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.

In ary halides halogen group is a ortho para director and a deactivatortowardselectrophilic substitution reactionwhy

Answer»

SOLUTION :(i) In aryl halidesthe strong-1effectof thehalogensdecreases the electrondensityofbenzeneringthereby deactivatingit forelectrophilicattack .
(II) the pressureof lonepairon halogensis involvedin resonancewith `PI-`ELECTRONS of thehalogengroupis anorthoparadirection and adeactivator.
2.

In aqueous solutions lithium is best reductiant. Why ?

Answer»

Solution :Lithium catino is small and its hydration ability is HIGH . The stronger REDUCTION ability of lithium is ALSO reflected in least STANDARD potenstial of `LI^+ //Li`
3.

In aqueous solutions lithium is best reductant. Why?

Answer»

Solution :LITHIUM cation is small and its HYDRATION ability is HIGH.
The stronger reduction ability of lithium is ALSO reflected in least STANDARD potential of `Li^(+)//Li`.
4.

Inaqueous solution, the ionization constants for carbonic acid are K_(1)=4.2xx10^(-7) and K_(2)=4.8xx10^(-11). Select the correct statement for asaturated 0.034 M solution of the carbonic acid.

Answer»

The concentrations of `H^(+)` and `HCO_(3)^(-)` are APPROXIMATELY equal.
The concentration of `H^(+)` is double that of `CO_(3)^(-)`.
The concentration of `CO_(3)^(2-)` is 0.034 M.
The concentration of `CO_(3)^(2-) ` isgreater than that of `HCO_(3)^(-)`.

Solution :`H_(2)CO_(3)hArr H^(+)HCO_(3)^(-), K_(1)=4.2xx10^(-7)`
`HCO_(3)^(+)hArrH^(+)+CO_(3)^(2-),K_(2)=4.8xx10^(-11)`
As `K_(2)lt lt K`, concentrations of `H^(+)` and `HCO_(3)^(-)` will be approximately equal.
5.

In aqueous solution of Al_(2)(SO_(4))_(3)concentration ofAl^(+3) is 1.8 M then what is the concentration of SO_(4)^(-2) ?

Answer»



Solution :`Al_(2)(SO_(4))_(3) RARR 2Al^(+3)+3SO_(4)^(-2)`
`2Al^(+3) = 1.8`
`Al^(+3) = (1.8)/(2) = 0.9`
`SO_(4)^(-2) : 3 xx 0.9`
`=2.7M`
6.

In aqueous solution, hydrogen perioxide oxidises H_(2)S to

Answer»

SULPHUR
SULPHURIC acid
Caro's acid
Marshall's acid

Solution :(a) `H_(2)S+H_(2)O_(2) to 2H_(2)O+S`
7.

In aqueous alkaline solution, two electron reduction ofHO_2^- gives

Answer»

`HO^(-)`
`H_(2)O`
`O_(2)`
`O_(2)^(-)`

SOLUTION :
8.

In aqueous alkaline solution two electron reduction of HO_(2)^(-) gives

Answer»

`HO^(-)`
`H_(2)O`
`O_(2)`
`O_(2)^(-)`

Solution :`HO_(2)^(-)rarrOH^(-)`
O.N of each O atom in `HO_(2)^(-)` is -1 while that in `OH^(-)` is -2 in other WORDS total O.N of two O atoms in`HO_(2)^(-)` is -2 while that of two O atoms in `OH^(-)` -4 THEREFORE to balance O atoms on EITHER side add `2E^(-)` to lL.H.S and one doing so we have
`HO_(2)^(-)+2e^(-)rarr 2OH^(-)`
Since the reaction is occuring in basic medium therefore to balance H atoms add one `H_(2)O` molecules to L.H.S and one `KO^(-)` ion to R.H.Sof eqn (ii) Thus the balanced EQUATION is
`HO_(2)^(-)+H_(2)O+2e^(-)rarr3OH^(-)`
9.

In any subshell, the maximum number of electrons having same value of spin quantum number is

Answer»

`SQRT(L(l+1))`
`l+2`
`2l+1`
`4l+2`

ANSWER :C
10.

In any fuel, the percentage by volume of iso-octane in a mixture of iso-octane and n-heptane which will knock under same conditions as the fuel being tested, is called

Answer»

Cracking
Iodine number
Aromatisation
Octane number

Answer :D
11.

In Antarctica, ozone depletion is due to the formation of the following compound

Answer»

Acrolein
peroxyacetyl NITRATE
`SO_2` and `SO_3`
CHLORINE nitrate

Answer :D
12.

