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51101.

40 ml of a hydrocarbon undergoes combustion in 260 ml oxygen and gives 160 mlof CO_(2). If all volumes are measured under similar conditions of temperature and pressue, the formula of the hydrocarbon is

Answer»

`C_(3)H_(8)`
`C_(4)H_(8)`
`C_(6)H_(14)`
`C_(4)H_(10)`

Solution :`C_(4)H_(10)+(13)/(2)O_(2)rarr4CO_(2)+5H_(2)O`
`40 ml-260 ml-160 ml`
Discussion
51102.

40 ml of 0.1 Mammonia is mixed with 20 ml of 0.1 M HCl. What is the pH of the mixture ? ( pK_(b) of ammoniasolution is 4.74)

Answer»

4.74
2.26
9.26
`5.00`

Solution :40 ml of 0.1 M `NH_(3)` solution `= 40xx0.1` millimole
= 4 millimoles
`NH_(4)OH+HCl rarr NH_(4)CL + H_(2)O`
2 millimole of HCl will NEUTRALIZE 2 millimoles of `NH_(4)OH` to form 2 millimoles of `NH_(4)Cl`.
`NH_(4)OH` left = 60 ml
`:. [NH_(4)OH]=(2)/(60)M, [NH_(4)Cl]=(2)/(60)M`
`pOH=pK_(b)+LOG .([NH_(4)Cl])/([NH_(4)OH])`
`=4.74 + log .(2//60)/(2//60)=4.74`
`:. pH=14-4.4.74=9.26`
Discussion
51103.

40 ml gaseous mixture of CO, CH_(4) and Ne was exploded with 10 ml of oxygen. On cooling, the gases occupied 36.5 ml. After treatment with KOH the volume reduced by 9 ml and again on treatment with alkaline pyrogallol, the volume further reduced, percentage of CH_(4) in the original mixture is

Answer»

a. `22.4`
B. `77.5`
c. `7.5`
d. `15`

Solution :`underset(x)(CO)+(1)/(2)underset((x)/(2))(O_(2))rarrunderset(x)(CO_(2))`
`underset(y)(CH_(4))+underset(2y)(2O_(2))rarrunderset(y)(CO_(2))+2H_(2)O`
On passing through KOH, volume decrease due to `CO_(2)` absorption
`impliesx+y=9….(1)`
Final volume after reaction =
`CO_(2)+Ne+"left over "O_(2)`
`=(x+y)+(40-x-y)+(10-x//2-2y)`
`36.5=50-(x)/(2)-2y`
`x+4y=27....(2)`
solving, `y=6,x=3`
`IMPLIES %" of "CH_(4)=(6)/(40)xx100=15%`
Discussion
51104.

40 mL of 0.05 M Na_(2)CO_(3). NaHCO_(3). 2H_(2)O (sesquicarbonate ) is titrated against 0.05 M HCl. X and of HCl is used when phenolphthalein is the indicator in two separate titrations , hence (y-x) is :

Answer»

80 mL
30 mL
120 mL
none of these

Solution :N//A
Discussion
51105.

40 gm of a carbonate of an alkali metal or alkaline earth metal containg some insert impurities was made to react with excess HCl solution. The liberated CO_(2) occupied 12.315 litre at 1 atm and 300K. The corrrect option is:

Answer»

Mass of IMPURITY of 1gm and METAL is Be
Mass of impurity is 3gm and metal is Li
Mass of impurity is 6GM and metal is Li
Mass of impurity is 2gm and metal is Mg

Answer :B
Discussion
51106.

40 gm NaOH, 106 gm Na_(2)CO_(3) and 84 gm NaHCO_(3) is dissolved in water and the solution is made 1 lit, 20 ml of this stock solution is titrated with 1 N HCl, hence which of the followign statements are correct?

Answer»

The burette reading of HCl will be 40 ML, if phenolphthalein is used as indiator from the beginning
The burette reading of HCl will be 60 ml, if phenolphthalein is used as indicator form the beginning.
The burette readin of HCl will be 40ml, if methyl orange is used as indicator after the first end point
The burette reading of HCl will be 80 ml, if methyl ORGANE is used as indicator from the very beginning.

