Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Consider the following reactions, H_(2)(g)+I_(2)(g)hArr2HI(g) In each of the above reaction find out whether you have to increase (or) decrease the volume to increase the yield of the product.

Answer»

Solution :`H_(2)(g)+I_(2)(g)hArr2HI(g)`
ACCORDING to Le Chateliers princple increase in PRESSURE will shift the equilibrium a direction that has lesser number of moles.
In the above equilibrium, pressure and volume has no effect, since no of moles of product is equal to no. of moles of REACTANT.
Discussion
2.

Consider the following reactions, CaCO_(3)(s)iffCaO(s)+CO_(2)(g) In each of the above reaction find out whether you have to increase (or) decrease the volume to increase the yield of the product.

Answer»

Solution :`CaCO_(3)(s)iffCaO(s)+CO_(2)(G)`
In this EQUILIBRIUM no. of moles of product is greater than no. of moles of reactant. SO increase of pressure will shift the equilibrium towards the backward reaction. To increase the YIELD of product, pressure should be decrease and hence VOLUME increases
Discussion
3.

Consider the following reactions, CaCO_(3)(s)hArrCaO(s)+CO_(2)(g) In each of the above reaction find out whether you have to increase (or) decrease the volume to increase the yield of the product.

Answer»

SOLUTION :`CaCO_(3)(s)hArrCaO(s)+CO_(2)(g)`
In this equilibrium no. of moles of product is greater than no. moles of reactant. So increase of PRESSURE will shift the equilibrium towards the backward reaction. To increase the YIELD of product, pressure should be decrease and hence volume increases.
Discussion
4.

Consider the following reactions. Br_(2)+OH^(-)rarrBrO_(3)^(-)+Br^(-)+H_(2)O (unbalanced). select the correct statement

Answer»

EQUIVALENT weight of `Br_(2)` when it is reduced to `BR^(-)` is 80
Equivalent weight of `Br_(2)` when it is oxidised to `BrO_(3)^(-)` is 96
Net equivalent weight of `Br_(2)` is 96
It is a DISPROPORTIONATION reaction

Solution :For disproportionate reactions,
`E=E_("red")+E_("ox")=(160)/(2)+(160)/(10)=80+16=96`
Discussion
5.

Consider the following reactions, (a) H_2(g)+ I_2(g)hArr 2HI(g) (b) CaCO_3(s) hArr CaO(s)+ CO_2(g)(c ) S(s) + 3F_2(g) hArr SF_6(g) In each of the above reaction find out whether you have in increase (or)decrease the volume to increase th yield of the product.

Answer»

SOLUTION :(a) `H_2(g) + I_2(g) hArr 2HI(g)`
In the above equilibrium reaction, volume of gaseous molecules is equal on both sides. So INCREASE or DECREASE the volume will not effect the equilibrium and there will be no change in the yield of product.
(b) `underset(0(V))(CaCO_2(s)) hArr underset(0(V))(CaO(s))+ underset(1(V))(CO_2(g))`
Volume is greater in product side. By decreasing the PRESSURE, volume willincrease thus, to get more of product `CO_2`, the pressure should be decreased or volumeshould be increased.
`S(s) + underset(3V)(3F_2(g)) hArr underset(IV)(SF_6(g))`
Volume is lesser in product side. So by increasing the pressure, equilibrium shifts to the product side.
Discussion
6.

Consider the following reactions: 2Na + 2NH_(3) rightarrow 2NaNH_(2) + H_(2)(g)……………..(i) 2NaNH_(2) + C rightarrow Na_(2)CN_(2) + 2H_(2)(g)…………….(ii) Na_(2)CN_(2) + C rightarrow 2NaCN……………….(iii) 51 gm dry ammonia gas is passed over excess heated sodium to form sodamide NaNH_(2) which is further reacted with carbon (excess) to finally form NaCN. Find the total volume of H_(2)(g) evolved at 0.5 atm, 273 K.

Answer»

201.6 L
100.8 L
403.2 L
50.4L

Answer :A
Discussion
7.

Consider the following reactions, (a) H_(2)(g)+(g)+I_(2)(g)harr2HI(g) (b) CaCO_(3)(s)harrCaO(s)+CO_(2)(g) (c) S(s)+3F_(2)(g)harrSF_(6)(g) In each of the above reaction find out whether you have to increase (or) decrease the volume to increase the yield of the product.

Answer»

Solution :(a) `H_(2)(g)+I_(2)(g)harr2HI(g)`
In the above equilibrium reaction, volume of gaseous molecules is EQUAL on both sides. So increase or decrease the volume will not affect the equilibrium and there will be no CHANGE in the yield of product.
(b) `CaCO_(2)(s)harrCaO(s) +CO_(2)(g)`
Volume is greater in product side. By decreasing the pressure, volume will increase thus, to get more of product `CO_(2)`, the pressure should be decreased or volume should be increased.
(c) `S(s)+underset(3V)(3F_(2)(g))harrunderset(1V)(SF_(6)(g))`
Volume is lesser in product side. So by increasing the pressure, equilibrium shifts to the product side.
Discussion
8.

