100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS Page 1950 InterviewSolution

97451.	A coil of radius R carries a current `I`. Another concentric coil of radius `r (r lt lt R)` carries current `(1)/(2)`. Initially planes of the two coils are mutually prependicular and both the coil are free to rotate about common diameter. They are released from rest from this position. The masses of the coils are M and m respectively. `(m lt M)`. During the subsequeny motion let `K_(1)` and `K_(2)` be the maximum kinetic energies of the two cells respectively. and let U be the magnetic of maximum potential energy of magnetic interaction of the system of the coils. Choose the correct options.A. `(K_(1))/(K_(2))=(M)/(m)((R)/(r))^(2)`B. `K_(1)=(Umr^(2))/(mr^(2)+MR^(2))K_(2)=(UMR^(2))/(mr^(2)+MR^(2))`C. `U=(mu_(0)pi l^(2)r^(2))/(4R)`D. `K_(2)gt gt K_(1)`
Answer» Correct Answer - B::C::D `tau_(1)=tau_(2)rArr (int tau dt)/(I) =omega rArr omega prop (1)/(I)` `K.E =(1)/(2) I omega^(2) prop (1)/(I)` `(K_(1))/(K_(2))=((mr^(2))/(2))/((MR^(2))/(2))rArr (K_(1))/(K_(2))=(m)/(M)((r)/(R))^(2)` `K_(1)+K_(2)=U rArr K_(2)(m)/(M)((r)/(R))^(2)+K_(2)=U` `K_(2)=(UMR^(2))/(mr^(2)+MR^(2))rArr K_(1)=(Umr^(2))/(mr^(2)+MR^(2))` `U=(I)/(2)pi r^(2)(mu_(0)I)/(2R)=(mu_(0)I^(2)pi r^(2))/(4R)`

Discussion

97452.

Distinguish between wireline and wireless communication.

Answer»

Wireline communication	Wireless communication
It is a Point-to-point communication	It i a broadcast mode communication
It uses mediums like wire, cable and optical fibers.	it uses free space as a communication medium
These systems cannot be used for long distance transmission as the are connected.	These systems can be used for long distance transmission.
Ex. telephone, intercom and cable TV.	Ex. mobile, radio or TV broadcasting and satellite communication.

Discussion

97453.	A proton enters a magnetic field of flux density 1.5 Wb/m2 with a speed of 2 x 107 m/s at angle of 30°v with the field. The force on the proton will be ………(a) 0.24 x 10-12 N (b) 2.4 x 10-12 N (c) 24 x 10-12 N (d) 0.024 x 10-12 N
Answer» (b) 2.4 x 10^-12 N F = Bqv sin θ = 1.5 x 1.6 x 10^-19 x 2 x 10⁷ x sin 30°= 2.4 x 10^-12 N

97453.

A proton enters a magnetic field of flux density 1.5 Wb/m2 with a speed of 2 x 107 m/s at angle of 30°v with the field. The force on the proton will be ………(a) 0.24 x 10-12 N (b) 2.4 x 10-12 N (c) 24 x 10-12 N (d) 0.024 x 10-12 N

Answer»

(b) 2.4 x 10^-12 N

F = Bqv sin θ = 1.5 x 1.6 x 10^-19 x 2 x 10⁷ x sin 30°= 2.4 x 10^-12 N

Discussion

97454.	Find force in newton which mass A exerts on mass B if B is moving towards right with 3ms-2.(All surfaces are smooth)
Answer» The correct answer is 5N.

Discussion

97455.	Find the reading of spring balance as shown in figure. Assume that M is in equilibrium. (All surfaces are smooth).
Answer» The reading of spring balance 6N.

Discussion

97456.	Why are substance such as platinum and palladium often used for carrying out electrolysis of aqueous solutions?
Answer» Substances such as platinum and palladium are often used for carrying out electrolysis of aqueous solutions due to the following reasons`:` `a.` Chemical inertness `:` Platinum and palladium are chemically inert. `b.` Adsoptive capacity `:` Platinum and palladium have good absorptive capacity for hydrogen. The adsorbed hydrogen is very active and hence platinum and palladium be pronounced catalytic activity.

Discussion

97457.	Explain the terms sorption and desorption.
Answer» [Hint : Sorption is used to describe the process when adsorption and absorption take place simultaneously. Desorption : Removal of adsorbate from the surface of adsorbent.]

