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.

How much water should be added to 100 ml of1MH_(2)SO_(4) solution to make it exactly 0.1 N ?

Answer»

1200 ml
400 ml
1800 ml
1900 ml

Solution :100 ml of 1M `H_(2)SO_(4)=100` ml of `2N H_(2)SO_(4)`
Applying `N_(1)V_(1)=N_(2)V_(2)`
`2xx200=0.1xxV_(2)`
or `V_(2)=(1xx200)/(0.1)=2000` ml
Amt. of water to be ADDED `=2000-100`
`=1900` ml.
2.

How much volume of oxygen will be required for complete combustion of 40 mL of acetylene (C_(2)H_(2)) and how much volume of carbon dioxide will be formed? All volumes are measrued at NTP.

Answer»

Solution :`{:(2C_(2)H_(2),+,5O_(2),=,4CO_(2),+,2H_(2)O),("2vol",,"3vol",,"4vol",,),(40mL,,(5)/(2)xx40mL,,(4)/(2)xx40mL,,),(40mL,,100mL,,80,,):}`
So, for complete combustion of 40mL of acetylene, 100 mL of OXYGEN are required and 80mL of CARBON dioxide is formed.
3.

How much urea must be dissolved in 10^(-2)m^(3) water to yield a pressure is 2.03xx10^(5)Nm^(-2) at 300K?

Answer»

48.30 g
60.0 g
24.15 g
48.03 g

Solution :OSMOTIC PRESSURE `=2.03xx10^(5)NM^(-2)`
`pi=CRT`
`2xx10^(5)=m/60xx1000=80.186xx60xx0.01g` in `0.01m^(3)=48.3gimplies1J=1Nm`
4.

How much urea (molar mass = "60 g mol"^(-1)) should be dissolved in 50 g of water so that its vapour pressure at room temperature is reduced by 25%. Calculate molality of the solution obtained.

Answer»


Solution :If `p^(@)=100mm, p_(s)=75MM`. Hence, `=(100-75)/(100)=(w_(2)//60)/(50//18+w_(2)//60).` This gives `w_(2)=55.556g`
`"Molality of the solution "=(55.556//"60 mol")/(50 g)xx1000g KG^(-1)=18.52m`
5.

How much time would it take in minutes to deposit 1.18 g of metallic copper on a metal object when a current of 2.0 A is passed through the electrolytic cell containing Cu^(2+) ions? [Molar mass of Cu = 63.5 g mol^(-1), 1F = 96,500 Cmol^(-1)]

Answer»

SOLUTION :`m = Z xx I xx t`
`1.18 = (63.5)/(2 xx 96500) xx 2 xx t`
`t = (1.18 xx 2 xx96500)/(2 xx 63.5)`
`= 1793.23` SEC
`= (1793.23)/(60) = 29.88` min.
6.

How much type of colloidal systems can be formed on the basis of physical state of dispersion medium and dispersion phase ?

Answer»

Two
Three
Nine
EIGHT 

SOLUTION :Eight
7.

How much time would be needed to deposit 0.25 g of metallic nickel. (Atomic mass = 58.5) on a metal object using a current of 1A during electroplating?

Answer»

Solution :`Ni^(++) + 2e^(-) to Ni`
2F or `2 xx 96500` C of electricity is required to deposit 58.5 g of Ni.
58.5 g of Ni REQUIRES `2 xx 96500` C
0.25 g Ni requires `=(2 xx 96500 xx 0.25)/58.5 = 824.8` C
Amount of electricity = CURRENT in amperes `xx` time in SECONDS
Time in seconds `=("Amount of electricity")/("Current in amperes")`
`=824.8/1 = 824.8 sec = 13.74` MINUTES.
8.

How much time in minutes passes between 99% to 99.9% completion of a first order reaction with half life 0.3010"min"^(-1)?

Answer»


SOLUTION :`(0.693/0.3010)xxt_(1)=2.303xxlog(100/1)t_(1)=2,0.693/0.310xxt_(2)=2.303xxlog(100/0.1)t_(2)=3`
`DELTA=(t_(2)-t_(1))=(3-2)=1,Deltat=1`
9.

How much time (in hours ) would it take to distribute one avogardo number of wheat grains if 10^20 grains are distributed each second ?

