Explore topic-wise InterviewSolutions in .

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

51351.

24 g of magnesium in the vapour state absorb 1200 kJ of energy. If IE_1 and IE_2 of magnesium are 750 and 1450 kJ mol^(-1) respectively, the final composition of mixture is

Answer»

0.69 MOL `MG^(+)` and 0.31 mol `Mg^(2+)`
0.59 mol `Mg^(+)` and 0.41 mol `Mg^(2+)`
0.49 mol `Mg^(+)` and 0.51 mol `Mg^(2+)`
0.29 mol `Mg^(+)` and 0.71 mol `Mg^(2+)`

Solution :Molar MASS of Mg =24 g `mol^(-1)`
`:.` Amount of Mg=1 mol
THUS, 750 kJ of energy is used up to convert 1 mol of Mg to `Mg^(+)`
Energy used to convert `Mg^(+)` to `Mg^(2+)`
`=1200-750 =450 kJ`
`:.` Amount of magnesium converted to `Mg^(2+)`
`=(450 kJ)/(1450) =0.31 mol`
`:. ` Amount of magnesium converted to `Mg^(+)`
=1-0.31 =0.69 mol
Discussion
51352.

23g of Na will react with ethanol to give

Answer»

ONE MOLE of oxygen
one mole of `H_(2)`
`(1)/(2)` mole of `H_(2)`
None of these

Solution :`Na+C_(2)H_(5)OH rarr C_(2)H_(5)ONa+1//2 H_(2)`
23gof Na gives `H_(2) = 0.5 MOL`.
Discussion
51353.

23g of sodium will react with methyl alcohol to give

Answer»

one mole of oxygen
`22.4 dm^(3)` of hydrogen as at N.T.P
1 mole of `H_(2)`
11.2 L of hydrogen gas at N.T.P

Solution :`{:(CH_(3)OH+NA,rarr,CH_(3)ONA,+(1)/(2)H_(2)),("" 23g,,,((1)/(2)mol)),(""(1 mol),,,):}`
`23g` of `Na -= 1 mol Na -= (1)/(2) mol H_(2)`
`= (1)/(2) xx 22.4 = 11.2 L` of `H_(2)` at N.T.P.
Discussion
51354.

2,3-dimethylhexane contains......tertiary......secondary and...primary carbons respectively.

Answer»

2,2,4
2,4,3
4,3,2
3,2,4

Answer :A
Discussion
51355.

2,3-Dimethyl-2-butene can be prepared by heating which of the following compounds with a strong acid ?

Answer»

`(CH_(3))_(2)C=CH-CH_(2)-CH_(3)`
`(CH_(3))_(2)CH-CH_(2)-CH=CH_(2)`
`(CH_(3))_(2)CH-UNDERSET(CH_(3))underset(|)(CH)-CH=CH_(2)`
`(CH_(3))_(3)C-CH=CH_(2)`

SOLUTION :
Discussion
51356.

2,3-Dimethyl-2-butene can be prepared by heated which of the following compounds with a strong acid ?

Answer»

`(CH_3)_3C-CH=CH_2`
`CH_3)_2C=CH-CH_2CH_3`
`(CH_3)_2CH-CH_2-CH=CH_2`
`(CH_3)_2 CH-undersetunderset(CH_3)|CH-CH=CH_2`

SOLUTION :
Discussion
51357.

2,3-Butanediol has the 2R, 3R configuration Identify the correct statement among the following.

Answer»

2R, 3S - is its ENANTIOMERS
2S, 3R - is its enantiomers
2S, 3S - is its enantiomers
2S, 3R - Is its diastereoisomer

Solution :2S. 3S - is its enantiomers , 2S, 3R - is its diastereoisomer
Discussion
51358.

.^(227)Ac has a half-line of 22 year with respect to radiioactive decay. The decay follows two parallel paths, one leading to .^(227)Th and the other leading to .^(223)Fr. The percentage yiedls of these two daughter nuclides are 2% and 98.0% respectovely. What is the rate constant in "year"^(-1), for each of the separate paths?