97457.

Explain the terms sorption and desorption.

Answer»

[Hint : Sorption is used to describe the process when adsorption and absorption take place simultaneously.

Desorption : Removal of adsorbate from the surface of adsorbent.]

Discussion

97458.	'Chemisorption is highly specific.' Illustrate with an example.
Answer» [Hint : As it involves chemical bonding between adsorbent and adsorbate.]

Discussion

97459.	Name two compounds used as adsorbent for controlling humidity.
Answer» Silica gel, Alumina gel

Discussion

97460.	'Adsorbents in Finely divided form are more effective. Why ?
Answer» [Hint : Due to their more surface area in finely divided form.]

Discussion

97461.	Statement-1:All colloidal dispersions give very low osmotic pressure and show very small freezing point depression or boiling point elevation. Statement-2:Tyndall effect is due to scattering of light from the surface of colloidal particles.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True.
Answer» Correct Answer - B Colloidal particles have high molar mass, so their mole fraction is very less causing low colligative properties.

Discussion

97462.	Arsenic (III) sulphide forms a sol with a negative charge. Which of the following ionic substances should be most effective in coagulating the sol?A. `KCI`B. `MgCI_(2)`C. `AI_(2)(SO_(4))_(3)`D. `NaPO_(4)`
Answer» Correct Answer - a `As_(2)S_(3)` sol is a negatively charged colload `AI^(3+)` has greater charge, threfore `AI_(2)(SO_(4))_(3)` is most effective in the congulation of `As_(2)S_(3)` sol

Discussion

97463.	Match the compounds listed in column I with characteristic listed in column II. `{:("Column-I","Column-II"),((A)"Cyclic silicates",(p)"Tetrahedral hybridisation"),((B)"Single chain silicates",(q)"Si-O bonds are 50% ionic and 50% covalent"),((C )"Pyro silicates",(r)"General formula is")(SiO_3)_n^(2n),((D)"Sheet silicates (two dimensional ",(s)"Two oxygen atoms per tetrahedron are shared "),(,(t)"Square planar geometry"):}`
Answer» Correct Answer - A`to`p,q,r,s ; B`to`P,Q,R,S ; C`to`P,Q ; D`to`P,Q (A)Two oxygen atoms per tetrahedron are shared forming rings.`(SiO_3)_n^(2n-)`.Hybridisation of each Si is `sp^3`. (B)Two oxygen atoms per tetrahedran are shared forming a chain of tetrahedron, `(SiO_3)_n^(2n-)`, Hybridisation of each Si atom is `sp^3`. (C )One oxygen atom per tetrahedron is shared . `Si_2O_7^(2-)` .Hybridisation of each Si atom is `sp^3`. (D)Three oxygen atom per tetrahedron is shared . `(Si_2O_5)_n^(2-)` .`sp^3` hybridisation .

Discussion

97464.	If hydrolysis of any one of the ions will occur, after the dissolution of a sparingly soluble salt, then - (A) solubility of salt decreases. (B) solubility of salt increases (C) there will be no effect on solubility (D) question is absurd
Answer» Correct option: (B) solubility of salt increases Explanation: Dissolution equilibrium shift towards right side due to hydrolysis of cation or anion.

97464.

If hydrolysis of any one of the ions will occur, after the dissolution of a sparingly soluble salt, then - (A) solubility of salt decreases. (B) solubility of salt increases (C) there will be no effect on solubility (D) question is absurd

Answer»

Correct option: (B) solubility of salt increases

Explanation:

Dissolution equilibrium shift towards right side due to hydrolysis of cation or anion.

Discussion

97465.	The effect of sting of an ant can be neutralized by rubbing the area with1. Soap powder2. Baking soda3. Epsom salt4. Common salt
Answer» Correct Answer - Option 2 : Baking soda The correct answer is Baking soda Ant sting or bee sting contains methanoic acid. Methanoic acid is also known as formic acid. Bee or ant injects formic acid into our skin while biting. Injection of acid by ant or bee results in pain at the place of bite. Rubbing baking soda over the skin gives relief from pain due to ant or bee sting. Baking soda, which is a base, neutralizes the effect of acid injected by bee or ant. Another base, such as zinc carbonate (Calamite solution) is also used in case of ant or bee sting. Neutralization When a solution of acid is mixed with the solution of base, both of them neutralize each other and a third substance (called salt) is formed. Such a phenomenon is called neutralization or neutralization reaction. The solution formed because of the mixing of the solution of acid and base is neither acidic nor basic in nature. Such a solution is known as a neutral solution. Acid + Base → Salt + Water Salt formed because of neutralization reaction may be acidic or basic in nature. The acidic or basic nature of salt depends upon the strength of acid and base. In neutralization reaction heat is evolved. Reactions in which heat is evolved are known as exothermic reactions. Thus, the neutralization reaction is an exothermic reaction.