Answer»

0.1673
1.673
16.73
167.3

Solution :Time TAKEN to DISTRIBUTE `10^20` GRAINS = 1 second
Time taken to distribute `6.022 XX 10^23`grains
` = (1)/(10^20) xx 6.022 xx 10^23 ` seconds
` = (6.022 xx 10^23)/(10^20 xx 60 xx 60) ` hours
= 1.673 hours
10.

How much time (in hours) is required for a current of 2 amp to decompose electrolytically 18g of water.

Answer»


ANSWER :26.8
11.

How much time is required for completion of a zero order reaction?

Answer»

`(2[R_(o)])/(K)`
`[R_(o)]/(2K)`
`([R_(o)])/(K)`
`(K)/([R_(0)])`

ANSWER :C
12.

How much sulphur is present in organic compoundif on analysis 0.53 gmof thiscompound gives 1.158 gm of BaSO_(4).

Answer»

`10%`
`15%`
`20%`
`30%`

Solution :Percentage of sulphur
`= (32)/(233) xx (" MASS of" BaSO_(4))/(" mass of ORGANIC compound") xx 100`
`= (32)/(233) xx (1.158)/(0.53) xx 100 = 30%`.
13.

How much sulpher is present in the rubber used for tyres ?

Answer»

`5 %`
`10 %`
`20 %`
`30 %`

SOLUTION :`5 %`
14.

How much sulphur is added during vulcanization to get rubber for tyre ?

Answer»

`30 %`
`5%`
`55%`
`3 %`

SOLUTION :`5 %`
15.

How much sulpher is present in the rubber used for bettery case ?

Answer»

`5%`
`30 %`
`20 %`
`50 %`

SOLUTION :`30 %`
16.

How much sugar (C_(12)H_(22)O_(11)) will be required if each person on the earth is given 100 molecules of sugar? The population of the earth is 3xx10^(10).

Answer»


ANSWER :`170.43xx10^(-11)G`
17.

How much sodium chloride be dissolved in water to make 1 litre of 0.1 F solution ?

Answer»


Solution :Gram formula MASS NaCI=23+35.5=58.5 G `("Formula mass")^(-1)`
VOLUME OS solution = L
Formality of solution = 0.1 F=0.1 (Formula mass)/LITRE
`"Formality (F)"=("Mass os NaCI/Gram formula mass")/("Volume of solution in litres")`
`"0.1 (Formula mass/litre)"=W/(58.5 "g (Formula mass)"^(-1)xx1L)`
`W=0.1 ("Formula mass/L")xx58.5 g("Formula mass")^(-1)xx1L`
` W=0.1xx58.8 g= 5.85 g`.
18.

How much sodium acetate should be added to a 0.1 M solution of CH_(3)COOH to give a solution of pH = 5.5 (pK_(a) of CH_(3)COOH = 4.5)

Answer»

0.1 M
0.2 M
`1.0 M`
`10.0 M`

SOLUTION :`pH = pK_(a) + log.(["Salt"])/("Acid") , 5.5 = 4.5 + log.(["Salt"])/("Acid")`
`log.(["Salt"])/("Acid") = 5.5 - 4.5 = 1`
`(["Salt"])/(0.1) = "antilog 1" = 10, ["Salt"] = 1` .
19.

How much should the pressure be increased in order to decrease the volume of a gas 5% at a constant temperature ?

Answer»

<P>0.05
`5.26 %`
0.1
`4.26 %`

Solution :PV = a constant
New volume =0.95 V
`therefore P_(1)V=P_(2)xx0.95 V`
`P_(2)=(P_(1))/(0.95)=1.0526P_(1)`
`therefore` INCREASE in pressure `=1.0526P_(1)-P_(1)`
`=0.0526P_(1)`
`=5.26%" of "P_(1)`
HENCE, (B) is the correct answer.
20.

How much quantity of electricity present on 1 electron in coulomb ?

Answer»

`6.02xx10^(23)`
`6.02xx10^(19)`
`1.602xx10^(-19)`
`0.1602xx10^(-20)`

SOLUTION :`0.1602xx10^(-20)`
21.

How much quantity of electricity is required to liberated 1 g equivalent weight of an element ?

