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 many grams of NaOH must be present in one litre of the solution of give it a pH = 12?

Answer»

0.20 g/lit
0.4 g/lit
4 g/lit
0,10 g/lit

Answer :B
2.

How many grams of NaOH are equivalent to 100 ml of 0.1 N oxalic acid

Answer»

0.2
2
0.4
4

Solution :`N=(W_(B)xx1000)/("EQ. WT."xxV),W_(B)=(Nxx"Eq. wt"xxV)/(1000)`
`=(0.1xx40xx100)/(1000)=0.4`
3.

How many grams of NaBr must be added to 270g of water to lower the vapour pressure by 3.125mm Hg at which vapour pressure of water is 50mm Hg (Na=23,Br=80) ? Assume 100% ionisation of NaBr.

Answer»

<P>

SOLUTION :`NaBrhArrNa^(+)+Br^(-)`
van'tHoff FACTOR `i=2`
let moles of `NaBr=2n_(1)`
mole of `H_(2)O(mn_(2))=(270)/(18)=15`
`(DeltaP)/(P_("solvent")=X_("solute")=(n_(1)i)/(n_(1)i+n_(2))`
`(3.125)/(50)=(2n_(1))/(2n_(1)+15)`
`(2n_(1)+15)/(2n_(1))=(50)/(3.125)=16`
`1+(15)/(2n_(1))=16`
`(15)/(2n_(1))=15`
`n_(1)=0.5`
`therefore NaBr` taken `=0.5` mol `=0.5xx103=51.5g`.
4.

How many grams of NaCl can be added to 785 mL of 0.0015 M AgNO_3before a precipitate forms? K_(sp) (AgCl) = 1.8 xx 10^(-10) .

Answer»

SOLUTION :`5.5 XX 10^(-6) G`
5.

How many grams of Na_(2)CO_(3) should be dissolved in 250 g of water to prepare 0.1 m solution?

Answer»


Solution :`"0.1 m solution MEANS 0.1 mole of "Na_(2)CO_(3)" is present in 1000 g of WATER"`
`therefore " 250 g of water should contain "Na_(2)CO_(3)=(0.1)/(4)"Mole = 0.025 mole "=0.025xx106 g = 2.65 g`
6.

How many grams of methyl alcohol should be added to I 0 litre tank of water to prevent its freezing at 268 K ? (K _(f) for water is 1.86 K k g mol ^(-1))

Answer»

`880.07g`
`899.04g`
`886.02g`
`868.06g`

Solution :`DELTA T _(f ) = K _(f ) m`
where m = molality
`273 -268 =1.86 xx(W)/( M X v)`
` 5=1 .86 xx(w)/( 32xx10)`
`w= ( 5xx32xx10)/(1.86)`
`= 560.2 ~~ 868 .06g`
7.

How many grams of KCI should be added to 1.0 kg of water to lower its freezing point to -8^(@)C. Calculate the degree of molality of BaCI_(2) in this solution. (K_(b) "for water "=0.52 K m^(-1))

Answer»


Solution :In solution, KCI dissociates as :
`KCIoverset(("AQ"))toK^(+)CI^(-)("aq")therefore" Van't Hoff factpr"(i)=2`
`DeltaT_(f)=0^(@)C-(-08^(@)C)=8^(@)C=8C=8K, *K, K_(f)=1.86" K KG mole"^(-1)`
`DeltaT_(f)=ixxK_(f)xxm or m=(DeltaT_(f))/(ixxK_(f))`
`m=((8K))/(2xx(1.86"K kg mol"^(-1)))=2.15" mol kg"^(-1)=2.15m`
`therefore"Mass of KCI in grams = (2.15 mol)XX(74.5" g mole "^(-1))=162.2 g.`
8.

How many grams of ice at 0^(@)C can be melted by the addition of 500 J of heat (The molar heat of fusion for ice is 6.02 kJmol^(-1))

Answer»

0.0831 G
1.50 g
3.01 g
12.0 g

Answer :B
9.

