Saved Bookmarks
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 50651. |
A bubble of air is underwater at temperature 15^@ C and the pressure 1.5 bar. If the bubble rises to the surface where the temperature is 25^@ C and the pressure is 1.0 bar, what will happen to the volume of the bubble? |
|
Answer» VOLUME will BECOME GREATER by a FACTOR of 1.6 |
|
| 50652. |
A bubble of a gas released at the botton of a lake increases to eight times its original volume when it reaches the surface. Assuming that atmospheric pressure is equivalent to the pressure exerted by a column of water 10 m height, the depth of the lake is |
|
Answer» 80 m |
|
| 50653. |
A brown ring is formed in the ring test for NO_(3)^(-) ion. It is due to the formation of |
|
Answer» `[Fe(H_(2)O)_(5)(NO)]^(2+)` |
|
| 50654. |
A brominated alkane an analysis gave 12.8% carbon and 2.1%H. If its vapour density is 93.95, what is its molecular formula? |
|
Answer» Solution :PERCENT weight of bromium = 100-(%C+%H) = 100-14.9 = 85.1 C : H: Br `=(12.8)/(12) : (2.1)/(1) : (85.1)/(80) = 1 : 2 : 1 ` Empirical formula is `CH_(2)Br` Molecular weight `= 2xxV.D. = 2xx93.95 = 187.9` ` n = (187.9)/(94) = 2` Molecular formula `=2xxCH_(2)Br = C_(2)H_(4)Br_(2)` |
|
| 50655. |
A+BrarrC+D, DeltaH=-10.000Jmol^(-1),DeltaS=-33.3mol^(-1)K^(-1) (i) At what temperature the reaction will occur spontaneously from left to right? (ii) At what temperature, the reaction will reverse? |
|
Answer» SOLUTION :`DeltaG=DeltaH-TDeltaS` At equilibrium, `DeltaG=0` so that `DeltaH=TDeltaS` (or) `T=(DeltaH)/(DeltaS)=(-10000Jmol^(-1))/(-33.3Jmol^(-1))=300.3K` (i) For spontaneity from left to RIGHT, `DeltaG` should be -ve for the given reaction. This will be so if `Tlt300.3K` (ii) For reverse reaction OT OCCUR. `DeltaG` should be `+ve` for FORWARD reaction. This will be so if `Tgt300.3K`. |
|
| 50656. |
A boy after swimming comes out from a poolcovered with a film of waterweighing 80 g .How much heat must be suppliedto evaporate this water ? (Delta_(v) H^(@)=40.79 kJmol^(-1)) |
|
Answer» `1.61 xx10^(2) KJ` WATER to be evaporated`= 80 g( 80)/( 18) MOLE` `=4.44` mole Heat required for evaporated of 1 moleof `H_(2)O =40.79kJ` `:. `Heat required for evaporated of 4.44 moleof`H_(2)O=40.79 xx 4.44 = 1.81 xx 10^(2) kJ` |
|
| 50657. |
A box contains some identical red coloured balls, labelled as A, each weighing 2g. Another box contains identical blue coloured balls, labelled as B, each weighing 5 g. Consider the combinations AB, AB_(2), A_(2)B and A_(2)B_(3) and show that law of multiple proportions is applicable. |
|
Answer» Solution :  MASS of B which is combined with fixed mass of A will be 2.5 g, 5G, 1.25gand 3.75g. They are in the ratio of `2:4 :1:3` which is a simple whole NUMBER ratio. Hence, the LAW of multiple proportions is applicable. |
|
| 50658. |
A bottle of NH_(3) gas and a bottle of dry HCl gas are connected through a long tube. The tube is opened simultaneously at both the ends. White fume of NH_(4)Cl is formed …………….. |
|
Answer» throughout the length of the TUBE Atomic pressure in HCl gas = 36.5 gm `mol^(-1)=M_(2)` According to Grapham.s gas diffusional Law,Whose atomic pressure is high its diffusion is slow.`(r=(1)/(sqrt(M)))` The atomic pressure of HCl (36.5 gm `"mole"^(-1)`) is more then the atomic pressure of `NH_(3)` (17.0 gm `"mole"^(-1)`) So, the diffusion of HCl is LESS. `THEREFORE` The diffusion of HCl is very less. `therefore` So, the FUME of `NH_(4)Cl` is formed `NH_(3)` bottle. |
|
| 50659. |
A bottle of H_(3)PO_(4) solution contains 70%(w//w) acid. If the density of the H_(3)PO_(4) solution required to prepare 1 L of 1N solution is : |
|
Answer» 90 mL |
|
| 50660. |
A bottle of commercial sulphuric acid (density = 1.787 gm/mL) is labelled 86 percent by weight. What is the molarity of the solution ? What volime of the acid is required to make it 1 litre of 0.2 M H_(2)SO_(4) ? |
|