Answer»

96,500 Faradays
96,500 Coulombs
96,500 AMPERES
96,500 gram

Answer :B
22.

How much quantity of electricity has to be passed through 200 mL of 0.5 M CuSO_(4) solution to completely deposite copper ?

Answer»

96500 C
`2 XX 9650 C`
`2 xx 96500 C `
`4 xx 96500 C`

Solution :200 ml of 0.5 M `CuSO_(4)` CONTAINS `CuSO_(4)` = 0.1 mole .
`Cu^(2+) + 2E^(-) to Cu`
To deposit 0.1 mole of `Cu^(2+)` , electricity required = `2XX 9650` C
23.

How much Pt is deposited on cathode when 0.80F current is pass through 1.0 M solution of Pt^(4+) ?

Answer»

1.0 MOLE
0.20 mole
0.40 mole
0.80 mole

Solution :`PT^(4+) +4E^(-) to Pt`
`4F=1` mole Pt
`THEREFORE 0.80F=(?)`
`=(1xx0.80)/(4)=0.20` mole.
24.

How much propanol is required for dehydration to get 2.24 litre of propene at N.T.P. If yield is 100%

Answer»

Solution :`C_(3)H_(8)O+H_(2)SO_(4)rarrC_(3)H_(6)+H_(2)O+H_(2)SO_(4)`
Molecular weight of propanol `=60`
from the equation given above we can see that from dehydration of `1` mole or `60` GRAM of propanol we get `1` mole (`22.4`lit) of PROPENE as PRODUCT.
`because 22.4` litre of `C_(3)H_(6)` can be get from dehydration of `60g` of propanol.
`therefore 1` litre of propene can be get from dehydration of `(60)/(22.4)g` of propanol
25.

How much PCl_(5) must be added to one litre volume reaction vessel at 250^(@)C in order to obtain a concentration of 0.1 mole of Cl_(2),K_(c) for PCl_(5)hArrPCl_(3)+Cl_(2) is 0.0414" mol dm"^(-3) at 250^(@)C.

Answer»


ANSWER :`0.341` MOLES `PCl_(5)`
26.

How much PCl_(5) must be added to one litre volume reaction vessel at 250^(@)C in order to obtain a concentration of 0.1 mole of Cl_(2),K_(c) for PCl_(5)hArrPCl_(5)+Cl_(2) is 0.0414" mol dm"^(-3) at 250^(@)C.

Answer»


ANSWER :`0.341` MOLES `PCl_(5)`
27.

How much oxygen is required for complete combustion of 560 g of ethene (Mw = 28 g/mol) ?

Answer»

6.4 KG
1.92 kg
2.8 kg
9.6 kg

SOLUTION :C2H4(G) + 3O2(g) `hArr` 2CO2(g) + 2H2O(ℓ)
`(560)/(28) mol `
28.

How much of sulphur is present in an organic compound, if 0.53g compound gave 1.158g of BaSO_(4) analysis?

Answer»

0.1
0.15
0.2
0.25

Answer :C
29.

How much of NaOH is required to neutralise 1500cm^(3) of 0.1 N HCl ? (Na = 23)

Answer»

40 G
4 g
6 g
60 g

Solution :`1500cm^(3)" of of 0.1 N HCl"-=(0.1)/(1000)xx1500="0.15 g eq."`
It will neutralise `NaOH=0.15" g eq."=0.15xx40g=6g`
30.

How much of NaOH is required to neutralise 1500 cm^(3)of 0.1 N HCI? (Na = 23)

Answer»

40 G
4 g
6 g
60 g

Solution :`1500 cm^(3)` of 0.1 N HCL `=0.1/1000 xx 1500 = 0.15` g eq.
It will NEUTRALISE `NaOH = 0.15 g eq. = 0.15 xx 40 g = 6 g`
31.

How much of NaOH is required to neutralise 1500 cm^(3) of 0.1N HCl? (Na= 23)

Answer»

4g
40g
60g
6g

Solution :As, `(W)/(E.W) = (V XX N)/(1000)`
E.W (NaOH) = 40M V= 1500 `cm^(3)`,
N= 0.1 N, W= ?
`rArr W= (1500 xx 0.1 xx 40)/(1000) rArr W= 6gm`
32.