97465.

The effect of sting of an ant can be neutralized by rubbing the area with1. Soap powder2. Baking soda3. Epsom salt4. Common salt

Answer» Correct Answer - Option 2 : Baking soda

The correct answer is Baking soda

Ant sting or bee sting contains methanoic acid.
Methanoic acid is also known as formic acid.
Bee or ant injects formic acid into our skin while biting.
Injection of acid by ant or bee results in pain at the place of bite.
Rubbing baking soda over the skin gives relief from pain due to ant or bee sting.
Baking soda, which is a base, neutralizes the effect of acid injected by bee or ant.
Another base, such as zinc carbonate (Calamite solution) is also used in case of ant or bee sting.

Neutralization
- When a solution of acid is mixed with the solution of base, both of them neutralize each other and a third substance (called salt) is formed.
- Such a phenomenon is called neutralization or neutralization reaction.
- The solution formed because of the mixing of the solution of acid and base is neither acidic nor basic in nature. Such a solution is known as a neutral solution.
- Acid + Base → Salt + Water
- Salt formed because of neutralization reaction may be acidic or basic in nature.
- The acidic or basic nature of salt depends upon the strength of acid and base.
- In neutralization reaction heat is evolved. Reactions in which heat is evolved are known as exothermic reactions. Thus, the neutralization reaction is an exothermic reaction.

Discussion

97466.	Why does bee-sting cause pain and irritation?
Answer» Bee sting has acid that cause pain and irritation. Baking soda being alkaline, Neutralises acid and given relief.

Discussion

97467.	For polyatomic molecules having ‘f ’ vibrational modes, the ratio of two specific heats, \(\frac{C_p}{C_v}\) is(A) \(\frac{1+f}{2+f}\)(B) \(\frac{2+f}{3+f}\)(C) \(\frac{4+f}{3+f}\)(D) \(\frac{5+f}{4+f}\)
Answer» (C) \(\frac{4+f}{3+f}\)

97467.

For polyatomic molecules having ‘f ’ vibrational modes, the ratio of two specific heats, \(\frac{C_p}{C_v}\) is(A) \(\frac{1+f}{2+f}\)(B) \(\frac{2+f}{3+f}\)(C) \(\frac{4+f}{3+f}\)(D) \(\frac{5+f}{4+f}\)

Answer»

(C) \(\frac{4+f}{3+f}\)

Discussion

97468.	If for a gas \(\frac{R}{C_V}\) = 0.67, this gas is made up of molecules which are(a) monoatomic(b) diatomic(c) Polyatomic(d) mixture of diatomic and polyatomic molecules
Answer» (a) monoatomic For a gas, we know \(\frac{R}{C_V}\)= γ – 1 or, 0.67 = γ – 1, or γ = 1.67 Hence the gas is monoatomic.

97468.

If for a gas \(\frac{R}{C_V}\) = 0.67, this gas is made up of molecules which are(a) monoatomic(b) diatomic(c) Polyatomic(d) mixture of diatomic and polyatomic molecules

Answer»

(a) monoatomic

For a gas, we know \(\frac{R}{C_V}\)= γ – 1

or, 0.67 = γ – 1, or γ = 1.67

Hence the gas is monoatomic.

Discussion

97469.	A gas consisting of rigid diatomic molecules (degrees of freedom r = 5) under standard conditions `(P_(0) = 10^(5)` Pa and `T_(0) = 273 K`) was compressed adiabatically `eta = 5` times. Find the mean kinetic energy of a rotating molecule in the final state.
Answer» In an adiabatic expansion, `TV^(gamma-1)=` constant `T_(0)V^(gamma-1)=T((V)/(5))^(gamma-1) gamma=1+(2)/(5)` `therefore T=(273).(5)^(2//5)` `lt(KE)_("rotational")gt` `= kT` `=1.38xx10^(-23)xx273xx(5)^(2//5)` `=7xx10^(-21)J` (approx).