Answer» Mass of `H_(2)SO_(4)=86 G` , Mass of the solution = 100 g Volume of 100 g of the solution `= ("Mass")/("Density") = (100)/(1.787) = 55.96 ML = 0.05596 L` Molarity of solution (M) `= (("Mass of "H_(2)SO_(4))/("Molecular mass of "H_(2)SO_(4)))/("Volume of solution in mL"/(1000))=((86g))/(("98 g mol"^(-1))xx("0.05596 L"))` `=15.68 "mol L"^(-1)=15.68 M` Step II. Volume of solution required `overset(("conc."))(M_(1)V_(1))-=overset(("dilute"))(N_(2)V_(2))` `15.68xxV_(1)=0.2xx1000,V_(1)=(0.2xx1000)/(15.68)=12.75mL`. |
|
| 50661. |
A bottle of H_2 O_2is labelled as 10 vol H_2O_2 112ml of this H_2O_2 solution is completely titrated with 0.04M acidified KMnO_4solution. The volume of KMnO_4consumed in litres is |
|
Answer» `10/(5.6)xx112=0.04xx5xxV_2` VOLUME of `KMnO_4`solution = 1 LIT |
|
| 50662. |
A bottle of ammonia and a bottle of HCl connected through a long tube are opened simultaneously at both ends. The white ammonium chloride ring first formed will be |
|
Answer» At the center of the tube `m_(NH_3) = 17, m_(HCL)= 36.5` `gamma_(NH_3) gt gamma_(HCl)` Hence white fumes first formed near hydrogen chloride. |
|
| 50663. |
A bottle of ammonia and a bottle of HCl connected through a long tube are opend simultaneously at both ends. The white ammonium chloride ring first formed will be |
|
Answer» At the CENTER of the tube |
|
| 50664. |
A bottle contains two immiscible liquids. They may be separated by |
|
Answer» VACUUM distillation |
|
| 50665. |
A bottle contaning ammonia and a bottle containing hydrogen chloride are connected throughalong tube are opened simultaneously at both ends. The white ammoniumchloride first formed will be : |
|
Answer» at the centre of the tube |
|
| 50666. |
A bottle is heated with mouth open to have a final temperature as five times its original value at 25^(@)C. The fraction of air originally present in the bottle that is expelled, is: |
|
Answer» 0.5 |
|
| 50667. |
(A) : Both H_(2) and D_(2) have same internuclear distance (R) : Enthalpy of bond dissociation is same for H_(2) and D_(2) |
|
Answer» A is TRUE but R is false |
|
| 50668. |
(A): Borax bead test is not suitable for Al(III) (R) : Al_2 O_3 is insoluble in water |
|
Answer» A and R are true, R EXPLAINS A |
|
| 50669. |
Abonding molecular orbital is produced by |
|
Answer» Destructive interference of WAVE FUNCTIONS |
|
| 50670. |
A bolttle of dry ammonia and a bottle of dry hydrogen chloride connected through a long tube are opened simultaneously at both ends, The white ammonium chloride ring first formed will be |
|
Answer» at the centre of the TUBE |
|
| 50671. |
(A) : Bohr.s orbits are called stationary orbits (R) : Electrons remain stationary in these orbitss |
|
Answer» Both (A) and (R) are true and (R) is the correct EXPLANATION of (A) |
|
| 50672. |
A bodyof mass x kg ismovingwithvelocityof 100 ms^(-1)itsde - brogliewavelengthis 6.62 xx 10^(-34)m^(-1)Hencex is … (h= 6.62 x 10^(-34) J s) |
|
Answer» `0.25 KG` |
|
| 50673. |
A body of mass x kg is moving with avelocity of 100 ms^(-1) , its de-Broglie wavelength is 6.62xx10^(-35) m . Hence x is (h=6.62xx10^(-34) Js) |
|
Answer» 0.25 kg |
|
| 50674. |
A body centred unit cell has an atom at the each vertex and at --------- of the unit cell. |
| Answer» SOLUTION :CENTRE | |
| 50675. |
A blue coloured mineral 'lapis lazuli' which is used as a semi-precious stone is a mineral of the following class : |
|
Answer» SODIUM aluminiosilicate |
|
| 50676. |
A 500g tooth paste sample 0.3 g fluoride. What is the concentration of fluoride in ppm? |
|
Answer» 50 ppm |
|
| 50677. |
A bivalent metal has an equivalent mass of 32. The molecular mass of the metal nitrate is |
|
Answer» 168 `:.` Atomic mass of `MM = 32 xx 2 = 64` The metal nitrate formed has the formula `M(NO_(3))_(2)` `:.` Molecular mass of the metal nitrate `= 64+28+96=188` |
|
| 50678. |
A bivalent metal has 37.2 equivalent weight. The molecular weight of its chloride is |
|
Answer» `216.6` |
|
| 50679. |