Answer»

Solution :We know,
`lambda_(av) = (0.693)/(22) = 3.15xx10^(-2) "year"^(-1)`
For the decay involving TWO PARALLEL paths,

We have `lambda_(AC) = lambda_("Th path") + lambda_("Fr path")`
`:. Lambda_(Ac) xx "Fraction of Th" = lambda_("Th path")` ....(1)
`lambda_(AC) xx "Fraction of Fr" = lambda_("Fr path")` .....(2)
or `lambda_(AC) xx (1 - "Fraction of Th") = lambda_("Fr path")` .....(3)
Thus by eqs. (1) and (3) we get
`lambda_(AC) =lambda_("Th path") + lambda_("Fr path")`
Thus by eqs. (1) and (3) we get
`lambda_(AC) = lambda_("Th path")+ lambda_("Fr path")`
Thus, Fractionlal yield of `Th = (lambda_("Th path"))/(lambda_("Ac path"))`
or `lambda_("Th path")=3.15xx10^(-2)xx(2)/(100)`
`=6.30xx10^(-4)yr^(-1)`
Also Fractional yeild of Fr`=(lambda_("Fr path"))/(lambda_(Ac path))`
`:. lambda_(Fr) = 3.15xx10^(-2) xx (98)/(100) =3.087xx10^(-2) yr^(-1)`
Discussion
51359.

22.71 Lit means ………. meter ?

Answer»

Solution :`22.71 L = 22.71 xx10^(-3)m^(3)`
1 LIT `= 1 (DM)^(3)`
`1 dm = 0.1 m = 1xx10^(-1)m`
`1 (dm)^(3)=(0.1 m)^(3)=1xx10^(-3)m^(3)`
`1L=1(dm)^(3)=1xx10^(-3)m^(3)`
`therefore 22.71 L =(22.71 L xx1xx10^(-3)m^(3))/(12)=22.71xx10^(-3)m^(3)` gas
Molar VOLUME `= 22.71 L = 22.71xx10^(-3)m^(3)`
Discussion
51360.

2.26 g of impure ammonium chloride were boiled with 100 mL of NaOH solution till no more ammonia was given off. The excess of NaOH solution left over required 30 mL 2N H_(2)SO_(4) for neutralisation. Calculate the percentage purity of the salt. (H=1, N=14, O=16, Na=23, S=32, Cl=35.5)

Answer»


SOLUTION :N//A
Discussion
51361.

22.44 kJ energy is required to convert 8 gm of gaseous atom of metal M to M_((g))^+ if I.E._1 of metal M = 374 kJ/mole. Select correct for above metal M

Answer»

0.6 MOLE gaseous ion (`M^+`) are formed
Same ENERGY can convert all `M_((G))^+` to `M_((g))^(2+)`
Atomic MASS of metal = 133.33
`3.613xx10^22` atoms of M are converted to `M_((g))^(+)`

Answer :C::D
Discussion
51362.

22.4 litres of water vapour at NTP, when condensed to water, occupies an approximate volume of:

Answer»

18 litres
1 litre
1 mL
18 mL

Answer :D
Discussion
51363.

Fr has half life period of

Answer»

21 secs
21minutes
21 hours
21 days

Answer :B
Discussion
51364.

2.20 g of an ammonium salt were boiled with 75 mL of NaOH till the emission of ammonia gas ceased. The excess of unused NaOH solution required 70 mL of N//2 sulphuric acid for neutralisation. Calculate the percentage of ammonia in the salt.

Answer»


SOLUTION :N//A
Discussion
51365.

2.2 moles of phosphorus penta chloride were taken in a closed vessel and dissociated into phosphorus trichloride and chlorine . At equilibrium , the total number of moles of the reactants and the products was 2.53 . The degree of dissociation is ,

Answer»

0.33
0.165
0.15
0.3

Answer :C
Discussion
51366.