Consider the following reaction: Zn+Cu^(2+)rarrZn^(2+)+Cu with reference to the above which one of the following is the correct statement ?

Answer»

ZN is reduced to`Zn^(2+)`ions
Zn is reduced to `Zn^(2+)ions`
`Zn^(2+)`ions are oxidised to Zn
`CU^(2+)` ions are oxidised to Cu

Solution :Zn LIES above Cu in the activity SERIES and is oxidised to `Zn^(2+)` ions by releasing ELECTRONS.
`ZnrarrZn^(2+)+32e^(-)`
Discussion
9.

Consider the following reaction, xMnO_(4)^(-)+yC_(2)O_(4)^(2-)+zH^(+)toxMn^(2+)+2yCO_(2)+(z)/(2)H_(2)O The values of x,y and z in the reactions are, respectively

Answer»

5, 2 and 16
2, 5 and 8
2, 5 and 16
5, 2 and 8.

Answer :C
Discussion
10.

Consider the following reaction xMnO_(4)^(-) +yC_(2)O_(4)^(2-) +zH^(-) rarr xMn^(2+) +2yCO_(2) +(z)/(2)H_(2)O

Answer»

`2,5 and 16`
`5, 2 and 8`
`5,2 and 16`
`2,5 and 8`

Solution :`2xx{{:(MnO_(4)^(-)RARR,MN^(2+)),(8H^(+),4H_2O),(5e^(-),):}:|{:(C_2O_4^(2-),rarr,2CO_2),(,,+2e^(-)):}}xx5`
Discussion
11.

Consider the following reaction xMnO_(4)^(-)+yC_(2)IO_(4)^(2-)+zH^(+)rarrxMn^(2+)+2yCO_(2)+z//2H_(2)O the value of x,y and z in the reaction are respectively

Answer»

5,2 and 8
5,2 and 16
2,5 and 8
2,5 and 16

Solution :Reduction half reaction
`MnO_(4)^(-)+8H^()+5e^(-)rarrMn^(2+)+4H_(2)O`
Oxidation half reaction `C_(2)^(+3)O_(4)^(2-)rarrCO_(2)`
balance C ATOMS `C_(2)O_(4)^(2-)rarr2CO_(2)`
balance O.N by adding `2e^(-)` to R.H.S of Eq (II) we have
`C_(2)O_(4)^(21-)rarr2CO_(2)+2e^(-)`
charge is automically balanced to cancel eletrons multily eq (i) by 2 and eq (iii) by 5 and adding we have
`2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+)rarr2Mn^(2+)+10CO_(2)+8H_(2)O`
Discussion
12.

Consider the following reaction : xMnO_(4)^(-)+yC_(2)O_(4)^(2-)+H^(+)toxMn^(2+)+2yCO_(2)+z/2H_(2)O The values of x, y and z in the reaction are, respectively :

Answer»

5, 2 and 16
2, 5 and 8
2, 5 and 16
5, 2 and 8

Solution :`2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+)to2Mn^(2+)+10CO_(2)+8H_(2)O`
Discussion
13.

Consider the following reaction x MnO_(4)^(-) + C_(2)O_(4)^(2-) + zH^(+) rarr x Mn^(2+) + 2y CO_(2) + (z)/(2)H_(2)O The value of x, y and z in the reaction are respectively

Answer»

5,2 and 8
5,2 and 16
2,5 and 8
2,5 and 16

ANSWER :D
Discussion
14.

Consider the following reaction Which of the following base will give best yield of the alkene II as the major product?

Answer»

`CH_3 O^(-)`
`C_2H_5O^(-)`
`(CH_3)_3CO^(-)`
`(C_2H_5)_3CO^(-)`

SOLUTION :When an alkene is treated with potassium tertiary butoxide INSTEAD of ethoxide, Hofmann product is favoured.
Discussion
15.

Consider the following reaction to answer the next three equestions. Q. If the product 'X' is treated with AlCl_(3) it undergoes an interamolecular friedel-craft reaction. The expected product P is

Answer»




SOLUTION :
Discussion
16.

Consider the following reaction, PCl_(5) (g) hArr PCl_(3) (g) + Cl_(2)(g) the forward reaction at constant temperature is favoured by

Answer»

INTRODUCING an inert GAS at constant VOLUME
introducing `Cl_(2)` gas at constant volume
introducing an inert gas at constant pressure
increasing the volume of the container

Solution :Le-chatelier.s principle
Discussion
17.

