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51601.

1,2-Dimethylcyclohexene undergoes only trans-addition with HBr in non-polar solvents but both cis-and trans-additions occur with aq. Acid. Explain.

Answer»

Solution :This sterospecific reaction proceeds through a bridged cation like the bromonium ion where `H` replaces `Br`. The intermediate may actually be a `pi` -complex. In the absence of a solvent that can satbilise a FREE `R^(oplus), Br^(Ө)` attacks the protonated complex from the opposite side resulting in trans (anti) addition.
In the case of a POLAR solvent such as `H_2 O`,the protonated complex collapses to the free `R^(oplus)`, which now REACTS from both sides (faces) and gives both `CIS-` and trans-additions.
.
Discussion
51602.

1,2-dimethyl benzene on ozonolysis gives

Answer»

GLYOXAL
METHYL glyoxal
dimethyl glyoxal
glycol

SOLUTION :
Discussion
51603.

1,2-Dibromopropane on treatment with X moles of NaNH_2 followed by treatment with ethyl bromide gave a pentyne. The value of X is

Answer»

One<BR>Two
THREE
Four

Solution :i) `CH_3CHBr - CH_2Br + 2NaNH_2 to CH_3C -= CH + 2NaBr + 2NH_3`
II) `CH_3 C -= CH + NaNH_2 to CH_3C -= C^(-) Na^(+) + NH_3`
iii) `CH_3C -= C^(-) Na^(+) + CH_3CH_2 Br to CH_3C -= C CH_2 CH_3 + NaBr`
Thus 3 MOLE of `NaNH_2` is used .
Discussion
51604.

1,2-Benzpyrene is

Answer»

a polynuclear hydrocarbon
CARCINOGENIC in nature
an aromatic hydrocarbon
both (a) and (B)

ANSWER :D
Discussion
51605.

_(1)^(1)H,_(1)^(2)Hand _(1)^(3)H will have the same

Answer»

MASS NUMBER
Chemical reactivity
Electron configuration
Nuclear RADIUS

ANSWER :C
Discussion
51606.

""_(1)^(1)H,""_(1)^(2)H and ""_(1)^(3)H differ in their

Answer»

ATOMIC radius
Position in the periodic table
CHEMICAL properties
PHYSICAL properties

ANSWER :D
Discussion
51607.

[H_(1)^(2)H and ""_(1)^(3)H] will have the same

Answer»

Mass number
Chemical reactivity
ELECTRONIC configuration
NUCLEAR RADIUS

Answer :C
Discussion
51608.

""_1^1H, ""_1^2H and ""_1^3H differ in their

Answer»

Atomic radius
POSITION in the PERIODIC table
Chemical PROPERTIES
Physical properties

ANSWER :D
Discussion
51609.

""_1^1H, ""_1^2H and ""_1^3Hwill have the same

Answer»

MASS NUMBER
CHEMICAL reactivity
Electronic configuration
Nuclear radius

Answer :C
Discussion
51610.

""_(1)^(1)H, ""_(1)^(2)H" and ""_(1)^(3)H will have the same

Answer»

MASS number
Chemical reactivity
Electron configuration
Nuclear radius

Answer :C
Discussion
51611.

1.17 g of an impure sample of oxalic acid was dissolved and made up to 200mL with water. 10 mL of this solution in acid medium required 8.5 mL of a solution of potassium permanganate containing 3.16 g per litre of oxidation. Calculate the percentage purity of oxalic acid.

Answer»


SOLUTION :N//A
Discussion
51612.

1.15 g of an organic compound was analysed by Kjeldahl.s method and the ammonia produced was collected in 30 mL of normal HCl solution. The excess of acid consumed 18.4 mL of normal sodium hydroxide solution for back titration. Calculate the percentage of nitrogen in the substance.

Answer»

SOLUTION :`N = 14.12%`
Discussion
51613.

1.15 g of an organic compound was analysed by Kjeldahl's method and the ammonia produced was collected in 30 mL of normal HCl solution. The excess of acid consumed 18.4 mL of normal sodium hydroxide solution for back titration. Calculate the percentage of nitrogen in the substance.

Answer»

SOLUTION :`N = 14.12%`
Discussion
51614.