Solution :`n_(NaOH)=n_(Na_(2)CO_(3))=n_(NaHCO_(3))=1`
a) In presence of Hph,
Eqts of `NaOH+(1)/(2)` EQ `N_(2)CO_(3)` = Eq HCl
`(200xx1xx1)+(1)/(2)(20xx1xx2)=1xxV`
`V_(HCl)=40ml`
c) In presence of methyl orange at first end point
`(1)/(2)` eq `NA_(2)CO_(3)+` eq `NAHCO_(3)=` eq HCl
`((1)/(2)xx20xx1xx2)+(20xx1xx1)=V_(HCl)XX1`
`V_(HCl)=40ml`
d) If methyl orange used
Eq `NAOH+` Eq `NAl_(2)CO_(3)+` Eq `NaHCO_(3)`
= Eq HCl
`(20xx1xx1)+(20xx1xx2)+(20xx1xx1)`
`=V_(HCl)xx1impliesV_(HCl)=80`
Discussion
51107.

4.0 g of NaOH are dissolved per litre. Find (i) molarity of the solution (ii) OH^(-)ion concentration (iii) pH value of the solution (At. Masses : Na = 23, O=16, H=1).

Answer»

Solution :(i) Calculation of molarity : Mass of NaOH DISSOLVED = 4.0 G/LITRE
Mol . Mass of NaOH = 40 `:.` Molarity of the solution `= ("Strength in g // litre")/("Mol . Mass") = (4.0)/(40) = 0.1 M`
(II) Calculation of the `OH^(-)` ion conc.
NaOH completely ionizes as : `NaOH rarr Na^(+) + OH^(-)" " :. [OH^(-)]=[NaOH]=0.1 M = 10^(-1)M`
(iii) Calculationof pH :
We know that `[H_(3)O^(+)][OH^(-)]=K_(w)=1.0xx10^(-14)`
`:. [H_(3)O^(+)]=(K_(w))/([OH^(-)])=(1.0xx10^(-14))/(10^(-1))=10^(-13)M" " :. pH = - LOG [H_(3)O^(+)]= - log 10 ^(-13) = 13`.
Discussion
51108.

4.0 g of argon (at mass = 40) in a bulb at a temperature of TK had a pressure P atm. When the bulb was placed in hotter bath at a temperature 50^(@) more than the first one, 0.8 g of gas had to be removed to get the original pressure. T is equal is

Answer»

510 K
200 K
100 K
73 K

Solution :According to ideal gas eqatio
`PV = w (RT)/(M) " or " WT = (PV)/(R) .M`
or `w_(1)T_(1) = w_(2) T_(2)` when P and V are constant
Hence `4 xx T = 3.2 (T + 50)`
or `T = 200 K`
Discussion
51109.

40 g f a gas occupies 20 dm^(3) at 300 K and 100 kPa pressure. If the pressure is changed to 50 lPa without changing to temperature, what would be its volume ?

Answer»

Solution :`P_(1)v_(1)=P_(2)v_(2)` at constant temperature.
`P_(1)=100 kPa, v_(1)=20 dm^(3), P_(2) =50 kPa, V_(2) ?`
`V_(2)=(P_(1)V_(1))/(P_(2)),v_(2)=(100xx20)/(50)=40 dm^(3)`
`:. V_(2)=40 dm^(3)`
Discussion
51110.

4-Oxobutanoic acid is reduced with Na-borohydride and the product is treated with aqueous acid. The final product is :

Answer»




ANSWER :A
Discussion
51111.

4-Pentenoic acid when treated with I_(2) and NaHCO_(3) gives:

Answer»

4,5-diiodopentanoic acid
5-iodomethyl-dihydrofuran-2-one
5-IODO-tetrahydropyran-2-one
4-pentenolyiodide

Solution :Iodo lactonization.
Discussion
51112.

4 mole of NH_(4) OH and 1 mole of H_(2) SO_(4) are mixed and dilute to 1 lit solution . The pK_(b) of NH_(4) OH is 4.8 . The pH of solution is

Answer»

`4.8`
`9.2`
`9.5`
`8.9`

ANSWER :B
Discussion
51113.

4-Methylbenzenesulphoic acid reacts with sodium acetate to give

Answer»




Solution :4-Methylbenzenesulphonic ACID is STRONGER than acetic acid THUS it will RELEASE acetic acid from sodiumm acetate
Discussion
51114.

4-methyl-pent-2-yne has how many sigma and pi bond?

Answer»

`15 sigma, 2pi`
`12 sigma, 2pi`
`13 sigma, 2pi`
`14 sigma, 2pi`

ANSWER :A
Discussion
51115.

4-methyl benzene sulphonic acid reacts with sodium acetate to give

Answer»




SOLUTION :H METHYL BENZENE sulphonic acid is STRONGER than acetic acid hence it will release acetic acid from sodium acetate
Discussion
51116.