Consider the following reaction, (i) CO_(3)^(2-)+H_(2)OhArrHCO_(3)^(-)+OH^(-) (ii) CO_(2)+H_(2)OhArrH_(2)CO_(3) (iii) NH_(3)+H_(2)OhArrNH_(4)OH (iv) HCl+H_(2)OhArrCl^(-)+H_(3)O^(+) Which of the pairs of reaction proves that water is amphoteric in character?

Answer»

(i) and (II)
(ii) and (III)
(iii) and (IV)
(i) and (iii)

ANSWER :C
Discussion
18.

Consider the following reaction How many different products (including stereoisomers) would be formed ?

Answer»


SOLUTION :
Discussion
19.

Consider the following reaction , H_3C-undersetundersetD|CH-undersetunderset(CH_3)|CH-CH_3+oversetdot(Br) to 'X' + HBr Identify the structure of the major product 'X'

Answer»

`CH_3-undersetundersetD|CH-undersetunderset(CH_3)|CH-oversetdot(CH_2)`
`CH_3-undersetunderset(D)|CH-undersetunderset(CH_3)|dotC-CH_3`
`CH_3-undersetunderset(D)|dotC-undersetunderset(CH_3)|CH-CH_3`
`CH_3-oversetdot(CH)-undersetunderset(CH_3)|CH-CH_3`

SOLUTION :Most stable FREE RADICAL (i.e., OPTION 'B') is the product.
Discussion
20.

Consider the following reaction H_3C-undersetunderset(D)(|)CH-undersetunderset(CH_3)(|)CH-CH_3+overset(.)(Br)to'X'+HBr Identify the structure of the major product 'X'.

Answer»

`H_3C-undersetunderset(D)(|)CH-undersetunderset(CH_3)(|)CH-OVERSET(.)CH_2`
`H_3C-undersetunderset(D)(|)CH-undersetunderset(CH_3)(|)overset(.)C-CH_3`
`H_3C-undersetunderset(D)(|)overset(.)C-undersetunderset(CH_3)(|)CH-CH_3`
`H_3C-overset(.)CH-undersetunderset(CH_3)(|)CH-CH_3`

ANSWER :B
Discussion
21.

Consider the following reaction , Fe^(3+) (aq) + SCN^(-) (aq) hArr [Fe (SCN)]^(2+) (aq) A solution is made with initial Fe^(3+) , SCN^(-) concentration of 1 xx 10^(-3) M and 8 xx 10^(-4) M respectively . At equilibrium [Fe(SCN)]^(2-) concentration is 2 xx 10^(-4) M . Calculate the value of equilibrium constant.

Answer»

SOLUTION :
`K_(eq) = ([FE (SCN)]^(2+))/([Fe^(3+)] [SCN^(-)]) = (2 xx 10^(-4) M)/(8 xx 10^(-4) M xx 6 xx 10^(-4) M) = 0.0416 xx 10^(4)`
`K_(eq) = 4.16 xx 10^(2) M^(-1)`
Discussion
22.

Consider the following reaction : CO_((g)) + 2H_(2(g)) hArr CH_(3)OH_((g)) Given : Delta_(r) H^(@) (CH_(3)OH, g) = -201 kJ/mol, Delta_(r) H^(@) (CO, g) = -114 kJ/mol S^(@) (CH_(3)OOH, g) = 240 J/K-mol , S^(@) (H_(2), g) = 29 JK^(-1)mol^(-1) S^(@) (CO, g) = 198 J/mol-K , C^(@)_(p,m) (H_(2)) = 28.8 J/mol-K C^(@)_(p,m) (CO) = 29.4 J/mol-K , C^(@)_(p,m) (CH_(3)OH) = 44 J/mol-K and ln ((320)/(300)) = 0.06, all data at 300 K Delta_(r) S^(@) at 320 K is :

Answer»

155.18 J/mol-K
150.02 J/mol-K
172 J/mol-K
none of these

Solution :`Delta_(r) S^(0) = S_(CH_(3)OH)^(0) - S_(CO)^(0) - 2S_(H_(2))^(0) - = - 16J mol - K`
`Delta_(r ) H^(0) = Delta_(f) H^(0) (CH_(3)OH) - Delta_(f)H^(0) (CO) - 2. Delta_(f)H^(0) (H_(2))`
`= -87kJ//mol`
`Delta_(r) S-(320)^(0) - Delta_(r) S_(300)^(0) = Delta_(r)C_(P) [T_(2) - T_(1)]`
where `Delta_(r) C_(P)^(0) = 44-29.4 -2 xx 28.8 = -43J//K - mol`
`Delta_(r) S_(320)^(0) = - 16 + (-43) "ln "(320)/(320) = - 18.58`
`Delta_(r)H_(320)^(0) = Delta_(r)H_(300)^(0) + Delta_(r) C_(P)^(0) [T_(2) - T_(1)]`
`= - 87 + ((-43) xx 20)/(1000) = - 87.86` kJ/mol
`Delta_(r) G_(320)^(0) = Delta_(r) H_(320)^(0) - T. Delta_(r) S_(320)^(0)`
`= -87.86 - (320 xx (-18.58))/(1000) = - 81.91 kJ//mol`
Discussion
23.