Discussion

97470.	Find the degrees of freedom of the following.1. A body is confined to move in a straight line2. A body moves in a plane3. A body moves in a space
Answer» 1. A body is confined to move in a straight line having 1 degree of freedom 2. A body moves in a plane 2 degree of freedom 3. A body moves in a space 3 degree of freedom

97470.

Find the degrees of freedom of the following.1. A body is confined to move in a straight line2. A body moves in a plane3. A body moves in a space

Answer»

1. A body is confined to move in a straight line having 1 degree of freedom

2. A body moves in a plane 2 degree of freedom

3. A body moves in a space 3 degree of freedom

Discussion

97471.

1. Fill in the blanksName of gasDegrees of freedomInternal energyCVCP?Monoatomic3......\(\frac{3}{2}R\)...........Diatomic rigid.......\(\frac{5}{2} RT\).......\(\frac{7}{2}R\)........ 2. What happens to the value of ratio of specific heat capacity, if we consider all rotational degrees of freedom of a 1-mole diatomic molecule?

Answer»

1.

Name of gas	Degrees of freedom	Internal energy	C_V	C_P	?
Monoatomic	3	\(\frac{3}{2}RT\)	\(\frac{3}{2}R\)	\(\frac{5}{2} R\)	\(\frac{5}{2} \)
Diatomic rigid	6	\(\frac{5}{2} RT\)	\(\frac{5}{2} R\)	\(\frac{7}{2}R\)	\(\frac{7}{2}\)

2. Total degrees of freedom = 3 (trans) + 3 (Rot) = 6

∴ C_V = 3R, C_P = 4R

Ratio of specific heat γ = \(\frac{4}{3}\)

Ratio of specific heat capacity decreases.

Discussion

97472.	Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0°C). Show that it is 22.4 litres.
Answer» PV = µRT or V = \(\frac{μRT}{P}\) V = \(\frac{1 \times 8.31 \times 273}{0.76 \times 13.6 \times 10^3 \times 9.8}m^3\) = 22.4 × 10^-3 m³ = 22.4 litre.

97472.

Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0°C). Show that it is 22.4 litres.

Answer»

PV = µRT or V = \(\frac{μRT}{P}\)

V = \(\frac{1 \times 8.31 \times 273}{0.76 \times 13.6 \times 10^3 \times 9.8}m^3\)

= 22.4 × 10^-3 m³ = 22.4 litre.

Discussion

97473.	A projectile fired at an angle of `45^(@)` travels a total distance R, called the range, which depends only on the initial speed v and the acceleration of gravity g. Using dimensional analysis, find how R depends on the speed and on g.
Answer» Correct Answer - `R=Kv^(2)/g` `R prop v^(a)g^(b) rArr [L]=[LT^(-1)]^(a)[LT^(-2)]^(b)` `rArr a+b=1, -a-2b=0` `rArr a=2, b=-1 rArr R prop v^(2)/g`

Discussion

97474.	The focal length of convex tens is `f` and the distance of an object from the principal focus is x. The ratio of the size of the real image to the size of the object isA. `((f+x))/(f)`B. `(f)/(x)`C. `sqrt((f)/(x))`D. `(f^(2))/(x^(2))`
Answer» Correct Answer - B `(v)/(u)=(f)/(u-f)` `= (f)/("distance of object from focus")=(f)/(x)`.

Discussion

97475.	The range of a projectile fired at an angle of `15^@` is 50 m. If it is fired with the same speed at an angle of `45^@` its range will beA. 25 mB. 37 mC. 50 mD. 100 m
Answer» Correct Answer - D `50=(V_(0)^(2)sin30^(@))/(g)` `R=(V_(0)^(2)sin90^(@))/(g)=100m`

Discussion

97476.	A ball is dropped vertically from `a` height `d` above the ground . It hits the ground and bounces up vertically to a height ` (d)//(2). Neglecting subsequent motion and air resistance , its velocity `v` varies with the height `h` above the ground asA. B. C. D.
Answer» Correct Answer - a

