the following question : Identify the two elements that belong to the same period and the two elements that belong to the same group. Give reason for your conclusion.

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SOLUTION :Element B and element D are in same period because their atoms have two shells.
Element A and element E are in the same group because their OUTERMOST shell has ONE electron.
3900.

Observe the given figure. What type of current is induced in the coil by doing the experiment related to this figure ? Give reason for your answer.

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Solution :• According to Lenz Jaw, the DIRECTION of induced current is such that it always opposes the cause which produce it.
• In the given FIGURE POINT B becomes north POLE due to induced current.
THUS, the direction of current induced is anticlockwise.
• As a result, galvanometer shows deflection towards left.