Consider the following reaction : CO_((g)) + 2H_(2(g)) hArr CH_(3)OH_((g)) Given : Delta_(r) H^(@) (CH_(3)OOH, g) = -201 kJ/mol, Delta_(r) H^(@) (CO, g) = -114 kJ/mol S^(@) (CH_(3)OOH, g) = 240 J/K-mol , S^(@) (H_(2), g) = 29 JK^(-1)mol^(-1) S^(@) (CO, g) = 198 J/mol-K , C^(@)_(p,m) (H_(2)) = 28.8 J/mol-K C^(@)_(p,m) (CO) = 29.4 J/mol-K , C^(@)_(p,m) (CH_(3)OH) = 44 J/mol-K and ln ((320)/(300)) = 0.06, all data at 300 K Delta_(r) H^(@) at 300 K for the reaction is :

Answer»

`-87` kJ/mol
87 kJ/mol
`-315` kJ/mol
`-288` kJ/mol

Solution :`Delta_(r) S^(0) = S_(CH_(3)OH)^(0) - S_(CO)^(0) - 2S_(H_(2))^(0) - = - 16J mol - K`
`Delta_(r ) H^(0) = Delta_(F) H^(0) (CH_(3)OH) - Delta_(f)H^(0) (CO) - 2. Delta_(f)H^(0) (H_(2))`
`= -87kJ//mol`
`Delta_(r) S-(320)^(0) - Delta_(r) S_(300)^(0) = Delta_(r)C_(P) [T_(2) - T_(1)]`
where `Delta_(r) C_(P)^(0) = 44-29.4 -2 xx 28.8 = -43J//K - mol`
`Delta_(r) S_(320)^(0) = - 16 + (-43) "ln "(320)/(320) = - 18.58`
`Delta_(r)H_(320)^(0) = Delta_(r)H_(300)^(0) + Delta_(r) C_(P)^(0) [T_(2) - T_(1)]`
`= - 87 + ((-43) xx 20)/(1000) = - 87.86` kJ/mol
`Delta_(r) G_(320)^(0) = Delta_(r) H_(320)^(0) - T. Delta_(r) S_(320)^(0)`
`= -87.86 - (320 xx (-18.58))/(1000) = - 81.91 kJ//mol`
Discussion
24.

Consider the following reaction : CO_((g)) + 2H_(2(g)) hArr CH_(3)OH_((g)) Given : Delta_(r) H^(@) (CH_(3)OOH, g) = -201 kJ/mol, Delta_(r) H^(@) (CO, g) = -114 kJ/mol S^(@) (CH_(3)OOH, g) = 240 J/K-mol , S^(@) (H_(2), g) = 29 JK^(-1)mol^(-1) S^(@) (CO, g) = 198 J/mol-K , C^(@)_(p,m) (H_(2)) = 28.8 J/mol-K C^(@)_(p,m) (CO) = 29.4 J/mol-K , C^(@)_(p,m) (CH_(3)OH) = 44 J/mol-K and ln ((320)/(300)) = 0.06, all data at 300 K Delta_(r) S^(@) at 300 K for the reaction is :

Answer»

152.6 J/K - mol
181.6 J/K - mol
`-16` J/K - mol
none of these

Solution :`Delta_(r) S^(0) = S_(CH_(3)OH)^(0) - S_(CO)^(0) - 2S_(H_(2))^(0) - = - 16J mol - K`
`Delta_(r ) H^(0) = Delta_(f) H^(0) (CH_(3)OH) - Delta_(f)H^(0) (CO) - 2. Delta_(f)H^(0) (H_(2))`
`= -87kJ//mol`
`Delta_(r) S-(320)^(0) - Delta_(r) S_(300)^(0) = Delta_(r)C_(P) [T_(2) - T_(1)]`
where `Delta_(r) C_(P)^(0) = 44-29.4 -2 xx 28.8 = -43J//K - mol`
`Delta_(r) S_(320)^(0) = - 16 + (-43) "ln "(320)/(320) = - 18.58`
`Delta_(r)H_(320)^(0) = Delta_(r)H_(300)^(0) + Delta_(r) C_(P)^(0) [T_(2) - T_(1)]`
`= - 87 + ((-43) xx 20)/(1000) = - 87.86` kJ/mol
`Delta_(r) G_(320)^(0) = Delta_(r) H_(320)^(0) - T. Delta_(r) S_(320)^(0)`
`= -87.86 - (320 xx (-18.58))/(1000) = - 81.91 kJ//mol`
Discussion
25.