Discussion

97477.	A small block slides without friction down an iclined plane starting form rest. Let `S_(n)` be the distance traveled from time `t = n - 1` to `t = n`. Then `(S_(n))/(S_(n + 1))` is:A. `((2n-1))/(2n)`B. `((2n+1))/((2n-1))`C. `((2n-1)/(2n+1))`D. `((2n)/(2n+1))`
Answer» Correct Answer - C Distance travelled in `t^(th)` second is `S_(t)=u+(a)/(2)(2t-1)` `u = 0` `S_(n)=(a)/(2)(2n-1)` `S_((n)/(n+1))=(a)/(2)[2n+1]` `(S_(n))/(S_(n+1))=((2n-1))/((2n+1))`

Discussion

97478.	Using the above table answer the following questions : (a) Compare the atomic sizes of B & C. (b) Select the letter/letters which represent alkali metals. (c) Identify the nature of bonds formed by elements of Group 1 and 14. (d) What would be the electronic configuration of element F ? (e) How does the metallic character vary on going from left to right in a period ?
Answer» (a) Atomic size of C will be smaller than the atomic size of B. (b) A & D are alkali metals. (c) Group 1 elements will form ionic bonds however elements of group 14 will form covalent bonds. (d) Electronic configuration of F is 2, 4. (e) Metallic character will decrease will decrease on going from left to right in a period.

Discussion

97479.	Why does a light ray incident on a rectangular glass slab immersed in any medium emerges parallel to itself?Explain using a diagram.
Answer» Hint— Draw the diagram and explain using laws of refractions at both the interfaces.

Discussion

97480.	Which statement/statement is/are correct? 1. An ammeter is connected in series in a circuit and a voltmeter is connected in parallel. 2. An ammeter has a high resistance. 3. A voltmeter has a low resistance.
Answer» (i) Ammeter is connected in series after the experiment wire. (i) Voltmeter is connected ein parallel between the two ends of the experiment wire.

Discussion

97481.	An object is placed in front of a concave mirror between the pole and the focus of the mirror. What is the nature of the image formed by the mirror ?
Answer» Image formed is erect, virtual and magnified.

Discussion

97482.	Which set of oxide does not used in slag formation during extraction of Copper.(a) FeO(b) Al2O3(c) ZnO(d) NiO(e) CaO(1) a, b(2) b, c, d(3) a, e(4) a, c
Answer» Correct option is (2) b, c, d During extraction of Cu following slag formation reaction takes place. \(FeO + \underset{Flux }{SiO_2} → \underset{ Slag}{FeSiO_3 } \)

97482.

Which set of oxide does not used in slag formation during extraction of Copper.(a) FeO(b) Al2O3(c) ZnO(d) NiO(e) CaO(1) a, b(2) b, c, d(3) a, e(4) a, c

Answer»

Correct option is (2) b, c, d

During extraction of Cu following slag formation reaction takes place.

\(FeO + \underset{Flux }{SiO_2} → \underset{ Slag}{FeSiO_3 } \)

Discussion

97483.	A bottle of cold drink has 200 mL liquid in which `CO_(2)` is 0.1 molar. If `CO_(2)` behaves as ideal gas the volume of `CO_(2)` at S.T.P. solution of cold drink isA. 22.4 LB. 0.224 LC. 2.24 LD. 0.448 L
Answer» Correct Answer - D Given, Molarity = 0.1 M Volume = 200 mL `because`Molarity=`("number of moles of solute")/("volume of solution(mL)")xx1000` `therefore` Number of moles`=(0.1xx200)/1000=0.02` Now, 1 mole of dissolved `CO_(2)` occupies = 22.4 l volume at STP.

Discussion

97484.	Cerussite is an ore ofA. NaB. CuC. PbD. Fe
Answer» Correct Answer - C Cerussite is an ore of lead (Pb). Its chemical formula is `PbCO_(3)`.

Discussion

97485.	Chalcopyrites is an ore ofA. galliumB. copperC. calciumD. magnesium
Answer» Correct Answer - B Chalcopyrites or copper pyrite is `CuFeS_2` . It is an ore of copper and iron .

Discussion

97486.	In the reaction, `2nR-Xunderset("Dry ether")overset(+2n Na)to` product The product obtained isA. 2n-alkeneB. n-sodium halideC. n-alcoholD. n-alkane
Answer» Correct Answer - D `2nR-Xunderset("Dry ether")overset(+2nNa)tounderset("Alkane")(R-R)+2NaX` The above given reaction is known as Wurtz reaction.