Solubility product of silver bromide is 5.0xx10^(-13). The quantity of potassium bromide (molar mass taken as 120 g mol^(-1)) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is :

Answer»

`2.5xx10^(-14)g`
`1.2xx10^(-9)g`
`1.2xx10^(-11)g`
`2.5xx10^(-9)g`

ANSWER :B
10.

How many grams of I_2 are present in a solution which require 40 ml of 0.11 N Na_2 S_2 O_3 to react with it ? S_2 O_3^(2-) + I_2 to S_4 O_6^(2-) + 2I^(-)

Answer»

12.7 G
25.4 g
0.558 g
4.25 g

Answer :C
11.

How many grams of H_2SO_4 are present in o.25 mole of H_2SO_4.

Answer»

2.45
24.5
0.245
0.25

Answer :B
12.

How many grams of gas would be adsorbed per gram of a substance at 8 atm by assuming Freundlich adsorption isotherm. (x)/(m) = kp^(1//m)"" and "" k = 10^(-2)atm^(-1//3)"" & "" n = 3.

Answer»

SOLUTION :`0.02 G`
13.

How many grams of H_(2) and O_(2) are produced during the electrolysis of water under a 1.30 amp of current for 5 hours ? What volumes of dry gases are produced at NTP ?

Answer»

SOLUTION :`{{:(H_(2)-0.245 G, O_(2)-1.94 g),(H_(2)-2.72 L, O_(2) - 1.36 L):}}`
14.

How many grams of copper will be replaced in 2L of a 1.50 M CuSO_(4) solution if the latter is made to react with 27.0 of aluminium (Cu=63.5, AI= 27.0)

Answer»

`190.50g`
`95.25g`
`31.75g`
`10.65g`

SOLUTION :2 L of 1.5 M `CuSO_(4)=3` MOL of `CuSO_(4)`
`underset("2 mol")(2Al)+underset("3 mol")(3CuSO_(4))to Al_(2)(SO_(4))_(3)+underset("3 mol")(3 CU)`
2 mol of Al = 3 mol of Cu
(27g = 1 mol of Al)
`=(3)/(2)xx6.3g` of Cu = 95.25 g
15.

How many grams of concentrated nitric acid solution should be used to prepare250 mL of 2.0M HNO_(3) ? The concentrated acid is 70% HNO_(3)

Answer»

`90.0` G conc. `HNO_(3)`
`70.0` g conc. `HNO_(3)`
`54.0` g conc. `HNO_(3)`
`45.0` g conc. `HNO_(3)`

SOLUTION :Molarity `(M) =(wt xx 1000)/( " mol . Wt x col (ml))`
`2= (wt)/(63) GM`
wt. of `70%` acid `=(100)/( 70) xx 31.5 =45` gm
16.

How many grams of concentrated nitric acid solution should be used to prepare 250 mL of "2.0 M HNO"_(3) ? The concentrated nitric acid is 70%HNO_(3)

Answer»

`"45.0 G conc HNO"_(3)`
`"90.0 g conc HNO"_(3)`
`"70.0 g conc HNO"_(3)`
`"54.0 g conc HNO"_(3)`

SOLUTION :250 ML of 2.0 M `HNO_(3)` will contain `HNO_(3)`
`=(2)/(1000)xx250=?0.5 MOL "=0.5xx63 g = 31.5g`
As concentrated `HNO_(3)` is `70%` , therefore, concentrated `HNO_(3)` required `=(100)/(70)xx31.5g=45g`
17.

How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M HNO_(3) ? The concentrated acid is 70% HNO_(3) .

Answer»

90.0 g conc. `HNO_(3)`
70.0 g conc. `HNO_(3)`
54.0 g conc. `NHO_(3)`
45.0 g conc. `HNO_(3)`

SOLUTION :MOLARITY `=("MOLES of" HNO_(3))/("Vol. of solution")xx1000`
`2=("mass"//63)/(100)xx63=31.5`
Mass of `HNO_(3)` required for 70% solution
`=(31.5xx100)/(70)=45.0g`
18.