Consider the following reaction: (CN_(2)) rarr CN^(-) +CNO^(-) The correct statement(s) is/are about the reaction

Answer»

It is disproportionation
Oxidation state of carbon charges from `+3` to `+2` and `+4`
Oxidation state of NITROGEN doesnot CHARGE
It is auto redox reaction

Solution :`(CN)_(2) RARR CN^(-) +CNO^(-)`
`UNDERSET(+3-3)((CN)_(2)) rarr underset(+2-3)(CN^(-)) +underset(+4-3-2)(CNO^(-))`
Discussion
26.

Consider the following reaction, {:(CHO""COO^(-)),("|"+OH^(-) rarr "|"),(CHO""CH_(2)OH):} Select the incorrect statement.

Answer»

It is not a DISPROPORTIONATION reaction.
It is intramolecular redox reaction.
`OH^(-)` is a reducing as well as oxidising agent.
`{:(CHO),("|"),(CHO):}`is a reducing aswell as oxidising agent.

Solution :One CHO is oxidised to `CO O^(-)` and oneCHO is REDUCED to `CH_(2)OH`.
Thus, it is not a disproportionation reaction. It isintramolecular redox reaction. Thus, options (a) and (b) are TRUEAND `{:(CHO),("|"),(CHO):}` isreducingas well as oxidising agent.
Thus, (d) is also true.Thus, (c) isincorrect.
Discussion
27.

Consider the following reaction CH_3- underset(D)underset(|)CH-underset(CH_3)underset(|)CH-CH_3 + Br to X + HBr Identify the structure of the major product 'x'

Answer»

`CH_3 - UNDERSET(D)underset(|)CH-underset(CH_3)underset(|)CH-CH_(2)^(*)`
`CH_3-underset(D)underset(|)CH-underset(CH_3)underset(|)C^(*)-CH_3`
`CH_3-underset(D)underset(|)C^(*)-Cunderset(CH_3)underset(|)H-CH_3`
`CH_3-overset(*)CH-underset(CH_3)underset(|)CH-CH_3`

Solution :Free RADICAL SUBSTITUTION
Discussion
28.

Consider the following reaction: CH_(3)-underset(D)underset(|)(CH)-underset(CH_(3))underset(|)(CH)-CH_(3)-Br rarr X+HBr Identify the structrue of the major product X

Answer»

`H_(3)C-underset(D)underset(|)(CH)-underset(CH_(3))underset(|)(CH_(3))-overset(.)(C )H_(2)`<BR>`H_(3)C-underset(D)underset(|)(CH)-underset(CH_(3))underset(|)overset(.)(C)-CH_(3)`
`H_(3)C-underset(D)underset(|)overset(.)(C)-underset(CH_(3))underset(|)(CH)-CH_(3)`
`H_(3)C-overset(.)(CH)-underset(CH_(3))underset(|)(C)-CH_(3)`

SOLUTION :`Br` is less reactive and more selective. Thus the formation of `3^(@)` RADICAL will be the MAJOR product.
Discussion
29.

Consider the following reaction . C_6H_4 Cl_2 (X) + CH_3Cl overset(AlCl_3)to C_6H_3Cl(CH_3)(Y)" Single isomer" The correct statement (s) concerning X and Y is /are

Answer»

X is a non-polar dichloride isomer
X is more reactive than benzene in the same reaction
Y gives three ISOMERS of dimethylation product
X and Y have different positions of CHLORINE ATOMS

Solution :
Discussion
30.

Consider the following reaction between zinc and oxygen and choose the corret options out of the options given below : 2Zn (s) + O_(2)(g)rarr 2ZnO(s), DeltaH= - 693.8 kJ mol^(-1)

Answer»

The enthalpy of two molesof ZnO is less than the TOTAL enthalpyof two moles of Zn and one mole of oxygen by 693.8kJ
The enthalpy of two enthalpy of two molesof ZnO is more than the totalenthalpy of two moles of Zn and one mole of oxygen by 693.8 Kj
693.8 kJ `mol^(-1)` energy is evolvedin the reaction
`693.8 kJ mol^(-1)` energy is absorbed in the reaction.

SOLUTION :`DeltaH = H_(p) -H_(R)`. A negative value of `DeltaH `SHOWS that `H_(r)GTH _(p) `or `H_(p)lt H_(R)`, i.e.,enthalpy of two molesof ZnOis less than the enthalpyof two moles of zinc and one moleof oxygen by 693.8kJ , As `H_(R)gt H_(P)`, 693.8 kJ mol^(-1)` of energyis evolvedin the reaction.
Discussion
31.