Both 22 grams of CO_(2) and 8 grams of methane contain the following carbon atoms

Answer»

`3 xx 10^(23)`
`12 xx 10^(23)`
`6 xx 10^(23)`
`1.5 xx 10^(23)`

Answer :C
Discussion
51367.

22 g of CO_(2) contains ……….. molecules of CO_(2).

Answer»


SOLUTION :44 g of `CO_(2)` contains `6.023 xx 10^(23)` molecules.
`:.`22 g of `CO_(2)` will contain = `(6.023xx10^(23))/44xx22 = 3.0115 xx 10^(23)`
Discussion
51368.

22 g of a gas occupies 11.2 litres of volume at STP. The gas is

Answer»

CH4
CO2
NO
CO

Solution :22 g of a GAS OCCUPIES 11.2 LITRES.
11.2 litres is occupied by 22 g of a gas.`:.` Molar volume 22.4 LITRE will be occupied by`22/11.2xx22.4 = 44` g
`:.` The gas is C`O_(2)` .
Discussion
51369.

2.18g of an organic compound containing sulphur produces 1.02 g of BaSO_(4). The percentage of sulphur in the compound is

Answer»

0.0726
0.0898
0.1
0.0642

Solution :% of `S=(32)/(233)XX(1.02)/(2.18)xx100=6.42%`
Discussion
51370.

2.16 grams of Cu on reaction with HNO_(3) followed by ignition of the nitrate gave 2.7 gm of copper oxide. In another experiment 1.15 gm of copper oxide upon reduction with hydrogen gave 0.92 gm of copper. This data illustrate the law of

Answer»

MULTIPLE Proportions
Definite Proportions
Reciprocal proportions
Conservation of mass

Answer :B
Discussion
51371.

21 Mol of FeSO_(2) (atomic weight of Fe is 55.84 g mol^(-1)) is oxidized to Fe_(2)(SO_(4))^(3) calculate theequivalent weight of ferrous ion

Answer»

55.84
27.92
18.61
111.68

Solution :Since OXIDATION of `Fe^(2+)rarrFe^(3+)` is a one electron CHANGE therefore equivalent wt of `Fe^(2+)` = ATOMIC wt of `Fe^(2+)` =55.84
Discussion
51372.

2.1 g of metal carbonate on thermal decomposition gave 1 g of metal oxide as residue. Determine theequivalent weight of metal.

Answer»

Solution :Let the EQUIVALENT weight of metal=x
Equivalent weight of metal oxide =x+8
Equivalent weight of metal CARBONATE =x+30
When weight RATIO is related to equivalent weight ratio for two substances `(W_(i))/(W_(2))=(E_(1))/(E_(2)), (2.1)/(1)=(x+30)/(x+8)`
SOLVING , x = 12
Equivalent weight of metal = 12
Discussion
51373.

20ml of hydrogen peroxide solution is added to excess of acidified KI solution. The iodine so liberated requires 20ml of 0.1M hypo solution. Calculate the molarity and percentage strength of H_2O_2solution.

Answer»

SOLUTION :0.05M, 0.17 (w/v)%
Discussion
51374.

20mL of each 0.1M NH,OH and IM NH CI aqueous solutions form a buffer of pH 8.3. Calculate the equilibrium constant for NH_4 OH hArr NH_(4)^(+)+ OH^(-).

Answer»

SOLUTION :`2 XX 10^(-5)`
Discussion
51375.

20mL fo 0.2M MbnSO_(4) are completely oxidiz3d by 16 mL of KMnO_(4) of unknown normaliity each fromingMn^(4+) oxidation state. Find out the normality and molarity of KMnO_(4) solution.

Answer»


ANSWER :`0.5N, 0.167M`
Discussion
51376.