How many grams of chlorine can be produced by the electrolysis of molten NaCl with a current of 1.00 A for 15 min? Also calculate the number of chlorine molecules liberated.

Answer»


SOLUTION :`Q=900C`
`2CL^(-)toCl_(2)+2E^(-),""2xx96500" C liberate "Cl_(2)=1`mole
`therefore900C` will liberate `Cl_(2)=(1)/(2xx96500)xx900=4.66xx10^(-3)"mole"=(4.66xx10^(-3))xx71g=0.331g`
`=(4.66xx10^(-3))xx6.02xx10^(23)` molecules `=2.80xx10^(21)` molecules.
19.

how many grams of cobalt metal will be deposited when solution of cabalt (II) chloride is electrolyzed with a current of 10 amperes for 109 minutes (1Faraday=96,500 C, Atomic mass of Co=59u)

Answer»

4
20
40
0.66

Solution :`w=Zit`
`thereforew=(E_(B)xxIt)/(96500) ""(thereforeE_(B)=(M_(B))/(Z)=(59)/(2))`
`w=(59)/(2)xx(10xx109xx60)/(96500)`
w=19.9
20.

How many grams of CH_(3)OH should be added to water to prepare 150 ml solution of 2.0 M CH_(3)OH ?

Answer»

9.6
2.4
`9.6xx10^(3)`
`2.4xx10^(3)`

Solution :MOLARITY `=("Wt. of " CH_(3)OH xx 1000)/("Mol. Wt." xx "VOLUME")`
`2.0=("Wt. of " CH_(3)OH xx 1000)/(32xx150)`
`:.` Wt. of `CH_(3)OH=(2.0xx32xx150)/(1000)=9.6 G`
21.

How many grams of calcium oxalate should be dissolved in water to make one litre of saturated solution? K_sp of CaC_2O_4 is 2.5 x 10^-9 and its molecular weight is 128 u

Answer»

`6.4*10^-3 G`
`8.0*10^-3 g`
`1.28*10^-3 g`
`6.4*3.2^-3 g`

ANSWER :A
22.

How many grams of CaO are required to neutralise 852g of P_(4)O_(10)?

Answer»

Solution :The REACTION will be:
`6CaO+P_(4)O_(10)rarr2Ca_(3)(PO_(4))_(2)`
`852g P_(4)O_(10) m3 "mol" P_(4)O_(10)`
1 MOLE of `P_(4)O_(10)` neutralises 6 MOLES of `CAO`
`therefore 3 "moles of" P_(4)O_(10)` will neutralise 18 moles of CaO.
`therefore "mass of" CaO=18xx56=1008g`
23.

How many grams of CaF_2must be dissolved in 100 mL of water at 25^@Cto make the solution saturated? K_(sp)(CaF_2) = 3.9 xx 10^(-11)

Answer»

SOLUTION :`1.66 XX 10^(-3)G `
24.

How many grams of CaC_(2)O_(4) (molecular weight = 128) on dissolving water will give a saturated solution [K_(sp)(CaC_(2)O_(4)) = 2.5 xx 10^(-9) mol^(2) I^(-2)]

Answer»

`0.0064 G`
`0.1280g`
`0.0128 g`
`1.2800 g`

Solution :Solubility of `CaC_(2)O_(4) = sqrt(K_(sp)) = sqrt(2.5 XX 10^(-9))`
`= 5 xx 10^(-5) mol L^(-1)`
`= 5 xx 10^(-5) xx 128 = 640 xx 10^(-5) = 0.0064 g`.
25.

How many grams of CaC_(2)O_(4) will be dissolved in 1L of distilled water to make its saturated solution ? (K_(sp)" of " CaC_(2)O_(4)=2.5xx10^(-9)and mol. wt. of CaC_(2)O_(4)=128) :

Answer»

0.0064 G
0.0128 g
0.0032 g
0.0640 g

ANSWER :A
26.