Consider the following reaction between zinc and oxygen and choose the correct options out of the options given below : 2Zn_((s)) + O_(2(g)) to 2 ZnO_((s)) , Delta H = -693.8 "kJ mol"^(-1)

Answer»

The enthalpy of two moles of ZnO is less than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ.
The enthalpy of two moles of ZnO is more than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ.
`693.8 "kJ mol"^(-1)` energy is evolved in the reaction.
`693.9 "kJ mol"^(-1)` energy is absorbed in the reaction.

SOLUTION : For the given reaction,
`2Zn_((s)) + O_(2(g)) to 2ZnO_((g)) , Delta H = - 693.8 "kJ mol"^(-1) Delta H = H_(P) - H_(R) `
A -ve value of `Delta H` shows that `H_R gt H_P "or" H_P lt H_R` i.e., enthalpy of two moles of ZnO is less than the enthalpy of two moles of zinc and one mole of oxygen by 693.8kJ.
As `H_R gt H_P, 693.8 "kJ mol"^(-1)` of energy is evolved in the reaction.
Discussion
32.

Consider the following reaction at 1000^(@)C . (A) Zn(s) + 1/2 O_2(g) rarr ZnO(s) deltaG^(o) = -360 kJ mol^(-1) (B) C(gr) + 1/2 O_2(g) rarr CO(g) deltaG^(o) = -460 kJ mol^(-1) Choose the correct statement at 1000^(@)C .

Answer»

ZINC can be OXIDIZED by CARBON monoxide
zinc oxide can be reduced by graphic
both STATEMENTS (A) and (B) are true
both statements (A) and (B) are false

Solution :N//A
Discussion
33.

Consider the following reaction :Ag^(+ ) (aq)+Cl^(-)(aq) rarrAgCl(s) When 10.0mLof 1.0 MAgNO_(3) solution is added to 10 mL of 1.0 mL NaCl solution at 25^(@)C in a calorimeter , a white ppt. of AgCl is formed and the temperature of the aqueous mixture mixture risesto 32.6^(@)C. Assuming that the specificheatof the aqueous mixture is 4.18 J //g//^(@)C, thatthe density of the mixture is 1.0 g mL^(-1) and that the calorimeter itself absorbsa negligible amountofheat, calculate the value of enthalpy change accompanying the processin kJ mol^(-1) ofAgCl.

Answer»

Solution :Mass of the mixture solution`=`VOLUME `xx` Density `= 20 mL xx1 g mL^(-1) =20g`
Riseis temperature`= 32.6 -25 = 7.6^(@)C`
Heat evolved `= m xxC xx Delta t = 20xx 4.18 xx 7.6 =640J`
Moles of`Ag^(+) =( 1)/( 1000) xx 10 =10^(-2)`
`:. ` Moles ofAgCl `= 10^(-2)`
`:. ` Heat evolved per MOLE of AgCl ` ( 640 )/( 10^(-2)) = 64000 J MOL^(-1) = 64kJ mol^(-1)`
Discussion
34.

Consider the following reaction A + 2B rArr C +D If W_(A)/W_(B) = 0.5, which condition will make B a limiting reagent and A to be present in excess?

Answer»

`(M_(B)/M_(A)) LT 1`
`M_(B)/M_(A) GT 1`
`(M_(B)/M_(A)) = 1`
B will ALWAYS be LIMITING reagent

Answer :B
Discussion
35.

Consider the followingreaction2Fe_(2)O_(3) + 3C(s) to 4Fe(s) + 3Co_(2)(g) DeltaH^(@) ofFe_(2)O_(3)andCO_(2) are -820kJ//"mol" and-390 kJ//"mol" respectively The reaction respectively .

Answer»

endothermic
exothermic
SPONTANEOUS at HIGH temperature
spontancousat LOW temperature

Answer :a,C
Discussion
36.

Consider the following reactioin, 2Al(s) + 6HCl(aq) to 2Al^(3+)(aq) + 6Cl^(-)(aq) + 3H_(2)(g) Which of the following statements is incorrect?

Answer»

6 L HCl (AQ) is consumed for EVERY `3LH_(2)`(g) produced
33.6 L `H_(2)`(g) is produced regardless of temperature and pressure for every MOLE of Al that reacts
67.2 L `H_(2)` L `H_(2)`(g) at STP, is produced for every mole of AI that reacts
11.2 L `H_(2)` (g) at STP, is produced for every mole of HCl (aq) consumed

Solution :2 moles of Al produce `=3xx22.4L`of `H_(2)` GAS
`therefore`1 mole of Al will produce `=(3xx22.4)/(2)=33.62L` of `H_(2)` gas
Discussion
37.