4-Methyl-1,3-pentadiene react with HBr, The possible products obtained are

Answer»

4-Bromo-2-methyl-2-pentene
4-Bromo-4-methyl-1-pentene
4-Bromo-4-methyl-2-pentene
All the above

Solution :CUMULATIVE DIENES
Discussion
51117.

4 litres of water are added to 2 litres 6 molar hydrochloric acid solution. What is the molarity of the resulting solution ?

Answer»


Solution :TOTAL volume of the solution after MIXING `=(4+2)=6L`
By APPLYING molarity equation :
`overset(("dilute"))(M_(1)V_(1))-=overset(("conc."))(M_(2)V_(2))`
`M_(1)xx6=6xx2orM_(1)=(6xx2)/(6)=2M`.
Discussion
51118.

4 L of 0.02 M aqueous solution of NaCl was diluted by adding one litre of water.The molality of the resultant solution is :

Answer»

`0.004`
`0.008`
`0.012`
`0.016`

ANSWER :D
Discussion
51119.

4-hydroxyl phenol reacts with acidified potassium dichromate to give …………….

Answer»

QUINOL
CYCLOHEXANONE
PHENOL
P-Benzoquinone

Solution :P-Benzoquinone
Discussion
51120.

4-hydroxy phenol reacts with acidified potassium dichromate to gives ..............

Answer»

SOLUTION :p-Benzoquinone
Discussion
51121.

4- heptanoneoverset(KMnO_(4)//H^(+)//Delta)rightarrowA+B. Identify A and B

Answer»

ETHANOIC ACID pentanoic acid
Ethanal andbutanone
Butanoic acid and PROPANOIC acid
Acetic acid andpentanoic acid

Solution :
Discussion
51122.

4^(@) hydrogen is not possible, give reasons…..

Answer»

CARBON has FOUR covalent bond.
Centre carbon is attach with other four carbon and so there is no valence for attachement of HYDROGEN
`1^(@), 2^(@)` and `3^(@)` carbon are stable but `4^(@)` carbon is unstable.
Free RADICAL is formed with `4^(@)` carbon.

Solution :Centre carbon is attach with other four carbon and so there is no valence for attachement of hydrogen
Discussion
51123.

4 grams of each pure huydrochloric acid and pure caustic soda are together dissolved in water. What weight of sodium chloride is obtained?

Answer»

Solution :The STOICHIOMETRIC equation is given as
`HCl+NaOH to NaCl+H_(2)O`
1 mole of HCl =1 mole of NaOH
36.5 grams of HCl=40 grams of NaOH
NaOH is the limiting reagent. Therefore, the WEIGHT of NaCl obtained is LIMITED by NaOH
1 mole of NaOH=1 mole of NaCl
10 grams of NaOH=58.5 grams of NaCl
4 grams of NaOH=?
The weight of sodium CHLORIDE obtained `=58.5 xx (4)/(40)=5.85"grams"`
Discussion
51124.

4 grams of copper chloride on analysis were found to contain 1.890 g of copper (Cu) and 2.110 g of chlorine (CI). What is the empirical formula of copper chloride?

Answer»


ANSWER :`CuCl_(2)`
Discussion
51125.

4 grams of each pure hydrochloric acid and pure caustic soda are together dissolved in water. What weight of sodium chloride is obtained?

Answer»


ANSWER :5.85g
Discussion
51126.

4 grams of an ideal gas occupies 5.6035 litres of volume at 546 K and 2 atm, pressure. What is its molecular weight ?

Answer»

4
16
32
64

Answer :B
Discussion
51127.

A container of 1 L capacity contains a mixture of 4 g of O_(2) and 2 g H_(2) at 0 .^(@)C . What will be the total pressure of the mixture ?(a) 50 . 42 atm(b) 25 . 21 atm(C) 15 . 2 atm(d) 12 . 5 atm

Answer»

<P>

SOLUTION :Calculation of TOTAL pressure :
ACCORDING to Dalton.s law of partial pressure, the total pressure of the gaseous mixture is given by
`P_(mixture) = P_(O_2)+P_(H_2) = 2.80 + 22.4 = 25.2` atm
Discussion
51128.

4 gm NaOH dissolve in 250 ml water and solution is prepared. Find molarity of the solution.

Answer»




ANSWER :0.4 M
Discussion
51129.