How many gramsof CaC_2O_4 saturation will dissolve in one litre of saturated solution ? ( K_sp of CaC_2O_4 is 2.5 xx 10^_9 mol^-2 and its molecular weight is 128).

Answer»

0.0064g
0.0128g
0.0032 g
0.0640 g

Answer :A
27.

How many grams of barium chloride (BaCl_(2)) are needed to prepare 100cm^(3) of 0.250 mL of a 0.50 M BaCl_(2) solution?

Answer»


SOLUTION :Molecular mass of `BaCl_(2)=137+71=208u`.
For 1000 `CM^(3)` of 1 M `BaCl_(2)` so. Mass of `BaCl_(2)` needed `=(208)/(1000)x100xx0.25g=5.20g`
28.

How many grams of a non volatile solute having a molecular mass of 90 g/mole are to be dissolved in 97.5 g water in order to obtain relative lowering in the vapour pressure of 2.5 percent:

Answer»

25
18
12.5
9

Answer :C
29.

How many grams of a dibasic acid (Mol. Wt.=200) should be present in 100mL of its aqueous solution to give decinormal strength?

Answer»


ANSWER :1
30.

How many grams atoms of S are present in 0.25 g of S

Answer»

2.5
32
4
6

Answer :A
31.

How many gram of KMnO_(4) are contained in 4 litres of 0.05N solution? The KMnO_(4) is to be used as an oxidant in acid medium. ( Mol. wt. of KMnO_(4) = 158)

Answer»

1.58g
15.8 g
6.32g
31.6g

Answer :C
32.

How many gram of KMnO_(4)are contained in 4 litre of 0.05 N solution . The KMmO_(4) is to be used as an oxidant in acidic medium ?

Answer»

1.58 g
15.8 g
6.32 g
31 .6 g

Answer :3
33.

How many gram of HCl will be present in 150 ml of its 0.52 M solution

Answer»

2.84 GM
5.70 gm
8.50 gm
3.65 gm

Solution :`M = (W)/(mxx V(L)), 0.52 = (w)/(36.5xx0.15), w = 2.84 gm`
34.

How many gram atoms of oxygen are present in 11.2 litres of CO_2at NTP ?

Answer»


ANSWER :1 GRAM ATOM
35.

How many gm of the oxidising agent gets reduced in the reaction of 65.4 gm of Zn with concentrated nitric acid ?

Answer»

126
252
130.8
65.4

Solution :`Zn + 4HNO_(3) to Zn(NO_(3))_(2) + 2H_(2)O + 2NO_(2)`
36.

How many gm of H_(2)SO_(4) is present in 0.25 gm mole of H_(2)SO_(4)

Answer»

`24.5`
`2.45`
`0.25`
`0.245`

SOLUTION :`N=(W)/(m) , w=n XX m =0.25xx98 =24.5 GM`
37.

How many gm. of ethanol is required in the reaction with Na metal in order to give 560 ml. dihydrogen gas at STP ? [mole. mass of ethanol = 46 gm mol^(-1)]

Answer»

1.15
2.3
4.6
11.5

Solution :`underset("(?)")(underset(46 "gram")(C_(2)H_(5)OH))+Na to C_(2)H_(5)ONA + underset(560)(underset(11200 "mili")((1)/(2)H_(2)))`
`(560 XX 46)/(11200)= 2.3` gram Na is NEEDED.
38.

How many gm of bromine will react with 21 gm C_3H_6

Answer»

80
160
240
320

Solution :`undersetundersetunderset"42 GMS""1 MOLE""PROPENE"(CH_3-CH)=CH_2+undersetunderset"160 gms""1 mole"(Br_2)to CH_3-underset"1,2-Dibromo propane"(undersetunderset(Br)(|)CH_2-undersetunderset(Br)(|)CH_2)`
`because` 24 gms of propene reacts with 160 gms of bromine
`because` 21 gms of propene `160/42xx21=80` gms
39.