Consider the following processes : {:(,DeltaH(kJ //mol) ),(1//2A rarr B ,+ 150 ),(3 B rarr 2C + D,-125),(E+Ararr2D ,+ 350 ):} For B +D rarrE +2C , DeltaHwill be

Answer»

`525 kJ //mol`
` -175 kJ //mol`
` -325kJ // mol`
` 325 kJ //mol`

Solution :`2 XX `Eqn. (i) `+`Eqn. (ii) -Eqn. (iii) GIVES the REQUIRED EQUATION.
`:. DetlaH =2( 150) + ( - 125) -350 =- 175 kJ //` mol.
Discussion
38.

Consider the following process of decay, ._(92)^(234)U rarr ._(90)^(230)Th + ._(2)^(4)He, t_(1//2) = 250000 yr ._(90)^(230)Th rarr ._(88)^(226)Ra + ._(2)(4)He, r_(1//2) = 80000 yr ._(88)^(226)Ra rarr ._(86)^(222)Rn + ._(2)^(4)He, t_(1//2) = 1600 yr After the above process has occurred for a long tiem, a state is reached where for every two thorium atoms formed form ._(92)^(234)U one decomposes to from ._(88)^(226)Ra and for every two ._(88)^(226)Raformed, one decomposes. The ratio of ._(90)^(230)Th to ._(88)^(226)Ra will be:

Answer»

`(25000)/(80000)`
`(80000)/(1600)`
`(250000)/(1600)`
`(250000xx1600)/(80000)`

SOLUTION :Follow RADIOACTIVE equillibrium.
`(N_(A))/(N_(B)) = (t_((1//2)A))/ (t_((1//2)B))`
Discussion
39.

Consider the following process : {:((1)/(2)A to B , Delta H =+ 150(kJ //mol)), (3 B to 2 C +D, Delta H= -125(kJ//mol)), (E +A to 2 D, Delta H = + 350(kJ//mol)):} For B + D to E + 2C, Delta Hwill be

Answer»

525 kJ/mol
`-175` kJ/mol
`-325` kJ/mol
325 kJ/mol

ANSWER :B
Discussion
40.

Consider the following points : (a) Cs is the strongest reducing agent is IA group element (B)" "Be(OH)_(2) is amphoteric (c ) The density of potassium is less than sodium (d) In alkali metals Li,Na,K and Rb, lithium has the minimum value of M.P. Correct statements are :

Answer»

`(a)& (G)` are correct
`(a),(B)&(C )` are correct
`(b) & (c )` are correct
`(b) (c) & (d)` are correct

Answer :D
Discussion
41.

Consider the following pairs of ions A) Sc^(+3) and Ti^(+4) B) Mn^(+4) and Fe^(+2) C) Fe^(+2) and Co^(+3) D) Cu^(+) and Zn^(+2) Among these pairs of ions, isoelectronic pairs would include

Answer»

ii, iii and iv
I, iii and iv
I, ii and iv
I , ii and iii

Solution :Iso ELECTRONIC PAIRS have same no. of electrons.
Discussion
42.

Consider the following orders : i) HF gtHCl gt HBr gt HI : Lewis basic character ii) CH_(4) lt "CCl"_(4) lt CF_(4) : Electronegativity of central 'C' -atom iii) Mg^(2+) lt K^(+) lt S^(2-) lt Se^(2-) : Ionic radius iv) Ni gt Pb gt Pt : Ionisation energy v) As^(5+) gt Sb^(5+) gt Bi^(5+) : Stable oxidation state vi) LiF gt NaF gt KF gt RbF : Lattice energy vii) F_(aq)^(-) gt Cl_(aq)^(-) gt Br_(aq)^(-) gt I_(aq)^(-) : Electrical consuctance viii) Li^(+) lt Mg^(2+) gt Al^(3+): Hydration energy ix) Cl gt Br gt F gt I : Electron affinity x) BeCl_(2) lt AlCl_(3) lt SiCl_(4): Lewis acidic character Then calculate value of |x-y|, where x and y are correct and incorrrect orders respectively.

Answer»


ANSWER :2
Discussion
43.

Consider the following order : (a) CH_(4)lt"C"Cl_(4)ltCF_(4):E.N. of central atom C (b) Mg^(2+)ltK^(+)ltS^(-2)ltSe^(-2): Ionic radius (c) Be_((aq))^(+2)gtMg_((aq.))^(+2)gtCa_((aq))^(+2): Ionic mobility (d) Be^(+2)gtLi^(+)gtAl^(3+): Hydrated size BegtLigtCs: Reducing power Fe_((aq))^(Θ)gtCl_((aq))^(Θ)gtBr_((aq))^(Θ): Electrical conductance at infinite dilute solution Then calculate value of |x-y|^(2), where x and y are correct and incorrect orders respectively.