In Antartica, ozone layer depletion is due to the formation of

Answer»

Peroxy acetyl nitrate
Acrolein
Chlorine nitrate
`SO_(2)` and `SO_(3)`

Answer :C
13.

In Antarctica ozone depletion is due to the formation of following compound

Answer»

ACROLEIN
`SO_2 and SO_3`
peroxy ACETYL nitrate
formaldehyde

Answer :D
14.

In another way to soften the hard water is by using a process called ___________

Answer»

Ion-exchange
Osmosis
Reverse Osmosis
Ionic Pressure

Answer :A
15.

In an universe other than earth all the quantum numbers have same value as on earth except .l. and .s.. Azimuthal quantum number can have values from 0 to (n + 1) in integral step and s (spinquantum no.) has values of (-1/2 , 0 & + 1/2) . If all rules like Aaufbau rule, Hund.s rule and Paulis exclusion rule follow in other universe also. Configuration of element with atomic number 29 will

Answer»

`1s^3 , 2s^3 , 1D^(14)`
`1s^3, 2s^3 , 2p^9 , 3s^3 , 3p^9 , 4s^2`
`1s^2 , 1p^6 , 2s^2 , 1d^(10) , 2p^9`
`1s^3 , 1p^6, 2p^9 , 2s^2 , 1d^(10)`

SOLUTION :`1s^3 , 1p^6 , 2s^3 , 1d^(14)`
16.

In an universe other than earth all the quantum numbers have same value as on earth except .l. and .s.. Azimuthal quantum number can have values from 0 to (n + 1) in integral step and s (spinquantum no.) has values of (-1/2 , 0 & + 1/2) . If all rules like Aaufbau rule, Hund.s rule and Paulis exclusion rule follow in other universe also. Atomic number of element in 4^(th) period and 5^(th) column will be

Answer»

89
71
73
75

Solution :71
17.

In an universe other than earth all the quantum numbers have same value as on earth except .l. and .s.. Azimuthal quantum number can have values from 0 to (n + 1) in integral step and s (spinquantum no.) has values of (-1/2 , 0 & + 1/2) . If all rules like Aaufbau rule, Hund.s rule and Paulis exclusion rule follow in other universe also. No. of element in third period of another universe will be

Answer»

27
32
36
42

Solution :`N= 1 , n=2 , n= 3 , n=4`
`l = 0 to 1 s , l= 1 to 1p , l = 2 to 1d`
`n=2 , l = 0 to 2 s, l =1 to 2P , l = 2 to 2d, l = 3 to 2f`
`n = 3 , l = 0 to 3s, l = 1 to 3 p , l = 2 to 3 d,`
`l = 3 to 3 f , l = 4 to 3g`
`n = 4 , l =0 to 4s , l = 1 to 4p , l=1 to 4d`
`l =3 to 4f , l = 4 to 4g, l = 5 to 4h`
Electronic configuration `to 1s, 1p , 2S , 1d, 2p , 3s , 3d , 4s , 3d`
For third PERIOD `3s^3 , 3d^(15) , 3p^9`,
Number of element = 27
18.

In an oxidation-reduction, MnO_(4)^(-) ion is converted to Mn^(2+), what is the number of equivalents of KMnO_(4) (mol. Wt.=158) present in 250 mL of 0.04 M KMnO_(4) solution ?

Answer»

0.02
0.05
0.04
0.07

Solution :In redox-reaction :
`8H^(+)+MnO_(4)^(-)+5e^(-) to Mn^(2+)+4H_(2)O`
CHANGE in oxidation state of `MnO_(4)^(-) =(+7)-(+2)=+5`
`therefore N_(KMnO_(4))=M_(KMnO_(4))XX5`
`=0.04xx5=0.20`
Number of equivalents `=(NV)/(1000)=(0.2xx250)/(1000)=0.05`
19.

In an ovan , due to insufficient supply of oxygen, 60% of the carbon is converted to carbon dioxide wehreas the remaining 40% is converted into carbon monoxide. If the heat of combustion of carbon to CO_(2)is 394kJ mol^(-1) while that of its oxidation to CO is 111 kJ mol^(-1), calculate the total heatproduced in the oven by burning 10 kgof coal containing 80% carbon by weight .Also calculate the efficiency of the oven.