20g of sample containing Ba(OH)_(2) is dissolved in 10 ml of 0.5 MHCl solution. The excess of HCl was then titrated against 0.2 M NaOH. The volume of NaOH used in the titration was 10 ml. Calculate the percentage of Ba(OH)_(2) in the sample. (Mol. wt. of Ba(OH_(2))=171)

Answer»

Solution :Calculation of volume of HCl used in titration between NAOH and HCl.
`V_(NaOH) =10 ML, M_(NaOH) = 0.2 M, M_(HC) = 0.5 M, V_(HCl)`=?
`M_(NaOH) xx V_(NaOH) = M_(HCl) xx V_(HCl), V_(HCl) = (10 xx 0.2)/0.5 = 4 ml`
This is the volume of HCl left unused when excess of HCl is added to `Ba(OH)_(2)` solution.
Total volume of HCl added = 10 ml
Volume of HCl used to react with `Ba(OH)_(2) = 10-4 = 6` ml
`2HCl + Ba(OH)_(2) to BaCl_(2) + 2H_(2)O`
M = No. of moles x `1000/("volume of solution") rArr 0.5 = "moles" xx 1000/6`.
`(0.5 xx 6)/1000` = moles of HCl Moles of HCl used = 0.003 moles.
Observing the molar ratio of HCl and `Ba(OH)_(2)`.
Moles of `Ba(OH)_(2)` REACTED `=1/2` x moles of HCl reacted `=1/2 xx 0.003 = 0.0015` moles.
Weight of `Ba(OH)_(2)` reacted = no. of moles x mol. WT. =` 0.0015 xx 171 = 0.2565` g
Percentage of `Ba(OH)_(2)` in the sample `=(wt. of Ba(OH)_(2) "reacted")/("Total weight of sample") xx 100` = 1.28 %
Discussion
51377.

20CC of hydro carbon were exploded with excess of oxygen. After explosion and cooling a contraction of 20cc was noted on addition of KOH another contraction of 40CC was noted. The molecular formula of hydrocarbon is

Answer»

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

Solution :Let formula of hydrocarbon = CXHY
`CxHY+(X+(y)/(4))O_(2)rarrxCO_(2)+(y)/(2)H_(2)O`
`{:("1 mol",(x+(y)/(4))"mol","x mol"),("1 vol",(x+y//4)"vol","x vol"),(20C C,20(x+(y)/(4))C C,20xC C):}`
`CO_(2)` PRODUCED = contraction on addition of KOH
= 40 CC implies 20x = 40 implies x = 2
`[20C C+20(x+(y)/(4))C C]-20xC C=30C C`
`y=2`
Formula of hydrocarbon =`C_(2)H_(2)`
Discussion
51378.

2.0g of a metallic carbonate on decomposition gave 1.5g of metallic oxide. The equivalent mass of metal is

Answer»

58
29
5.8
2.9

Solution :`("weight of METAL carbonate")/("weight of METALLIC OXIDE") = ("EQ. mass of metal + EQL. Mass of" CO_(3)^(2-))/("eq mass of metal + eq mass of" O^(2-))`
Discussion
51379.

200cc of ozone diffused in 15 min through a porous membrane. How much time does 150cc of oxygen take to difffuse, under similar conditions ?

Answer»

Solution :`(V_(O_3) xx t_(O_2))/( V_(O_2) xx t_(o_3)) = sqrt((M_(o_2))/(M_(O_3)))= >( 200 xx t_(o_2))/( 150 xx 15 ) = sqrt((32)/(48))`
`t_(O_2) = ( 15 xx 150 )/(200) =sqrt((32)/(48))=9.135 ` MIN
Discussion
51380.

200 mL of water is added to 500 mL of 0.2 M solution. What is the molarity of the diluted solution ?

Answer»

0.5010 M
0.2897 M
0.7093 M
0.1428 M

Solution :`M_(1)V_(1)-=M_(2)V_(2)`
`0.2Mxx500-=M_(2)xx700`
`M_(2)=((0.2M)xx500)/(700)=0.1428M`
Discussion
51381.

200ml of pure oxygen is subjected to electric discharge, 15% of oxygen is converted into ozone. The volume of ozonized oxygen is

Answer»

20 ML
30 ml
190 ml
80 ml

ANSWER :C
Discussion
51382.