How much of Fe can be theoretically obtained by the reaction of 1 kg of Fe_(2)O_(3)?

Answer»


SOLUTION :`underset(2xx56+3xx16=160g)(Fe_(2)O_(3))-=underset(2xx56=112g)(2Fe)`
33.

How much moles of oxidizing agents are reduced by complete reaction of 63.5 gm of Cu with concentrated HNO_(3) solution ? (Atomic weight of Cu=63.5 gm mol^(-1))

Answer»

8
4
2
1

Answer :C
34.

How much minimum volume (in ml) of ((5)/(51)) M aluminium sulphate solution should be added to excess calcium nitrate to obtain atleast 1 gm of each salt in the reaction Al_(2) (SO_(4))_(3) + 2Ca(NO_(3))_(2) rarr 2Al(NO_(3))_(3) + 3CaSO_(4)

Answer»


ANSWER :25 ML
35.

How much NaOH is required to prepare 10% w/w 500 gram solution of NaOH ?

Answer»

0.5 GRAM
100 gram
5.0 gram
50 gram

Solution :`%(W)/(w)=(100xx"mass of solute")/("mass of solution")`
`THEREFORE 10=(100xx"mass of solute")/(500)`
`therefore` Mass of solute `= (10xx500)/(100)=50` gram.
36.

How much molecular mass of NaCl is obtained experimentally using colligative properties?

Answer»

SOLUTION :As CONCENTRATION of particles in the solution is doubled, value of COLLIGATIVE property is double than expected value and hence molecular mass is HALVED, i.e. OBSERVED molecular mass = 58.5/2 = 29.25
37.

How much methanol should be added to water, to make 150 mL. 2 M solution of CH_(3)OH ?

Answer»

9.6 gram
2.4 gram
`9.6xx10^(3)` gram
`4.3xx10^(2)` gram

Solution :1000 mL 1 M `CH_(3)OH=32` gram
`therefore 150 mL 2M CH_(3)OH= ?`
`= (32xx150xx2)/(1000)=9.6` gram.
38.

How much marble of 96.5% purity would be required to prepare 10 litres of carbon dioxide at STP when the marble is acted upon by dilute hydrochloric acid?

Answer»


Solution :`underset(100g)(CaCO_(3))+2HCL rarrCaCl_(2)+H_(2)O+underset("22.4 at STP")(CO_(2))`
10 L of `CO_(2)` at STP will be obtained from pure `CaCO_(3)=(100)/(22.4)xx10=44.64g`
`"Impure marble required "=(100)/(96.5)xx44.64=46.26g`
39.

How much marble mainly [CaCO_(3)] " of " 96 . 5%purity would be required to prepare 10 litres of CO_(2) at STP when the marble is acted upon by dilute HCl ?

Answer»


SOLUTION :N/A
40.

How much litres of dihydrogen gas will be produced at STP in the reaction of ethanol with 12 gram of Mg ? (Mg = 24 gram//mol)

Answer»

11.2 LITER
2.24 liter
22.4 liter
5.6 liter

Solution :11.2 liter
41.

How much is the oxidising power of the MnO_4^(-) (1M)|Mn^(2+) (1M) couple decreased if the H^+ concentration is decreased from 1M to 10^-4 M at 25^@C

Answer»

Solution :The HALF cell REACTION is `MnO_4^(-) +8H^+ +5e=Mn^(2+)+4H_2O` Let us suppose that only the `H^+` concentration deviates from 1mole`//` litre
`E_(MnO overline4. Mn^(2+))=E_(MnO_4^-,Mn^(2+))- 0.0591/2 log""([Mn^(2+)])/([MnO_4^-][H^+]^8)`
`E-E^@=-0.0118 log""((1))/((1)(10^-4)^8)=-0.38` volt
The couple `MnO_4^-//Mn^(2+)` has thus moved down by 0.38 volt from its standard value i.e to a position of less oxidising POWER.
42.

How much ice will separate if a solution containing 25g of ethylene glycol [C_(2)H_(4)(OH)_(2)] in 100g of water is cooled to -10^(@)C, ? K_(f)(H_(2)O)=1.86

Answer»


ANSWER :`25.05g`
43.