Answer»

Solution :The reactions taking place in the oven are
`C(s) + O_(2)(g) rarr CO_(2)(g) + 394KJ mol^(_1)`
`C(s) + (1)/(2) O_(2)(g) rarr CO(g) + 111kJ mol^(-1)`
Carbon present in 10 kg coal `= (80)/( 100) xx 10 kg= 8 kg`
Carbon converted to `CO_(2)= ( 60)/( 100) xx 80 kg= 4.8 kg `
Carbon converted to `CO = ( 40)/( 100 ) xx 8 kg = 3.2 kg`
12 g. i.e., 0.012kg of carbon on COMBUSTION to `CO_(2)` produce heat `= 394 kJ`
`:. ` 4.8kg of carbon on oxidation to `CO_(2)` produce heat`= ( 394)/( 0.012) xx 3.2 = 157600 kJ`
0.012kg of carbon on oxidationto CO produce heat `= 111kJ`
`:.` 3.2 kg of carbon on oxidation to CO will produce heat `= ( 111)/( 0.012) xx 3.2 kJ = 29600 kJ`
`:. ` Total heat produced `= 157600+ 29600 = 1,87,200 kJ`
If oven were `100%` efficient, all carbon would have been converted to `Co_(2)`
Heat produced from 8 kg carbon would have been`= ( 394)/( 0.012) x 8 =262,667kJ`
`:. ` % efficiency `= ( 187,200)/( 262,667)xx 100= 71.3 %`
20.

In an organic compound, the phosphorus is estimated as:

Answer»

<P>`Mg_(2)P_(2)O_(7)`
`Mg_(3)(PO_(4))_(2)`
`H_(3)PO_(4)`
`P_(2)O_(5)`

Answer :A
21.

In an organic compound of molar mass 108 g mol^(-1), H and N atoms are present in 9 : 1 : 3.5 by weight. Molecular formula can be:"1.C_(6)H_(8)N_(2)"2.C_(7)H_(10)N3.C_(5)H_(6)N_(3)4.C_(4)H_(18)N_(3)

Answer»

`C_(6)H_(8)N_(2)`
`C_(7)H_(10)N`
`C_(5)H_(6)N_(3)`
`C_(4)H_(18)N_(3)`

Answer :A
22.

In an organic compound containing C and H, the % C is 3 times to that of % H. What is the compound ?

Answer»

`C_(6)H_(6)`
`C_(6)H_(12)`
`C_(2)H_(6)`
`CH_(4)`

Answer :D
23.

In an organic compound, phosphorus is estimated as ............ .

Answer»

`Mg_(2)P_(2)O_(7)`
`Mg_(3)(PO_(4))_(2)`
`H_(3)PO_(4)`
`P_(2)O_(5)`

ANSWER :A
24.

In an ore theonly oxidizable material in Sn^(2+) This is titrated with a dichromate solution containing 2.5 of K_(2)Cr_(2)O_(7) in 0.50 litre A 0.40 g sample of the ore required 10.0 cm^(3) of titrant to reach equivalenc point calculate the prcentage of tin in the re (k=39.1, Cr = 52 Sn =11.87)

Answer»

Solution :Step 1 To write balanced chemical EQUATION f the redox reaction
`Sn^(2+)rarr Sn^(4+)+2 E^(-)[xx3`
`Cr_(2)O_(7)^(2-)+14 H^(+)+6 e^(-) rarr 2Cr^(3+)+7 H_(2)O`
`Cr_(2)O_(7)^(2-)+3 Sn^(2+)+14 H^(+) rarr 2Cr^(3+)+3 Sn^(4+)+7 H_(2)O`
Now 0.5 L of the solution cotains `K_(2)Cr_(2)O_(7)=2.5 g`
`therefore` 10 mL of solution will contain `K_(2)Cr_(2)O_(7)=(2.5)/(500)xx10 g=(2.5)/(500xx(210)/(294)` mole
From the balanced redox reaction
1 mole of `K_(2)Cr_(2)O_(7)` oxidises `Sn^(2+)=3` moles
`therefore (2.5)/(500)XX(10)/(294)` mole of `K_(2)Cr_(2)O_(7)` will oxidise `Sn^(2+)=3xx(2.5)/(500)xx(10)/(294) mol =3x(2.5)/(500)xx(10)/(294)xx118.7 g`
Step 2 To determine the % age of `Sn^(2+)` in the ore
Now 0.40 g of the or contain `Sn^(2+) =0.06 g`
`therefore %"of"Sn^(2+) "in theare" = 0.06/0.40xx100=15%`
25.