200 ml of O_2 gas maintained at 700mm pressure and 250ml of N_2 gas maintained at 720mm pressure are put together in one litre flask. If the temperature is kept constant, the final pressure of the mixture in mm is

Answer»

450 
320 
632 
316 

Answer :B
Discussion
51383.

200 mL of a solution of mixture of NaOH and Na_(2)CO_(3) was first titrated with phenolphthalein and N//10HCl. 17.5 mL of HCl was required for the end point. After this methyl orange was added and 2.5 mL of same HCl was required for next end point. Find out amounts of NaOOH and Na_(2)CO_(3) in mixture.

Answer»


ANSWER :`NaOH=0.06g PER 200 mL`,
`Na_(2)CO_(3)=0.0265 g per 200 mL`;
Discussion
51384.

200 ml of an aqueous solution of a protein contains 1.26 g of protein. At 300 K, the osmotic pressure of this solution is found to be2.52xx 10^(-3)bar.The molar mass of protein will be (R=0.083 Lbar mol ^(-1) K^(-1))

Answer»

`62.22` KG `mol^(-1)`
`12444 g mol ^(-1)`
`300 g mol ^(-1)`
none of these

Solution :`pi = CRT`
`pi = (W)/(M xxV) xx RT`
` therefore M = (WRT)/(PIV) = (1.26 xx 0.083 xx 300)/(2.52 xx 10 ^(-3) xx 0.2)= 62.22 kg mol ^(-1)`
Discussion
51385.

200 mL of 3 N HCl were mixed with 200 mL of 6 N H_(2)SO_(4) solution. The final normality of H_(2)SO_(4) in the resultant solution will be:

Answer»

9 N
3 N
6 N
2 N

Solution :N//A
Discussion
51386.

20.0 L of 0.2 M weak acid (pK_(a)=5.0) is titrated against 0.2 M strong base. What is the pH at the equivalence point ?

Answer»

`5.0`
`7.0`
`9.0`
`11.0`

Solution :At the equivalence POINT, solution will contain salt of weak ACID and strong base. Its solution on hydrolysis will be basic with `pH gt 7`. Value of 9 may be CHOSEN as the base is strong.
Discussion
51387.

200 joules of heatwas supplied to a system at constant volume. It resulted in the increase in temperature of the system from 298 to 323 K. What is the change in internal energy of the system ?

Answer»

400 J
200 J
50 J
150 J

Solution :At constant volume, `DeltaV=0`
`w=-PDeltaV=0""DeltaU=q=200J`
Discussion
51388.

20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and8.0 g magnesium oxide. What will be the percentage of purity of magnesium carbonate in the sample?

Answer»

60
84
75
96

Solution :`MgCO_(3) overset(Delta)(to) MgO + CO_(2)UARR`
84 g of `MgCO_(3)`gives 40 g of MgO. (100% purity)
20g of `MgCO_(3)` will give `=(40xx20)/84=9.52g`
9.52g of MgO is given by 100% pure `MgCO_(3)`.
8.0 g of MgO will be given by=`(100xx8)/9.52=84.03%`
Discussion
51389.

20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample ? (At. Wt of Mg = 24)

Answer»

96
84
60
75

Solution :`UNDERSET(84g)(MgCO_(3))overset(HEAT)(rarr)underset(40g)(MgO)+CO_(2)`
84 g of `MgCO_(3)` yield upon heating MgO = 40 g
20 g of `MgCO_(3)` yield upon heating `MgO=((40g)xx(20g))/((84g))`
`=9.52g`
Weight of MgO ACTUALLY formed = 8.0g
% purity of `MgCO_(3)=((8.0g))/((9.52g))xx100=84%`.
Discussion
51390.

20 volume. of H_(2)O_(2) is equal to

Answer»

`20% H_(2)O_(2)` by MASS
`6% H_(2)O_(2)` by mass
`1.764N`
`3.571N`

SOLUTION :V.S. = `Nxx5.6impliesN=(20)/(5.6)=3.5171N`
Discussion
51391.