Discussion

97487.	Table `IfDelta_(r)=\|{:(,(2r), x, N(N+1)),(,(6r^(2)-1),y, N^(2)(2N+3)),(,(4r^(3) - 2Nr), z, N^(3)(N+1)):}\|` where N `in` natural numbers. Then `sum_(r=1)^(N) Delta_(r)` isA. NB. `N^(2)`C. ZeroD. None of these
Answer» Correct Answer - 3 `underset(r=1)overset(N)sumDelta_(r)=\|{:(,2N((N+1))/(2), x, N(N+1)),(,6N((N+1)(2N+1))/(6),y, N^(2)(2N+3)),(,4[(N(N+1))/(2)] -2N.(N(N+1))/(2), z, N^(3)(N+1)):}\|=0`

Discussion

97488.	Umesh has two bulbs having resistances of \( 15 \Omega \) and \( 30 \Omega \). He wants to connect them in a circuit, but if he connects them one at a time the filament gets burnt. Answer the following.
Answer» When Umesh connects the bulbs one at a time, the filaments of the bulbs get burnt. This means the heat energy generated in the filament is beyond what it can tolerate. Hence, we need to reduce the heat generated across the filament. Now, the heat generated in a resistor is proportional to the amount of current flowing through it. Thus, we need to reduce the current flowing through the bulbs. A. If bulbs are connected in series, the effective resistance of the circuit increases, and hence current through the bulbs decreases. In this way, both the bulbs can be saved from burning. B. Following are the characteristics of series connection: The effective resistance of the circuit is the sum of individual resistances of the resistors present in the circuit. The effective resistance of the series circuit is always larger than the greatest resistance in the circuit. Current through various electrical components connected in series is the same. The source voltage gets divided across the electrical components connected in series. C. The effective resistance of the circuit = 15Ω + 30Ω = 45Ω

97488.

Umesh has two bulbs having resistances of \( 15 \Omega \) and \( 30 \Omega \). He wants to connect them in a circuit, but if he connects them one at a time the filament gets burnt. Answer the following.

Answer»

When Umesh connects the bulbs one at a time, the filaments of the bulbs get burnt. This means the heat energy generated in the filament is beyond what it can tolerate. Hence, we need to reduce the heat generated across the filament. Now, the heat generated in a resistor is proportional to the amount of current flowing through it. Thus, we need to reduce the current flowing through the bulbs.

A. If bulbs are connected in series, the effective resistance of the circuit increases, and hence current through the bulbs decreases. In this way, both the bulbs can be saved from burning.

B. Following are the characteristics of series connection:

The effective resistance of the circuit is the sum of individual resistances of the resistors present in the circuit.
The effective resistance of the series circuit is always larger than the greatest resistance in the circuit.
Current through various electrical components connected in series is the same.
The source voltage gets divided across the electrical components connected in series.

C. The effective resistance of the circuit = 15Ω + 30Ω = 45Ω

Discussion

97489.	Calculate the percentage composition in terms of mass of solution obtained by mixing `300 g` of a `25%` and `400 g` of a `40%` solution by mass.
Answer» Correct Answer - `31.9%` Mass of `25% sol A =(25)/(100)xx300=75g` mass of `40% sol B (40)/(100)xx400=160g` `%` composition of `A` `=(75xx100)/(75+160)=(75xx100)/(235)=31.9%` `%` composition of `B=100-31.9 = 68.1%`

Discussion

97490.	Standard enthalpy of vapourisation `DeltaH^(@)` forwater is `40.66 KJ mol^(-1)`.The internal energy of vapourisation of water for its 2 mol will be :-A. `+43.76`B. `+40.66`C. `+37.56`D. None of these
Answer» Correct Answer - D `H_(2)O(l)toH_(2)O(g)` `DeltaH^(@)=DeltaE^(@)+Deltan_(g)RT` `40.66=66DeltaE^(@)+(1xx8.314xx373)/(1000)` `DeltaE^(@)=+37.56KJ//mol` for 2 mole `DeltaE^(@)=2xx37.26=75.12KJ`

Discussion

97491.	What will be the molality of solution prepared by dissolving 3.7 g propanoic acid 80 g benzene ?A. 0.77 mB. 0.625 mC. 0.045 mD. 46.25 m
Answer» Correct Answer - B moles of acid `=(3.7)/(74)=0.05` molality of solution `=(w)/(m)xx(1000)/(w_(("ing")))=(0.05)/(80)xx1000` =0.625