1.13 g of an ammonium sulphate were treated with 50 mL of normal NaOH solution and boiled till no more ammonia was given off. The excess of the alkali solution left over was titrated with normal H_(2)SO_(4). The volume required was 30 mL. Find out the percentage of NH_(3) in the salt.

Answer»


SOLUTION :N//A
Discussion
51615.

112ml. of oxygen at STP is subjected to liquefication. The mass of liquid oxygen obtained is

Answer»

0.64 g
0.16 g
0.32 g
0.96 g

SOLUTION :`PV = (WRT)/(M) implies W = (1 xx 0.112 xx 32)/(0.0821 xx 273)`.
Discussion
51616.

11.2L of O, at STP has the same mass as

Answer»

11.2L of METHANE at STP
22.4L of Methane at STP
33.6L of Methane at STP
44.8L of Methane at STP

ANSWER :B
Discussion
51617.

""_(11)^(22)Na contains

Answer»

22 protons
11 NEUTRONS 
22 neutrons
11 nucteons 

ANSWER :B
Discussion
51618.

1120 ml of ozonised oxygen (O_2+O_3) at 1 atm and passing the mixture through alkaline pyrogallol solution is :

Answer»

896 ml
224 ml
448 ml
672 ml

Answer :A
Discussion
51619.

11.2 g carbon reacts completely with 19.63 litre of O_(2) at NTP. The cooled gases are passed through 2"litre" of 2.5N NaOH and Na_(2)CO_(3) in solution. CO does not react with NaOH under these conditions.

Answer»


ANSWER :`NaOH=1.68N, Na_(2)CO_(3)=0.82N`;
Discussion
51620.

11.2 L of oxygen STP and 8 grams of calcium are allowed to react. What volume of which chemical 3 is left unreacted?

Answer»

Solution :The balanced chemical EQUATION is
`2Ca+O_(2)to 2CaO`
2 moles of Ca=1 moles of `O_(2)`
(2 x 40) grams of Ca=22.4 L of `O_(2)` at STP
8 grams of Ca=2.24 L of `O_(2)` at STP
The limiting REAGENT is calcium.
Volume of `O_(2)` at STP required=2.24 L
Volume of `O_(2)` left unreacted =11.2-2.24
=8.96 L at STP
Discussion
51621.

11.2 g carbon reacts completely with 19.63 litre O_(2) at NTP. The cooled gases are pased through 2 litre of 2.5 N NaOH solution. Calculate concentration of remaining NaOH and Na_(2)CO_(3) in solution. (CO does not react with NaOH under these conditions.)

Answer»

Solution :Let X MOLES of carbon be converted into CO and y moles of carbon be converted into `CO_(2)`.
`underset(x)(C )+(1)/(2)underset(x//2)(O_(2)) to CO`
`underset(y)C+underset(y)(O_(2)) to CO_(2)`
Total VOLUME of oxygen used `=(x)/(2)xx22.4+yxx22.4`
=19.63
11.2x+22.4y=19.63
`x+y=(11.2)/(12)`, i.e., 12x+12y=11.2
Solving eqs. (i) and (ii), we get
x=0.11, y=0.82
Number of moles of `CO_(2)` formed =0.82
Number of milliequivalents of NAOH solution through which `CO_(2)` is massed `=NxxV=2.5xx2000=5000`
Number of milliequivalents of `CO_(2)` passed`=0.82xx2xx1000`
=1640
`2NaOH+CO_(2) to Na_(2)CO_(3)+H_(2)O`
Number of milliequivalents of `Na_(2)CO_(3)=1640`
`N_(Na_(2)CO_(3))=(1640)/(2000)=0.82`
Number of milliequivalents of remaining NaOH
=5000-1640=3360
Normality of remaining `NaOH=(3360)/(2000)=1.68`
Discussion
51622.

1.11 gm of CaCl_(2) is added to water forming 500 ml solution. 20 ml of this solution is taken and diluted 10 folds. Find moles of Cl^(-) ions in 2 ml of diluted solution :

Answer»

`8XX10^(-6)`
`4xx10^(-6)`
`12XX10^(-8)`
`5XX10^(-6)`

Answer :A
Discussion
51623.

1.11g CaCl_(2) is added to water forming 500ml of solution 20ml of this solution is taken and diluted 10 folds Find moles of Cl ions in 2ml of diluted solution .