Answer»


SOLUTION :CORRECT`rArr(a),(b)rArrx=2`
INCORRECT`rArr(c), (d),(e),(F)rArry=4 |x-y|^(2)=4`
Discussion
44.

Consider the following nine phase transformations. H_(2)O(s,1atm,273 K)rarr H_(2)O(l,1 atm, 3273 K) H_(2)O(s,1atm,300K)rarr H_(2)O(l,1 atm, 300 K) H_(2)O(s,1 atm,200 K)rarrH_(2)O(l,1 atm, 200 K) H_(2)O (s,0.5 atm, 273 K)rarrH_(2)O(l,0.5 atm 273 K) H_(2)O(s,2 atm, 273 K)rarrH_(2)O (l,2 atm,273 K) H_(2)O(l, 1 atm, 400 K)rarrC_(6)H_(6)(g),1atm, 353K) C_(6)H_(6)(l,1 atm, 400 K)rarrC_(6)H_(6)(g,1atm,400 K) C_(6)H_(6)(l,1atm,300 K)rarrC_(6)H_(6)(g,1atm,300K) C_(6)H_(6)(l,2atm,327 K)rarrC_(6)H_(6)(g,2atm,323 K) Given : T_(nbp)of C_(6)H_(6)(l)=353 K Now a four digit number abcd is fined as follows: Digit a It is number of phase transformations for which DeltaS_(total)=0 Digit b It is number of phase transformations for which DeltaS_(total)lt0 Digit c It is number of phase transfprmation for which DeltaS_(total)gt0 Digit d It is number of phase transformation for which DeltaHgt0 Determine the number abcd [Note:If a =1, b=2,c=3 and d=4, then answeris 1234]

Answer»


ANSWER :2439
Discussion
45.

Consider the following molecules underset(I)(O_(2)),O_(2)underset(II)(("AsF"_(6))),underset(III)(KO_(2)) Choose the correct answer.

Answer»

The correctdecreasing bond ORDER is`II gt I gt III`.
The correct DECREASING order of bond length is `III gt II gt I`.
The bond strength of I is less than that of III.
Bond dissociation energy is highest in case of III.

Solution :In `O_(2)(AsF_(6)), O_(2)^(+)` has bond order 2.5 .
`O_(2)` has bond order 2.0 .
In `KO_(2), O_(2)^(-)` has bond order 1.5
Higher the bond order, smaller is the bond length.
HENCE, the order is `O_(2)(AsF_(6)) gt O_(2) gt KO_(2)`.
Discussion
46.

Consider the following molecules or ions : (i) CH_(2) Cl_(2)(ii) NH_(4)^(+) (iii) SO_(4)^(2-) (iv) ClO_(4)^(-) (v) NH_(3) sp^(3) hybridisation is involved in the formation of

Answer»

(i), (II), (v) only
(i) , (ii) only s
(ii) only
(i), (ii), (iii), (IV) and (v)`

Solution :All the GIVEN ions/molecules involve involve `sp^(3) ` hybridisation .
Discussion
47.

Consider the following list or reagents: Acidified K_(2)Cr_(2)O_(7), alkaline KMnO_(4),CuSO_(4),H_(2)O_(2),CI_(2),O_(3),FeCI_(3),HNO_(3) and Na_(2)S_(2)O_(3). The total number of reagents that can oxidise aqueous iodide to iodine is:

Answer»


ANSWER :7
Discussion
48.

Consider the following liquid - vapour equilibrium. Liquid hArr Vapour Which of the following relations is correct ?

Answer»

`(dlnP)/(DT)=-(DeltaH_V)/(RT)`
`(dlnP)/(dT^2)=-(DeltaH_V)/T^2`
`(dlnP)/(dT)=(DeltaH_V)/(RT^2)`
`(DLNG)/(dT^2)=(DeltaH_V)/(RT^2)`

ANSWER :C
Discussion
49.

Consider the following liquid -vapour equilibrium Liquid hArrVapour Which of the following relations is correct.

Answer»

`(dlnP)/( DT^(2))= - ( DeltaH _(v))/( T^(2))`
`(dlnP)/( dT)=( DeltaH _(v))/( RT^(2))`
`(dlnS)/( dT^(2))=( DeltaH _(v))/( RT^(2))`
`(dlnS)/( dT)=- ( DeltaH _(v))/( RT)`

SOLUTION :It REPRESENTS Clausius-Clapeyron euation.
Discussion
50.

Consider the following isothermal of a gas: which gas law is illustrated by these diagrams?

Answer»

SOLUTION :Boyle.s LAW
Discussion