In an ore the only oxidisable material is Sn^(2+). This ore is titrated with a dichromate solution containing 2.5 g K_(2)Cr_(2)O_(7) in 0.50 litre. A 0.40 g of sample of the ore required 10.0 cm^(3) of the titrant to reach equivalent point. Calculate the percentage of tin in ore. (K=39.1, Cr=52, Sn=118.7)

Answer»


Solution :Mol. Mass of `K_(2)Cr_(2)O_(7)=2xx39.1+2xx52+7xx16`
=78.2+104.0+112.0
=294.2
Eq. mass of `K_(2)Cr_(2)O_(7)=(294.2)/(6)=49.03`
NORMALITY of `K_(2)Cr_(2)O_(7)` solution `=(2.5)/(49.03)xx(1000)/(500)=(5)/(49.03) N`
`10 mL (5)/(49.03) N K_(2)Cr_(2)O_(7)-=10 mL (5)/(49.03)N` stannous ION
Eq. mass of `Sn^(2+)=(118.7)/(2)=59.35`
Amount of Sn in the sample `=(5)/(49.03)xx(59.35)/(1000)xx10`
=0.0605 g
Percentage of Sn in the ore `=(0.0605)/(0.40)xx100=15`
26.

In an oil drop experiment, the following charges (in arbitrary units) were found on a series of oil droplets : 2.30 xx 10^(-15) , 6.90 xx 10^(-15) , 1.38 xx 10^(-14) , 5.75xx10^(-15) , 3.45 xx10^(-15) , 1.96 xx 10^(-14) . The experimental value suggests not possible magnitude of the charge on the electron is in the same arbitrary unit)

Answer»

`2.30xx10^(-15)`
`1.38xx10^(-14)`
`6.90xx10^(-15)`
`1.15xx10^(-15)`

Solution :`1.15 xx 10^(-15)` , which is the largest NUMBER which divides all the given experimental VALUES of the charge.
The smallest charge noted is `2.3 xx 10^(-15)` , but this charge doesn.t seem as unit charge, doesn.t DIVIDE others into EVEN number of times.
Actually , it REPRESENTS the charges of 2 electrons.
27.

In an octahedral structure, the pair of d-orbitals ind""^(2)sp^(3) hydridiation is

Answer»

`d_(x^(2) - y^(2)),d_(Z^(2))`
`d_(XZ) ,d_(x^(2) - y^(2))`
`d_(z^(2)) ,d_(xz)`
`d_(XY) ,d_(yz)`

Answer :A
28.

In an isothermal reversible compression of an ideal gas the sign of q, DeltaS and w are respectively

Answer»

`+,-,-`
`-,+,-`
`+,-,+`
`-,-,+`

SOLUTION :During COMPRESSION, energy of the SYSTEM increases, in isothermal condition, to MAINTAIN temperature constant, heat is liberated from the system. Hence q is negative. During compression entropy decreases.During compression work is done on the system, hence w is positive
29.

In an irreversible process taking place at constant T and P and in which only pressure- volume work is being done, the change on Gibb's free energy (dG) and changing in entropy (dS), stisfy the criteria

Answer»

`(dS)_(V,s) lt 0, (dG)_(T,P) lt 0`
`(dS)_(V, E) GT 0 (dG)_(T, P) lt 0`
`(dS)_(V, E) = 0, (dG)_(T,P) = 0`
`(dS)_(V, E) = 0, (dG)_(T, P) gt 0`

SOLUTION :`(dG)_(T,P) lt 0` for SPONTANEOUS
`(dS)_(V, E) gt 0` for spontaeous
30.

In an irreversible process taking placeat constant T and Pand in which only pressure volume work is being done, the change in Gibbs free energy (dG) and the change in entropy ( dS), satisfy the criteris

Answer»

`( dS) _(V,E) gt 0 , (dG) _(T,P) lt 0`
`( dS) _(V,E) = 0 , (dG) _(T,P) = 0`
`( dS) _(V,E) = 0 , (dG) _(T,P) gt 0`
`( dS) _(V,E) lt 0 , (dG) _(T,P) lt 0`

SOLUTION :For irrev., i.e.,SPONTANEOUS PROCESS , `d S gt 0, dG lt0`.
31.

In an ionic compound A^+ X^- , the radii of A^+ and X^- ions are 1.0 pm and 2.0 pm respectively. The volume of the unit cell of the crystal AX will be

Answer»

`"27 pm"^3`
`"64 pm"^3`
`"125 pm"^3`
`"216 pm"^3`

Solution :`r_(A^+)/r_(X^-)=1/2`=0.5 . As it LIES in the RANGE 0.414-0.732, AX has octahedral structure like that of NaCl.
Hence, edge length (a)=`2(r_+ +r_(-))`
=2(1+2)pm =6 pm
Volume of the UNIT cell=`a^3`=`"(6 pm)"^3="216 pm"^3`
32.