Discussion

97492.	In an aqueous solution ethylene glycol has the mass percentage (% w/w) 30% then the mole fraction of ethylene glycol will be :-A. 0.428B. 0.124C. 0.11D. 0.889
Answer» Correct Answer - C 30g ethylene glyocol `(HO-CH_(2)-CH_(2)-OH)` is in 100 mL solution.`:."moles of" C_(2)H_(6)O_(2)=(30)/(62)=0.484` and moles of `H_(2)O=(70)/(18)=3.89` `X_(C_(2H_(6_O_(2)))=(0.484)/(0.484+3.89)=(0.484)/(4.347)=0.11`

Discussion

97493.	What will be the mass percentage of resulting solution prepared by mixing 15% (w/w) 500g aqueous solution of urea with 25% (w/w)400g aqueous solution of it :-A. 0.18B. 0.2C. 0.25D. 0.15
Answer» Correct Answer - B Amount of urea `= (15)/(100)xx500+(25)/(100)xx400` `=75+100=175g` `%w//w=(175)/(900)xx100=19.44%`

Discussion

97494.	Ships and AIRCRAFTS are often equipped with radio telephones. A) planes B) trains C) railroads D) highways E) boats
Answer» Correct option is A) planes

Discussion

97495.	Extraction of gold and silver involves leaching the metal with `CN^(-)` ion. Sliver is later recovered by :A. distillationB. zone refiningC. displacementD. liquation.
Answer» Correct Answer - C Silver is recovered by displacement with Zn. The process is known as leaching.

Discussion

97496.	Extraction of gold and silver involves leaching with `CN^(-)`ion.silver is later recovered by:A. distillationB. zone refiningC. displacement with `Zn`D. liquation
Answer» Correct Answer - C Mac arther forest process`//`cyanide process `Ag_(2)S +4NaCNoverset(O_(2))rarr2Na[Ag(CH)_(2)]+NaSO_(4)` `2Na[Ag(CN)_(2)]underset(Zn)rarrNa_(2)underset("Soluble complex")([Zn(CN)_(4)]+Ag(darr))` `Ag` extractes by displacement with `Zn`

Discussion

97497.	`CsBr` has bcc stucture with edge length `4.3` pm .The shortest interionic distance in between Cs and Br isA. 1.86 pmB. 7.44 pmC. 4.3 pmD. 3.72 pm
Answer» Correct Answer - D `CI^(-)` is pressent at conurs while `Cs^(+)` is present at Body centre, so minimum interatomic distance `=(sqrt(3a))/(2)=3.72"pm"`

Discussion

97498.	Give briefly the isolation of magnesium from sea water. Give equations for the steps involved.
Answer» In sea water, `MaCl` is present. `MgCl_(2) + Ca(OH)_(2) rarr Mg(OH)_(2) darr + CaCl_(2)` `underset ("Dry")(Mg(OH)_(2)) overset (Delta) rarr MgO + H_(2)O` `MgO + C + Cl_(2) overset (Delta) rarr MgCl_(2) + Co uarr` the anhydrous megnesium chloride is fused `NaCl` and anhydrous calcium chloride. The mixture is electrolysed at `700^@ C` in the presence of an inert gas in an electrolytic cell. Magnesium is discharged at the cathode. The purpose of the addition of `NaCl` and `CaCl_(2)` to anhydrous `MgCl_(2)` is to lower the fusion temperature and make fused mass a good conductor of electricity. Sodium chloride prevents the hydrolysis of magnesium chloride. `{:("At anode":,2Cl^(Ө)rarr Cl_(2) + 2e,,,),("At cathode":,Mg^(2+) + 2e^(Ө) rarr Mg,,,):}`.

Discussion

97499.	Although aluminium is above hydrogen in the electrochemical series, it is stable in air and water. Why ?
Answer» [Hint : Due to formation of inert oxide Al₂O₃.]

Discussion

97500.	Give reasons for the following. "Although aluminium is above hydrogen in the electrochemical series, it is stable in air and water".
Answer» When `Al` is exposed to air, it forms a thin invisible continuous resistant protective film of `Al_(2)O_(3)`.

Discussion

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