How much heat will be required at constant pressurent to form 1.28 kg of CaC_(2) from CaO(s) & C(s)? {:("Given :",Delta_(f)H^(@) (CaO","s),=,-152 kcal//mol),(,Delta_(f)H^(@) (CaC_(2)","s),=,-14 kcal//mol),(,Delta_(f)H^(@) (CO","g),=,-26 kcal//mol):}

Answer»

`+112` KCAL
`224` kcal
`3840` kcal
`2240` kcal

Answer :D
44.

How much heat is produced in buring a mole of CH_(4) under standard conditions if reactants and products are brought to 298 K and H_(2)O(l) is formed ? What is the maximum amount of useful work thatcan be accomplished under standard conditions by this system ? CH_(4)(g)+2O_(2)(g) to CO_(2)(g)+2H_(2)O(l) {:(DeltaHf^(0):-74.8,0,-393.5,-285.85kJ),(DeltaGf^(0):-50.8,0,-394.4,-236.8kJ):}

Answer»

Solution :`DeltaH^(@)=(-393.5)+(2xx-285.85)-(-74.8)-2xx0`
`=-890.4kJ`/MOLE
Now the free ENERGY CHANGE for a process `DeltaG` equals the maximum useful work that can be done by the system at constant TEMPERATURE and pressure.
`:. W_(max)=DeltaG=(2xx-236.8)+(-394.4)-(50.8)-2xx0`
`=-817.2kJ`/mole `CH_(4)`.
45.

How much heat is required to change 5 gram ice (0^@C) to steam at 100^@C ? Latent heat of fusion and vaporization for water are 80 cal/g and540 cal/g respectively . Specific heat of water is 1 cal/ g/k.

Answer»

7200 cal
3600 cal
1800 cal
900 cal

ANSWER :B
46.

How much heat is required to change 10 g of ice at 0^@Cto steam at 100^@C?

Answer»

SOLUTION :`Delta H("total") = Delta H_("FUSION") + Delta H_("heating") + Delta H_(vap)`
` = 10(80) + 1 XX 100 + 540 ) ` CAL
47.

How much gram of acid should be present in 100 mL volume of solution to obtain decimolar solution having 200 gm/mole molecular mass of diabasic acid ?

Answer»

20 gm
1 gm
2 gm
10 gm

SOLUTION :Decimolar means 0.1 M
`M=("MOLE")/("Liter")=(w)/(200)xx(1)/(100)xx1000 =0.1`
`therefore 0.1=(w)/(20) therefore 2=2` gm.
48.

How much H_(2)SO_(4) is required to prepare 0.1 M 1 liter H_(2)SO_(4) solution ? The density of 98% W/W H_(2)SO_(4) is 1.80 gm/mL.

Answer»

22.20 mL
5.55 mL
11.10 mL
16.65 mL

SOLUTION :`M=("% of solute"xx10xx"density")/("molecular WEIGHT")`
`M=(98xx1.80xx10)/(98)=18M`
`M_(1)V_(1)=M_(2)V_(2)`
`18xx V_(1)=0.1xx1000`
`V_(1)=(0.1xx1000)/(18)=5.55 mL`
49.

How much glucose, C_(6)H_(12)O_(6) must be dissolved per litre of the solution that yields osmotic pressure os 2.72 atmospheres at 298 K ? R=0.0821 L atm K^(-1).

Answer»


Solution :`"Molar mass of GLUCOSE "C_(6)H_(12)O_(6), M_(B)=6xx12+12+6xx16=180" G mol"^(-1)`
`V=1.0L, pi=2.72 ATM, R=0.0821" L atm K"^(-1)mol^(-1), T=298K, W_(B)=?`
`piV=n_(B)RT or n_(B)=(piV)/(RT),W_(B)/M_(B)=(piV)/(RT)`
`W_(B)=(M_(B)xxpixxV)/(RT)=((180" g mol"^(-1))xx(2.72" atm")xx(1L))/((0.0821"L atm K"^(-1)mol^(-1))xx(298 K))=20.0116 g.`
50.

How much faster would a reaction proceed at 25^@ than at 0^@C if the activation energy is 65 kJ:

Answer»

2 times
16 times
11 times
6 times

Answer :C