Equivalent weights 36. A bivalent metal has 12 equivalent weight. The molecular weight of its oxide is |
|
Answer» 16 |
|
| 50680. |
A binary solid, A^(+) B^(-) ( formula mass= 60) has aCsCl structure. The no. of unit cells present in 6 g ideal crystals of solid are |
|
Answer» `6.02xx10^(22)` |
|
| 50681. |
A binary salt AB (formula weight = 6.023 Y amu, where Y is an arbitray number) has rock salt structure with 1 : 1 ratio of A to B. The shortes A - B distance in the unit cell is Y^(1//3)nm. (a) Calculate the density of the salt in kg m^(-3) (ii) Given that the measured denstiy of the salt is 20 kg m^(-3), specify the type of point defect present in the crystal |
|
Answer» |
|
| 50682. |
A binary compound formed between the elements with atomic numbers 19 and 17 is expected to be |
|
Answer» a WATER soluble compound forming a CONDUCTING solution in water |
|
| 50683. |
A+BhArrC+D,K_(c) for this reaction is 10. If 1,2,3,4 mole/litre of A,B,C and D respectively are present in a container at 25^(@)C, the direction of reaction will be |
|
Answer» From LEFT to RIGHT |
|
| 50684. |
A benzene ring deactivated by strong & moderate electron withdrawing groups such a molecule is not electron rich enough to under go friedel-craft's reaction Friedel craft reaction also does not occur with NH_(2) group as it react with AlCl_(3) and produce deactivating group. Q. Which of the following sequence of reaction is correct for the synthesis of product |
|
Answer»
|
|
| 50685. |
A benzene ring deactivated by strong & moderate electron withdrawing groups such a molecule is not electron rich enough to under go friedel-craft's reaction Friedel craft reaction also does not occur with NH_(2) group as it react with AlCl_(3) and produce deactivating group. Q. Which of the following cannot be starting material for this compound Ph-underset(O)underset(||)(C)-CH_(2)-Ph |
|
Answer»
|
|
| 50686. |
A benzene ring deactivated by strong & moderate electron withdrawing groups such a molecule is not electron rich enough to under go friedel-craft's reaction Friedel craft reaction also does not occur with NH_(2) group as it react with AlCl_(3) and produce deactivating group. Q. Which of the following compound undergoes Friedel-crafts alkylation reaction |
|
Answer»
|
|
| 50687. |
A benzene ring deactivated by strong and moderate electron withdrawing group that is, any meta directing group, is not electron rich enough to undergo Friedel-Crafts ranctions. Friedel-Crafts reaction also do not occur with HN_(2) group as it react with AICI_(3) and produce deactivating group. Answer the following question : Which of the following cannot be starting for this compound Ph-underset(O)underset(||)(C)-CH_(2)-Ph ? |
|
Answer»
|
|
| 50688. |
(A) : Benzene prefers to participate in substitution reactions than addition reactions (R) : Addition products of benzene do not retain resonance stability |
|
Answer» Both A and R are TRUE, and R is CORRECT EXPLANATION of A |
|
| 50689. |
(A) Benzene on heating conc. H_(2)SO_(4) gives benzene sulphonic acid which heated with supherheated stem under pressure gives benzene. ( R) suplhonation is a reversible precess. |
|
Answer» If both (A) and ( R) are CORRECT and ( R) is CORRCT EXPLANATION of (A). |
|
| 50690. |
(A) : Benzene decolourises cold alkaline potassium permanganate solution (R) : Benzene is aromatic hydrocarbon |
|
Answer» Both A and R are TRUE, and R is CORRECT EXPLANATION of A |
|
| 50691. |
(a) [BeF_(4)]^(2-) exits, but [BeCl_(6)]^(4-) does not. Give reson. (b). Hydrated beryllium ion exists as [Be(H_(2)O)_(4)]^(2+), whereas hydrated magnesium ion exists as [Mg(H_(2)O)_(6)]^(2+). Give reason. |