In an ionic compound A^(+)X^(-) , the radii ofA^(+) and X^(-)ions are 1.0 pm and 2.0 pm rspecitvely . The volume of the unit cell of the crystal AX will be

Answer»

`27 "pm"^(3)`
` 64 "pm"^(3)`
` 125 "pm"^(3)`
` 216 "pm"^(3)`

Solution :` (r_(A^+))/(r_(X^(-))) = 1/2 = 0.5 `,As it lies in the range that of NaCl.
0.414 - 0.732 , AX has octahedral structure like that of NaCl.
Hence,edge LENGTH `= (a) =2 (r_(+) + r_(-))`= 2 (1 + 2) pm = 6 pm
Volume of the unit cell = ` a^(3) = ( 6 "pm")^(3) = 216 "pm"^(3)`
33.

In an ionic compound A^(+)B^(-) radius ofA^(+) is 88 pm while that ofB(-) is 200 pm. The coordination number ofA^(+)will be ……..

Answer»


ANSWER :0.707
34.

In an insulated container, 1 mole of a liquid of molar volume 100 mL is at 1 bar. If the liquid is steeply taken to 100 bar, the volume of liquid decreases by 1 mL. Find the work done and DeltaH.

Answer»


ANSWER :100 BAR ML: 9900 bar mL
35.

In an ionic compound A^+ B^-,radius of A^+ 88 pm while that of B^- is 200 pm. The coordination number A^+ will be ______

Answer»

SOLUTION :6(`r_+ //r_-`=0.44 which LIES in the RANGE 0.414-0.732)
36.

In an FCC unit cell a cube is formed by joining the centers of all the tetrahedral voids to generate a new cube. Then the new cube would contain voids as

Answer»

1 full tetrahdral VOID, 1 full octahedral void
1 full tetrahdral void only
8 full TETRAHEDRAL VOIDS and 1 full octahedral void
1 full Octahedral void by

Answer :A
37.

In an FCC unit cell a cube is formed by joining the centers of all the tetrahedral voids to generate a new cube . Then the new cube contains voids as :

Answer»

1 FULL TETRAHEDRAL VOID, 1 full OCTAHEDRAL void
1 full Tetrahedral void only
1 full Tetrahedral void and 1 full Octahedral void
1 full Octahedral void only

Answer :A
38.

In an experiment ethyliodide in ether is allowed to stand over magnesium pieces. Magnesium dissolves and product is formed (a) Name the product and write the equation for the reaction. (b) Why all the reagents used in the reaction should be dry? Explain. (c) How is acetone prepared from the product obtained in the experiment?

Answer»

Solution :(a) When ethyl IODIDE in ether is allowed to stand over magnesium pieces. the product formed will be Ethyl magnesium iodide.
`CH_3 - CH_2l + Mg overset("dry ether")to underset(("Grignard REAGENT"))underset("Ethyl magnesium iodide")(CH_3 - CH_2 MgI)`
(b) Grignard reagent are highly reactive compounds and react with any source of proton to give hydrocarbons. EVEN when water is there. it is sufficiently acidic to CONVERT it into the corresponding hydrocarbon. So it is necessary to avoid even traces of moisture with the grignard reagent as they are highly reactive. That is why the all reagents used in the reaction should be dry.
(c ) `CH_3 - MgI " to "CH_3 - underset(O)underset(||)C - CH_3`
METHYL magnesium iodide reacts with acetyl chloride to form acetone.
39.

In an experiment 5.0 g of CaCO_(3) on heating gave 2.8 g of CaO and 2.2 g of CO_(2). Show that these results are in accordance to the law of conservation of mass.

Answer»

Solution :CALCIUM carbonate decomposes on heating according to the following EQUATION.
`underset(5 g)(CaCO_(3)) to underset(2.9 g)(CAO) + underset(2.2 g)(CO_(2))`
The total mass of reactant (`CaCO_3`) taken at the BEGINNING of the reaction = 5 g.
The total mass of PRODUCTS (`CaO + CO_2`) obtained
`=2.8 + 2.2 = 5.0 g`
Since, the total mass of the products obtained in the reaction is the same as that of the reactant taken, the results are in accordance to the law of conservation of mass.
40.