4 g of O_2 and 2g of H_2 are confined in a vessel of capacity 1 litre at 0^@C. Calculate the partial pressure of each gas, and

Answer»

Solution :CALCULATION of partial pressures :
Volume of vessel = 1 L, `T = 0^@C = 273 K`
`:. "" PV = nRT " or " p = (nRT)/V`
`:.` Partial pressure of oxygen
`Po_2=(nRT)/V =( 0.125 xx 0.0821 xx 273)/1 = 2.80` atm
Similarly, partial pressure of hydrogen
`P_(H_2)= (nRT)/V= (1xx0.0821 xx 273)/1= 22.4` atm
Discussion
51130.

4 g of O_2 and 2g of H_2 are confined in a vessel of capacity 1 litre at 0^@C. Calculate the number of moles of each gas

Answer»

SOLUTION :CALCULATION of number of moles :
Number of moles = `("Mass in grams")/("GRAM molecular mass")`
`:. " Number of moles of " O_2 = 4/32 =0.125`
and, Number of moles of `H_2 =2/2 =1`
Discussion
51131.

4 g of NaOH are dissolved in 200 cm^3 of water. Find the molarity of the solution.

Answer»

Solution :Mass of NaOH dissolved = 4g
Molecular mass of NaOH = 23 + 16 +1 = 40
`THEREFORE` Moles of NaOH dissolved `=4/40 = 0.1`
Volume of water taken `=200 cm^(3) = 200 xx 10^(-3) L`
The molarity of the solution `=("No. of moles of NaOH")/("Volume in litres")`
`=0.1/(200 xx 10^(-3))`
`=0.5 mol L^(-1)`
Hence, the given solution is 0.5 M
Discussion
51132.

4 g. of methane at 380 torr and 273^@C occupies a volume of

Answer»

5.6 L 
11.2 L 
16.8 L 
22.4 L 

ANSWER :D
Discussion
51133.

4 g of a mixture of Na_(2)SO_(4) and anhydrous Na_(2)CO_(3) were dissolved in pure and volume made up to 250 mL. 20 mL of this solution required 25 mL of N//5 H_(2)SO_(4) for complete neutralisation. Calculate the percentage composition of the mixture.

Answer»


SOLUTION :N//A
Discussion
51134.

4 g of a mixture of NaCl and Na_(2)CO_(3) were dissolved in water and volume made up to 250 mL. 15 mL of this solution required 50 mL of N//10 HCl for complete neutralisation. Calculate the percentage composition of the original mixture.

Answer»


SOLUTION :N//A
Discussion
51135.

4g of mixture of Na_(2)CO_(3) and NaHCO_(3) on heating liberates 448 ml of CO_(2) at STP. The percentage of Na_(2)CO_(3) in the mixture is

Answer»

84
16
54
80

Answer :B
Discussion
51136.

4 g / L solutionof NaOH is given. The molarity of the solution is ........... M. (Molecular mass of NaOH =40 g / mol)

Answer»

160
10
0.1
4

Answer :A::C
Discussion
51137.

4 g Cu is dissolved in concentrated HNO_(3). On heating this solution 5 g oxide will be obtained Then the equivalent weight of Cu is.....

Answer»

23
32
12
20

Solution :4g CU give 5 g copper oxide
4 g Cu COMBINE with `(5-4)=1G` oxygen
`:.1 g O RARR Cu "" :. 8 g rarr (?)`
` = (8xx4)/(1) = 32 g Cu`
eq. wt `=32 "g eq. wt"^(-1)`
Discussion
51138.

3s orbital has 2 radial nodes

Answer»


ANSWER :T
Discussion
51139.

3O_(2) harr 2O_(3). Oxygen is ozonised until the partial pressures of both gases become equal at equilibrium. Which statements are correct if P_(eq) = 10 atm ?

Answer»

60% oxygen is ozonised by volume
mol. wt at equilibrium = 40
`K_(P)` = 0.2 atm
`K_(P)`= 0.5 atm

Solution :
PARTIAL PRESSURES are EQUAL `implies 1-ALPHA=(2alpha)/(3) implies alpha=(3)/(5)=60%`
`(D)/(d)=1+(n-1)alpha=1+((-1)/(3))alpha`
`(32)/(d)=(4)/(5), d=40`
`2P=P_(eqm) implies P=(10)/(2)=5`
`:. K_(P)=(P^(2))/(P^(3))=(1)/(P)=0.2 atm^(-1)`
Discussion
51140.

3N H_(3)PO_(4) aqu. solutionis given = .......... gm/litre.

Answer»

98
298
33
95

Solution :Equivalent weight `=("MOLECULAR mass")/("BASICITY")= (98)/(3)`
`N=("gm/lit")/("equ.wt.")= 3 XX (98)/(3) = 98`
Discussion
51141.