How many germs of K_(2)C_(7)O_(7) are required to oxidise 20.0g of Fe^(+) ion in FeSO_(4) "to" Fe^(3+) ions if the reaction is carried out in the acidic medium? Molar mass of K_(2)Cr_(2)O_(7) and FeSO_(4) are 294 and 152 respectively.

Answer»

6.45g
7.45g
8.45g
9.45g

Solution :Number of moles of `FeSO_(4)=((20g))/(152g mol^(-1))`
=0.132mol
Ionic equation for the OXIDATION reaction carried in the acidic medium may be WRITTEN as:
`Cr_(2)O_(7)^(2-)+6Fe^(2+)+14H^(+) to 2Cr^(3+)+6fe^(3+)+7H_(2)O`
6 mol. of `Fe^(2+) -= ` 1 mol of `K_(2)Cr_(2)O_(7)`
0.132 mol of `Fe^(2+)-=((1 mol))/((6 mol))xx(0.132 mol^(-1)`
= 0.022 mol . of `K_(2)Cr_(2)O_(7)`
MASS of `K_(2)Cr_(2)O_(7) = (0.022 mol) xx(294 g mol ^(-1))`
= 6.468 g.
40.

How many geometrical isomers of 2-Methyl-7-chloronobonmane are possible ?

Answer»


ANSWER :4
41.

How many geometrical isomers of this compound are possible:

Answer»

0
2
3
4

Answer :B
42.

How many geometrical isomers of this compound are possible ?

Answer»

0
2
3
4

Answer :B
43.

How many geometrical isomers are there for(a) [CO(NH_(3))_(2)Cl_(4)], octahedral (b) [AuCl_(2)Br_(2)] square planar (c) (CoCl_(2)Br_(2)]tetrahedral

Answer»

SOLUTION :TWO, CIS- and trans
(b) Two, cis- and trans
(c) No isomerism is exhibited.
44.

How many geometrical isomers arepossible for [Pt(NO_(2))(NH_(2)OH)(Py)]^(+)?

Answer»


SOLUTION :Mabcd, SQUARE PLANAR, TWO ISOMERS.
45.

How many geometrical isomers are possible in the following coordination entities ? (i) [Cr(C_(2)O_(4))_(3)]^(3-) (ii) [CoCl_(3)(NH_(3))_(3)]

Answer»

Solution :(i) NIL (ii) Two (fac and MER)
46.

how many geometrical isomers are posssible in the following coordination entitles ? (a) [Cr(C_(2)O_(4))_(3)]^(3-)(b) [Co(NH_(3))_(3)CL_(3)]

Answer»

SOLUTION :(a)`[Cr(C_(2)O_(4))_(3)]^(3-)`: No GEOMETRICAL isomerism is possible .
(b) `[Co(NH_(3))_(3)CL_(3)]`: Two geometrical ISOMERS : fac and mer
For details, consult section 7 .
47.

How many geometrical isomers are possible for [Pt(Py)(NH_(3))(Br)(Cl)] ?

Answer»

3
4
0
15

Answer :A
48.

How many geometrical isomers are possible in the following coordination entities ? [Cr(C_(2)O_(4))_(3)]^(3-)

Answer»

SOLUTION :There is no POSSIBILITY of GEOMETRICAL ISOMERISM.
49.

How many geometrical isomers are possible in the following coordination entities ? [Co(NH_(3))_(3)Cl_(3)]

Answer»

Solution :TWO GEOMETRICAL isomers are possible as GIVEN below :
50.

How many geometrical isomers are possible for the square planar complex [Pt(NO_(2))(py)(NH_(3))(NH_(2)OH)]NO_(2) (a) Four (b) Five (c ) Eight (d) Three .

Answer»

4
5
8
3

Answer :D