In an experiment, 2.847 g of pure MOCl_(3) was allowed to undergo a set of reactions as a result of which all the Cl was converted to AgCl. The weight of AgCl was 7.2g. Find at. Wt of M.

Answer»

35.52
47.72
65.2
80.42

Answer :B
41.

In an experiment 2.65 g of zinc displaced 2.58 g of copper from copper sulphate solution. If equivalent weight of copper is 31.75, calculate that of zinc.

Answer»


ANSWER :32.61
42.

In an experiment, 20 mL of a decinormal HCl solution was added to 15 mL of a decinormal AgNO_(3) solution. AgCl was precipitated out and excess of acid was titrated with N//20 NaOH solution. The volume of NaOH required was :

Answer»

10 mL
20 mL
30 mL
5 mL

Solution :N//A
43.

In an exothermic reaction, heat is evolved and system loses heat to the surroundings . For such system. (i) q_(p) will be negative(ii) Delta_(r)H will be positive (iii) q_(p) will be positive (iv) Delta_(r)H will be negative

Answer»

(i) , (II)
(iii), (IV)
(i) &(iv)
(ii) & (iii)

Answer :C
44.

In an exothermic reaction, heat is evolved and system loses heat to the surrounding.For such system

Answer»

`q_(p)` will be negative
`Delta_(r)H` will be ZERO
`q_(p)` will be positive
`Delta_(r)H` will be positive

Solution :In an exothermic REACTION, heat is evolved by the system and HENCE `q_(p)` and `DeltaH` are negative.
45.

In an exothermic reaction, heat is evolvedand system loses heat to the surrounding. For such system.

Answer»

<P>`q_(p)` will be NEGATIVE
`Delta_(r)H` will be negative
`_(p)`will be POSITIVE
`Delta_(r)H` will be positive

Solution :`q_(p)` and `Delta_(r)H` are `-ve` for exothermicreaction.
46.

In an exothermic reaction, heat is evolved, and system loses heat to the surrounding. For such system

Answer»

`q_p` will be negative
`Delta_(r) H` will be negative
`q_p` will be positive
`Delta_(r ) H` will be positive

Solution :Exothermic REACTIONS are those reactions which are accompanied by the EVOLUTION of HEAT.
e.g., `C_((s)) + O_(2(g)) to CO_(2(g)) + 393.5 "KJ"`
`H_(2(g)) + (1)/(2) O_(2(g)) to H_(2) O_((l)) 285.8 "kJ"`
`q_p and Delta_(r) H` are negative for exothermic reaction.
47.

In an exeperiment ethyliodide in ehter is allowed to stand over magnesium pieces. Magnesium dissolves and product is formed. Why all the reagents used in the reaction should be dry? Explain.

Answer»

SOLUTION :The GRIGNARD carbon is highly basic and reacts with acidic protons of polar SOLVENTS LIKE water to form an alkali so all reagents should be PURE and dry.
48.

In an exeperiment ethyliodide in ehter is allowed to stand over magnesium pieces. Magnesium dissolves and product is formed. Name the product and write the equation for the reaction.

Answer»

Solution :The PRODUCT formed is ethylmagnesium IODIDE (Grignard REAGENT)
`C_(2)H_(5)I+Mg overset("dry ETHER")to underset("Ethyl magnesium iodide")(C_(2)H_(5)MgI)`
49.

In an estimation of sulphur by Carius method, 0.468 of an organic sulphur compound gave 0.668 g of barium sulphate. Find the percentage of sulphur in the compound.

Answer»

SOLUTION :`"PERCENTAGE of sulphur" = (32)/(233) xx ("Mass of barium sulphate")/("Mass of compound") xx 100`
`=(32)/(233) xx ((0.668 G))/((0.468 g)) xx 100=19.60%`
50.

In an estimation od sulphur by carius method, 0.2175 g of a compound gave 0.5825 g of barium sulphate. Calculate the percentage of sulphur in the compound.

Answer»

Solution :MASS of the compound `=0.2175 g`
Mass of barium sulphate `=0.5825 g`
`{:(BaSO_(4),EQUIV,S),((137+32+64)" parts",,"32 parts"),(233 g,,32 g):}`
233g of barium sulphate contain SULPHUR `=32 g`
0.5825 g of barium sulphate contain sulphur `=32/233xx0.5825 g`
PERCENTAGE of sulphur `=("Mass of sulphur")/("Mass of compound")xx100=32/233xx0.5825/0.2175xx100=36.78`.