20 volume H_2O_2 Solution has a strength of about

Answer»

`30%`
`6%`
`3%`
`10%`

Solution :22.4 LITRE `O_2` at N.T.P. OBTAINED by 68 gm of `H_2O_2`
`THEREFORE` 1 litre `O_2` at N.T.P. obtained by `68/22.4` gm of `H_2O_2`
`therefore` 20 litre `O_2` at N.T.P. obtained by `68/22.4xx20` gm of `H_2O_2` =60.71 gm of `H_2O_2`
`therefore` 1000 ml `O_2` at N.T.P. obtained by =60.71 gm of `H_2O_2`
`therefore` 100 ml `O_2` at N.T.P. obtained by `=60/1000xx100`=6.71%
Discussion
51392.

20 % of N_(2)O_(4) " molecules are disociated in a sample of a gas at " 27^(@)C and 760 " torr . Calculate the density of the equilibrium mixture ".

Answer»

Solution :` {:(,N_(2)O_(4)(g),hArr,2NO_(2)(g),),("Intial",1 " mole",,,),("At.eqm.",1-0*2 = 0*8" mole",,0*4 "mole,",Total = 1*2 " moles" ):}`
If V is the volume of the vapour PER mole, volume of vapour before dissociation = V
` " Hence density " (D) propto1/V `
But density after dissociation
`D= (" Mol.wt.of " N_(2)O_(4))/2 = 92/2=46"" ("Theoretical density")`
Volume after dissociation = `1*2` V
` :. "Density (d)" propto 1/(1*2 V) `
` :.D/d = 1/V xx 1*2V =1*2OR d=D/(1*2)= 46/(1*2)= 38*3`
Alternatively, USE theformula directly,
` alpha = (D-d)/d`
Discussion
51393.

20 ml of the solution containg Na_(2)CO_(3) and NaHCO_(3) is titrated with 0.1MHCI using Phenolphthalein indicator the end point was 10ml. 20ml of the same solution is titrated with 0.1M HCI, the end point was 25ml with Methylorange indicaor from the begining. NaHCO_(3)+HCl rarr NaCl +H_(2)O +CO_(2) Na_(2)CO_(3)+2HCl rarr 2NaCl +CO_(2) +H_(2)O What amount of NaOHis required to convertNaHCO_(3) to Na_(2)CO_(3) in 1 litre of solution

Answer»

`2g`
`20g`
`1g`
`0.5g`

SOLUTION :`NaHCO_(3)+NaOH rarr Na_(2)CO_(3) +H_(2)O`
84g of `NaHCO_(3)` requires 40g of `NaOH`
`2.1g` of `NaHCO_(3)` requires?
`= (40)/(84) xx 2.1 = 1g`
Discussion
51394.

20 mL of x M HCl neutralises 5 mL of 0.2 M Na_(2)CO_(3) solution to phenolphthalein end-point. The value of x is

Answer»

`0.167M`
`0.133M`
`0.150M`
`0.05M`

Solution :With phenolphthalein,
EQ. `HCl=(1)/(2)` eq. `Na_(2)CO_(3)`
`20XX X xx1=(1)/(2)xx5xx0.2xx2impliesx=0.05`
Discussion
51395.

20 ml of the solution containg Na_(2)CO_(3) and NaHCO_(3) is titrated with 0.1MHCI using Phenolphthalein indicator the end point was 10ml. 20ml of the same solution is titrated with 0.1M HCI, the end point was 25ml with Methylorange indicaor from the begining. NaHCO_(3)+HCl rarr NaCl +H_(2)O +CO_(2) Na_(2)CO_(3)+2HCl rarr 2NaCl +CO_(2) +H_(2)O What is amount of NaHCO_(3) present in 1 litre of solution

Answer»

`2.1g`
`1.05g`
`8.1g`
`0.844g`

SOLUTION :Using METHYL orange,
`x +y = 25 XX 0.1 = 2.5, y = 2.5 - 2 = 0.5`
Wt of `NaHCO_(3)` in 20ml `=(0.5)/(1000) xx 84 = 0.042g`
wt of `NaHCO_(3)` in 1 lit `=(0.042)/(20) xx 1000 = 2.1g`
Discussion
51396.