Answer»

Solution :`(1.11)/(111) = 0.01 "mol" CaCl_(2)`
Moles of `CaCl_(2)` in 20mlsolution `= (0.01)/(500) xx 20 = (0.01)/(25)`
In `200ml` solution moles of `CaCl_(2) = (0.01)/(25)` [Note Dilution does not change moles of solute]
In `2ML` of DILUTE solution moles of `CaCl_(2)=(0.01)/(25/(2000))xx2=(0.01)/(2500)=8xx10^(-6)` .
Discussion
51624.

1.1 mole of A and 2.2 moles of B reach an cquilibrium in I lit container according to the reaction. A + 2B hArr 2C + D. If at equilibrium 0.1 mole of D is present, the equilibrium constant is:

Answer»

0.001
0.002
0.003
0.004

Solution :`underset(1.1-x)underset(1.1)(A)+underset(2.2-x)underset(2.2)(2B) harr underset(2x)underset(0)(C)+underset(x)underset(0)(D)`
at EQUILIBRIUM no. of moles D=0.1=x
`1""2""0.3""0.1`
`K_(c)=((0.2)^(2)xx0.1)/(1xx(2)^(2))=0.001`
Discussion
51625.

1.1 g of sample of copper ore is dissolved and Cu^(2+) (aq) is treated with KI the iodine thus liberated requried 12.12 cm^(3) of 0.1 M Na_(2)S_(2)O_(3) solution for tritration what is the percentage of copper in the ore ?

Answer»

Solution :The complete balanced equation for the redox reaction is
`2cu^(2+)+4I^(-)+2S_(2)O_(3)_^(2)rarrCu_(2)I_(2)+S_(2)O_(6)^(2-)+2I^(-)`
No of MOLES of `S_(2)O_(3)^(2-)`used =`(12.12)/(1000)xx0.1=1.212xx10^(-3)` moles
From the BALANCE equation
2 moles of `S_(2)O_(3)^(2-)` reduce `Cu^(2+)` =2 moles
`therefore 1.212 xx10^(-3)` moles of `S_(2)O_(3)^(2-)` will reduce `Cu^(2+)=1.212 xx10^(-3)` moles
Thus % AGE of Cu in the are =`(0.77)/(1.1)xx100=7%`
Discussion
51626.

10ml of mixture containing CO and N_(2) required 7 ml of oxygen to form CO_(2) and NO respectively on combution. Find the volume of N_(2) in the mixture.

Answer»

`7//2`
`17//2`
4 ML
7 ml

Solution :If VOL. of `CO=x`
`CO+(1)/(2)O_(2)rarrCO_(2)`
`underset(x)(CO)+(1)/(2)underset((x)/(2))(O_(2))rarrCO_(2)`
`underset((10-x))(N_(2))+underset(10-x)(O_(2))rarr2NO`
Vol. of `O_(2)` required = `(x)/(2)+10-x=10-(x)/(2)=7`
`IMPLIES (x)/(2)=3impliesx=6mLimplies"vol."N_(2)=4mL`
Discussion
51627.

10mL of a solution of H_(2)O_(2) of 10 V secolouries es 100 ml of KmnO_(4)solution acidified with dilute sulphuric acid ,the amount of KMnO_(4) in the given solution is (AW of k=39, Mn =55)

Answer»

1.125gm
0.155gm
0.56gm
0.28gm

Answer :C
Discussion
51628.

10ml of compound containing 'N' and 'O' is mixed with 30ml of H_(2) to produce H_(2)O (l) and 10ml of N_(2)(g). Molecular formula of compound if both reactants completely, is

Answer»

`N_(2)O`
`NO_(2)`
`N_(2)O_(3)`
`N_(2)O_(5)`

Solution :`C_(6)H_(5)OH(g)+7O_(2)(g)rarr6CO_(2)(g)+3H_(2)O(L)`
`V_(i)=30ml "excess"`
`V_(f)=0 ""("excess"-210) ""180`
`V_(i)=30+ "excess"`
`V_(f)=(180 +"excess"-210)`
`V_("cont")=60ml`
Discussion
51629.