3L Mixture of propane and butane on complete combustion at 298K and 1atm gave 10L CO_2. Calculate the composition of gas

Answer»

`1L C_(3)H_(8), 2L C_(4)H_(10)`
`2L C_(3)H_(8) 1L C_(4)H_(10)`
`1.5 L C_(3)H_(8), 1.5L C_(4)H_(10)`
`0.75L C_(3)H_(8), 2.25L C_(4)H_(10)`

ANSWER :B
Discussion
51142.

3Fe_((s)) + 4H_2O_((g)) undersetlarrto 3Fe_3O_(4(s)) + 4H_2O_((g)) formula of K_P and K_C is …………….

Answer»

SOLUTION :`{:(,3Fe_((s)) + 4H_2O_((g)) UNDERSETLARRTO,Fe_3O_(4(s)) + 4H_2O_((g))),("Current reaction:", 3Fe_((s)) + 4H_2O_((g)) undersetlarrto , Fe_3O_(4(s))+4H_(2(g))):}`
`K_c=[H_2]^4/[H_2O]^4`
`K_p=(p_(H_2))^4/(p_(H_2O))^4`
Discussion
51143.

3Cl_(2)+6NaOHrarr5NaCl+NaClO_(3)+3H_(2)O, chlorine gets

Answer»

Oxidised
Reduced
Both 1 & 2
None of these

Answer :C
Discussion
51144.

3CIO^(-)(aq)rarrCIO_(3)+2C^(-) is an exameof

Answer»

oxidation reaction
REDUCTION reaction
disproportionation reaction
DECOMPOSTION reaction

ANSWER :c
Discussion
51145.

3Ca_(3)(PO_(4))_(2).CaF_(2) is a part of enamel on teeth.

Answer»


ANSWER :T
Discussion
51146.

3C_(2)H_(2) hArr C_(6) H_(6) the above reaction is performed in a I lit vessel. Equilibrium is established when 0.5 mole of benzene is present at certain tempe-rature. If equilibrium constant is 4 lit^(2) "mole"^(-2) The total number ot moles of the substances present at cquilibrium,

Answer»

0.5
1
1.5
2

Solution :`3C_(2)H_(2(g)) HARR C_(6)H_(6(g))`
`K_(C)=([C_(6)H_(6)])/([C_(2)H_(2)]^(3)), 4=(0.5)/([C_(2)H_(2)]^(3)) implies [C_(2)H_(2)]^(3)=(1)/(8)`
`[C_(2)H_(5)]=0.5=(n)/(v) implies n=0.5`
Total no. of moles of `C_(2)H_(2)` & `C_(6)H_(6)` at equilibrium = 0.5+0.5=1
Discussion
51147.

3.92 g of ferrous ammonium suphate ar dissolved in 100 Ml water 20 ml of this solution requires 18 complete oxidation the weight of KMnO_(4) present in one litre of the solution is

Answer»

34.76 G
12.38 g
1.23 g
3.476 g

Solution :The redox reaction involving the oxidation of `Fe^(2+)` is
`MnO_(4)^(-)+5Fe^(2+)+8H^(+)rarr5Fe^(3+)+MN^(2+)+4H_(2)O`
Mol WT of ferrous ammonium suphate
`(NH_(4))SO_(4)FeSO_(4)6H_(23)O=392`
applyinhg molarity equation
`=(M_(2)V_(2))/(n_(2))`
or `(0.1xx20)/(5)=(M_(2)xx18)/(1)`
or `M_(2)=(0.1xx20)/(5xx18)=(M)/(45)`
Amount of `KMnO_(4)` present in one litre
=Molarity x mol wt
`1/45xx158=3.51` g
since 3.476 g isclose to 3.51 g therefore option (d) is correct
Discussion
51148.

38 cm^(2) N_(2) gas at 300K temperature and 96pa pressure. In 0.25g organic compound calculate the % of nitrogen. (1 atm .- 101.3 pa)

Answer»


ANSWER :0.1638
Discussion
51149.

3.7 g of a gas at 25°C occupies the same volume as 0.184 g of hydrogen at 17°C and at the same pressure. What is the molecular mass of the gas ?

Answer»


ANSWER :41.33
Discussion
51150.

3.65% W/V HCl solution has density 1.0365 gm/cc, then its concentration is

Answer»

1 M
1 m
1 N
0.5 M

Solution :`M=(10V%)/("M.wt.")=(10xx3.65)/(36.5)=1M=1N`
`m=(1000M)/(1000d-MM_(1))=(1000)/(1036.5-36.5)=1m`
Discussion