20 ml of nitric oxide combines with 10 ml of oxygen at STP to give NO_(2). The final volume will be

Answer»

30ml
20ml
10m
40ml

Answer :B
Discussion
51397.

20 mL of HCl having a certain normality neutralises exactly 1.0 g CaCO_(3) . The normality of acid is

Answer»

0.5 N
0.12 N
0.01 N
1.0 N

Solution :`underset((100 g))(CaCO_(3)) + underset((73 g))(2HCl) to CaCl_(2) + H_(2)O + CO_(2)` is
100 g of `CaCO_(3)` shall be NEUTRALISED by
`(73)/(100)=0.73 g ` HCl
`therefore 20` mL of HCl has HCl in equivalent `=(0.73)/(36.5)`
Hence Normality `=(0.73)/(36.5)XX(1000)/(20)=1.0 N`
Discussion
51398.

20 mL of hydrogen measured at 15^(@)C are heated to 35^(@)C. What is the new volume at the same pressure ?

Answer»

Solution :`{:("Given CONDITIONS","Final Conditions"),(V_(1)=20 mL,V_(2)=?mL),(T_(1)=15+273=288 K ,T_(2)=35+273=308 K):}`
By applying Charles' LAW,`(V_(2))/(308)=(20)/(288)" or " V_(2)=(20)/(288)xx308=21.38`
Volume of hydrogen gas at `35^(@)C=21.38 mL`
Discussion
51399.

20 ml of H_(2)O_(2) after acidification with dil H_(2)SO_(4), required 30 ml of N/2 KMnO_(4) for complete oxidation. Calculate the % of H_(2)O_(2) in gr/lit.

Answer»

10.75 g/lit
11.75 g/lit
12.75 g/lit
13.75 g/lit

Solution :MEQ. of `O_(2)="Meq. of "H_(2)O_(2)="Meq. of "KMnO_(4)`
`(W times 1000)/(34//2)=1/2 times 30`
`W=1/2 times (300 times 17)/1000=0.255 gr//20ml`
`N_(H_(2)O_(2))=0.255/17 times 1000/20=0.75`
STRENGTH `=0.75 times 17=12.75` gr/lit
Discussion
51400.

20 mL of a solution of H_2SO_4 neutralises 21.2 mL of 30% solution (w/v) of Na_2CO_3. How much water should be added to each 100 mL of the solution to bring down its strength to decinormal ?

Answer»


Solution :The corresponding equation is:
`underset(105.99 g)(Na_(2)CO_(3)) + underset(98.076 g)(H_(2)SO_(4)) to Na_(2)SO_(4) + H_(2)O + CO_(2)`
The amount of `Na_2CO_3` present in 21.2 mL solution.
`=3/100 xx 21.2 = 0.63 g`
`therefore 105.99 g` of `Na_(2)CO_(3)` react with `H_(2)SO_(4) = 98.076 g`
`therefore 0.636 g` of `Na_(2)CO_(3)` will react with `H_(2)SO_(4)`
`=98.076/105.99 xx 0.636 = 0.588 g`
This much `H_(2)SO_(4)` is present in 20 mL.
`therefore W=(NEV)/1000`
`therefore 0.588 = (N xx 49.038 xx 20)/1000`
(Eq. wt. of `H_(2)SO_(4) = (98.076)/2 = 49.038`)
`therefore N=0.599 = 0.6`
Therefore, the normality of the given `H_2SO_4` solution is 0.6 N. Suppose, v mL of water are REQUIRED to be added to 100 mL of it to make it decinormal `N/10`
`therefore 0.6 xx 100 = 1/10 xx (100 + v)` which GIVES v = 500 mL
Hence, 500 mL of water should be added to each 100 mL of the solution of `H_2SO_4` to bring down its strength to decinormal.
Discussion