10mL of a solution of H_2 O_2of 10 volume strength decolourises 100mL of KMnO_4solution acidified with dil H_(2)SO_(4)The amount of KMnO_4in the given solution is

Answer»

0.282g
0.564g
1.128g
0.155g

SOLUTION :Normality of `H_2O_2=10/(5.6)`
`W/(EW) =(NxxV(ml))/1000`
`w/(31.6)=(10xx10)/(5.6 xx1000)implies w = 0.564` g
Discussion
51630.

10ml of a gaseous hydrocarbon on combustion gives 40ml of CO_(2) and 50ml of H_(2)O vapour under the same conditions. The hydrocarbon is

Answer»

`C_(4)H_(6)`
`C_(6)H_(10)`
`C_(4)H_(8)`
`C_(4)H_(10)`

SOLUTION :`C_(x)H_(y)rarrxCO_(2)+(y)/(2)H_(2)O`
`10rarr40rarr50`
`impliesx=4impliesy=10`
Discussion
51631.

10mL of 0.2N KMnO_(4) solution in dilute sulphuric acid was decolorised by 40mL of dilute hydrogen peroxide. Calculate the normality of hydrogen peroxide.

Answer»


ANSWER :0.05N
Discussion
51632.

10ml of 0.1M CH_(3)COOH is mixed with 990 ml of 0.01 M NaCl solution. What is the change in pH of CH_(3)COOH solution

Answer»


SOLUTION :N//A
Discussion
51633.

1.0g sample of subtance A at 100^(@)C is added to 100 mL ofH_(2)O at25^(@)C. Using separate 100 mL portions of H_(2)O , the procedure is repeated with substance B and then with substance C. How will the final temperatures of the water compare ? {:("Substance","Specific heat"),(A,0.60 J g^(-1).^(@)C^(-1)),(B,0.40 J g^(-1).^(@)C^(-1)),(C,0.20 J g^(-1).^(@)C^(-1)):}

Answer»

`T_(C) gt T_(B)gt T_(A`
`T_(B) gt T_(A)gt T_(C)`
`T_(A) gtT_(B)gtT_(C)`
`T_(A) =T_(B)=T_(C)`

Solution :Specificheat is the amount of heat required to raise the temperature of 1 g of substance through `1^(@)C` or it is the heatreleased when temperature of 1 g of the substance FALLS through `1^(@)C`. Greater the SPECIFIC heat of the substance, greateris th heat released which is absorbed by water and hence greater is the risein temperature of water or greater is the temperature ofwater. As specific heats OFA, B and C are in the order `A gt B gt C`, therefore, `T_(A) gt T_(B) gt T_(C)`
Discussion
51634.

10g of pyrolusite on reaction with conc. HCl liberated 0.1 equivalent of Cl_(2). Percentage purity of pyrolusite sample-

Answer»

`87.0%`
`43.5%`
`21.75%`
`100%`

ANSWER :B
Discussion
51635.

10g of NH_(4)CI (mol.wt. 53.5) when dissolved in 1000g water lowered the freezing point by 0.637^(@)C. Calculate the degree of hydrolysis of the salt if its degree of dissociation is 0.75 . The molal depression constant of water is 1.86 K molality^(-1).

Answer»


ANSWER :`0.106;`
Discussion
51636.

10g of an organic substance when dissolved in two litres of water gave an osmotic pressure of 0.59 atm, at 7^(@)C. Calculate the molecular weight of the substance.

Answer»

Solution :`PI=("MOLES of solute")/("LITRES of solution")xxRT`
`0.59=(10xx0.082xx280)/(Mxx2)`
`M=(10xx0.82xx280)/(2xx0.59)="194.6 g/mol"`
`"Molecular WEIGHT = 194.6 g/mole"`
Discussion
51637.

1.0g ofa samplecontaining NaCl, KCl and some inert impurity is dissolvedin excess of water and treated with excess of AgNO_(3) solution. A 2.0 g precipitate to AgCl separate out. Also sample is 23% by mass in sodium. Determine mass percentage of KCl in the sample :

Answer»


Solution :Moles of `NaCl`in SAMPLE `=0.01` = moles of `AgCl` from `NaCl` in PRECIPITATE `=0.01`
Total moles of `AgCl` precipitate `=(2)/(143.5)=0.01393`
implies Moles of `AgCl` from `KCL = 0.00393` = moles of `KCl`
implies Mass of `KCl` in sample `=0.00393xx74.5=0.2928g`
Mass `%` of `KCl` in the sample `=29.28`
Discussion
51638.