Answer» Solution :a. `[BeF_(4)]^(2-)` EXISTS, but `[BeF_(6)]^(4-)` does not. This can be explained on the basis of valence ELECTRONIC configuration:  DUE to the absence of low-lying `d`-orbitals in beryllium, it cannot expand its coordination number beyond`4`, hence `[BeF_(4)]^(2-)` exists, but `[BeF_(6)]^(4-)` does not. b. Hydrated beryllium ion exists as `[Be(H_(2)O)_(4)]^(2+)`.  Due to the presence of only four orbitals of equivalent energy in `Be^(2+)` ion and due to the absence of low-lying `d-`orbitals in `Be^(2+)` ion, it can coordinate with four water molecules only, hence hydrated beryllium ion exists as `[Be(H_(2)O)_(4)]^(2+)`. Whereas in case of MAGNESIUM, due to the availability of low-lying d-orbitals of SUITABLE energy, `Mg` can expand its coordination number to 6  Hence, hydrated magnesium ion exists as `[Mg(H_(2)O_(6))]^(2+)` |
|
| 50692. |
(A): BeF_2 is predominently a covalent compound. (R) : Electronegativity difference between Be and F is too small |
|
Answer» Both (A) and (R) are TRUE and (R) is the correct EXPLANATION of (A) |
|
| 50693. |
A beam of specific kind of particles of velocity 2.1 xx 10^7 m/s is scattered by a gold (z = 79) nucleli, Find out specific charge (charge/mass) of this particle if the distance of closet approach is 2.5 xx 10^(-14) m. |
|
Answer» `4.84xx 10^7 ` C/kg `implies(q_2)/(m) = (2.5 xx 10^(-14)xx (2.1 xx 10^7)^2)/(2xx 9xx10^9 xx 79 xx 1.6 xx 10^(-19))` `implies4.84 xx 10^7` coulomb/kg |
|
| 50694. |
Assertion: Be and Al have similar properties. Reason : Cations of Be and Al have same polarising power |
|
Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
|
| 50695. |
A bcc lattice is made up of hollow spheres of X. Spheres of solid 'Y' are present in hollow spheres of X. The radius of 'Y' is half of the radius of 'X' . Calculate the ratio of the total volume of spheres of 'X' unoccupied by Y in a unit cell and volume of the unit cell ? |
|
Answer» Solution :Let the RADIUS of HOLLOW sphere X be r. As spheres X have BCC structure, edge length (a)=`"4r"/sqrt3 "" (therefore "for bcc", r=sqrt3/4 a)` `therefore` VOLUME of the unit cell =`a^3=("4r"/sqrt3)^3` Radius of sphere Y=`r/2` (Given ) `therefore` Volume of sphere `X=4/3pir^3` Volume of sphere `Y=4/3 pi(r/2)^3` `therefore` Volume of X unoccupied by Y in unit cell =`2xx[4/3pir^3-4/3pi(r/2)^3]` ( `because` bcc lattice has 2 spheres per unit cell ) `=2xx4/3pir^3(1-1/8)=2xx4/3pir^3xx7/8` `therefore "Volume of X unoccupied by Y in unit cell"/"Volume of unit cell"=(2xx4/3pir^3xx7/8)/((4r)/sqrt3)^3=(7pisqrt3)/64` |
|
| 50696. |
A bcc lattic is made up of hollow spheres of X. spheres of soldid 'Y,' are present in hollow spheres of X. The radius of 'Y'is half of the radius of 'X' . Calculate the ratio of the total volume of spherees of 'X' unoccupied by Y in a unitcell and volume of the unit cell ? |
|
Answer» Solution :Let the radius of hollow SPHERE X be r. As spheres, X have bccstructure, edge lenth `(a)= (4R)/(sqrt3)` Volume of the unit CELL = ` a^(3) = ( (4r)/(sqrt3))^(3)` Radius of shphere Y =` r/2 ` (Given ) Volume of Sphere X =` 4/3 PIR^(3)` volume of sphere Y ` 4/3 pi (r/2)^(3)` Volume of X unoccupied by Y in unit cell = `2 xx [ 4/3 pi r^(3)- 4/3 pi ( r/2)^(3) ]` ( bcc lattice has 2 spheres per unit cell ) ` 2xx 4/3 pi r^(3) (1-1/8) = 2xx 4/3 pir^(3) xx 7/8` `(" Volume of X unoccpied by Y in unit cell")/( " volume of unit cell")= (2xx 4/3 pir^(3) xx7/8)/(((4r)/sqrt3)^(3))= ( (7 pisqrt3))/64` |
|
| 50697. |
A basic lining is given to a furnace by using |
|
Answer» Calcined dolomite |
|
| 50698. |
A basic buffer contains equal concentration of base and its salt . The dissociation of base is 10^(-6) . Then the pH of the buffer solution is |
|
Answer» 6 |
|
| 50699. |
(a) Based on the nature of intermolecular forces, classify the following solids . silicon carbide, Argon (b) ZnO turns yellow on heating , why ? (c ) what is meant by groups 12-16 compounds . Give an example. |
|
Answer» (b) N/A (C ) N/A |
|
| 50700. |
(a) Based on the nature of intermolecular forces, classify the following solids. Benzene, Silver (b) AgCl shows Frenkel defect while NaCl does not. Give reason. (c ) What type of semiconductor is formed when Ge is doped with Al ? |
|
Answer» (B) Elecrical conductivity increases(c ) N/A |
|