1.0g magnesium atoms in vapour phase absorbs 50.0kJ of energyto convert all Mg into Mg ions. The energy absorbed is needed for the following changes : Mg_((g))rarr Mg_((g))^(+)+e, DeltaH=740kJ mol^(-1) Mg_((g))^(+) rarr Mg_((g))^(2+)+e, DeltaH=1450kJ mol^(-1) Find out the % of Mg^(+) and Mg^(2+) in final mixture.

Answer»


ANSWER :`MG^(+)=68.28%,Mg^(2+)=31.72%`
Discussion
51639.

10.875 g of a mixture of NaCI and Na_2CO_3 was dissolved in water and the volume was made up to 250 mL. 20.0 mL of this solution required 75.5 mL of N/10 H_2SO_4. Find out the percentage composition of the mixture.

Answer»

Solution :NaCl does not react with `H_2SO_4`. Therefore, whole of the sulphuric acid reacted with `Na_2CO_3` only.
Suppose, 10.875 g of the mixture contain x g of `Na_2CO_3`. This is present in 250 ML of the solution.
Since, `w =(NEV)/1000`
We have `x=(N xx 53 xx 250)/1000`
or `N =(1000 x)/(53 xx 250) = (1000 x)/(13250)`
According to the normality relation,
`underset(Na_(2)CO_(3) "sol")(N_(1)V_(1)) = underset(H_(2)SO_(4))(N_(2)V_(2))`
`=(1000 x)/(13250) xx 20.0 =1/10 xx 75.5`
or `x=(13250 xx 75.5)/(1000 xx 20.0 xx 10) = 5.0018 g`
Thus, 10.875 g of the sample contain 5.0018 g of `Na_(2)CO_(3)`
`therefore` MASS of NaCl present `=10.875 - 5.0018 = 5.8732 g`
Hence, the amount of `Na_2CO_3`in the sample `=5.0018/10.875 xx 100 = 45.994 %`
and the amount of NaCl in the sample `=5.6732/10.875 xx 100 = 54.006%`
Discussion
51640.

1.40 g of a metal when heated in a current of oxygen gave 1.93 g of the metal oxide. Calculate the equivalent weight of the metal.

Answer»

SOLUTION :MASS of the metal oxide = 1.08 G
Mass of the metal obtained = 1 g
Mass of oxygen that combines with 1 g of the metal = 1.08- 1 = 0.08 g
Hence, Equivalent WEIGHT of the metal
`("Mass of the metal")/("Mass of oxygen")xx 8 = 1/0.08 xx 8 = 100.0`
Discussion
51641.

10^(6) ml balloon contains He gas at 27^(@)C temperature and 1 bar then calculate of balloon. Take molecular mass of air is 28.84 gm/mol.

Answer»


ANSWER :1010 KG
Discussion
51642.

10^(5)g of a sample of water contained 16.2 8 of Ca(HCO_(3))_(2) Its hardness is

Answer»

200 ppm
300 ppm
60 ppm
100 ppm

Solution :Hardness in temrs of
`Ca(HCO_(3))_(2)=16.2xx(100)/(162)=10g`
degree of hardness `=(10)/(10^(5))xx10^(6)=100 ppm`
Discussion
51643.

1.05 g of a metal gives a oxidation 1.5g of its oxide. Calculate its equivalent mass.

Answer»

Solution :MASS of oxygen `=1.5 -1.05`
`=0.45 g`
`0.45g` of oxygen combines with `1.05 g` of METAL.
`therefore 8 g` of oxygen combines with `(8xx1.05)/(0.45)` g of metal `=18.66g` of metal
`therefore` equivalent mass of metal `=18.66 "g EQU"^(-1)` .
Discussion
51644.

10^5g of a sample of water contained 16.2 g of Ca(HCO_3)_2 its hardness is

Answer»

200 ppm
300 ppm
60 ppm
100 ppm

Answer :D
Discussion
51645.

1.03 g mixture of sodium carbonate and calcium carbone require 20 mL N HCl for complete neutralisation. Calculate the percentage of sodium carbonate and calcium carbonate in the given mixture.

Answer»

Solution :`underset(106)(Na_(2)CO_(3))+underset(2xx36.5)(2HCl) to 2NaCl +H_(2)O+CO_(2)`
Eq. mass 53 `"" `36.5
1 g eq. `""` 1 g eq.
`underset(100)(CaCO_(3))+underset(2xx36.5)(2HCl)to CaCl_(2)+H_(2)O+CO_(2)`
Eq. 50 `"" `36.5
1g eq `"" `1 g eq.
Let x go `CaCO_(3)` be PRESENT in the mixture
Mass of `Na_(2)CO_(3)` in the mixture =(1.03-x)g
No. of equivalents of `CaCO_(3)=(x )/(50)`
No. of g equivalents of `Na_(2)CO_(3)=((1.03-x))/(53)`
No. of g equivalents in 20 mL. N HCl`=("Normality " xx "VOL.")/(1000)`
`=(1xx20)/(1000)=(1)/(50)`
At equivalence point,
No. of g equivalents of `CaCO_(3)` + No. of g equivalents of `Na_(2)CO_(3)`=No. of GRAM equivalents of HCl
`(x)/(50)+(1.03-x)/(53)=(1)/(50)`
or x=0.50
`CaCO_(3)=0.50 g, % CaCO_(3)=(0.50)/(1.03)xx100=48.54`
`Na_(2)CO_(3)=0.53 g, % Na_(2)CO_(3)=(0.53)/(1.03)xx100=51.46`
Discussion
51646.

103 mL of carbon dioxide were collected at 27^(@)C and 763 mm pressre. What will be its volume if the pressure is changed to 721 mm at the same temperature ?

Answer»

Solution :`{:("GIVEN Conditions","Final Conditions"),(V_(1)=103 mL,V_(2)=?mL),(P_(1)=763 mm,P_(2)=721 mm):}`
By applying Boyle's Law since temperature is constant, `P_(2)xxV_(2)=P_(1)xxV_(1)`.
Substituting the corresponding values, we have
`721xxV_(2)=763xx103,V_(2)=(763xx103)/(721)=109 mL:.` Volume of carbon dioxide=109 mL.
Discussion
51647.

10^(21) molecules are removed from 440mg of CO_(2). It becomes

Answer»

366mg
8.3 milli mole
200 mg
4.1 milli mole

Solution :`10^(21)` molecules `(0.01)/(6)` MOLES = 73.3 mg
440mg = 0.01 moles
`THEREFORE` left over = `0.01-(0.0.1)/(6)` moles = 8.3 moles
`=440-73.3mg =366.7mg`
Discussion
51648.

102 g of solid NH_(4)HS is taken in the 2L evacuated flask at 57^(@). Following two equilibrium exist simultaneously NH_(4)(s)iffNH_(3)(g)+H_(2)S(g) NH_(3)(g)iff(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g) one mole of the solid decomposes to maintain both the equilibrium and 0.75 mol e of H_(2) was found at the equilibrium then find the equilibrium concentration of all the species and K_(C) for the both the reaction.

Answer»

Solution :Moles of `NH_(4)HS=(102)/(51)=2`
`NH_(4)(s)iffNH_(3)(g)+H_(2)S(g)`
`{:(2,0,0),(1,1-x,1):}`
`NH_(3)(g)IFF(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g) K_(C_(2))`
`1-x (x)/(2) (3x)/(2)`
Given that moles of `H_(2)=(3x)/(2)=0.75 iff x=(1)/(2)`
`K_(C_(2))=1/2 ((1-x))/(2)=1/8` [SINCE `V=2L`]
`K_(C_(2))=(((3x)/(4))^(3//2)((x)/(4))^(1//2))/(((1-x)/(2)))=(((3)/(8))^(3//2)((1)/(8))^(1//2))/(1/4)=(3)^(3//2)1/64xx4/1=(3)^(3//2)/(16)`
Discussion
51649.

10^(20) atoms of an element has a mass of 4 mg. What is the atomic mass of the element?

Answer»

SOLUTION :`10^(20)` atoms of element =4mg
`10^(23)` atoms of element =4g
`6XX10^(23)` atoms of element =?
ATOMIC mass of element is 24
Discussion
51650.

100mLOf top water containg Ca(HCO_(3))_(2)was titrated with 30mL if HCl were required calculate the tempporrary harness as parts of CaCO_(3) per 10^(6) parts of water .

Answer»

300ppm
150ppm
100pm
600ppm

Answer :A
Discussion