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.

51801.

1 mole of photons, each of frequency 250 s^(-1) would have approximately a total energy of

Answer»

1 ERG
1 joule
1 eV
1 MeV

Solution :`E = N_(A) HV = (6.02 xx 10^(23)) (6.6 xx 10^(-34) Js) (250 s^(-1)) = 10^(-7) "erg " = 1J`
Discussion
51802.

1 mole of photon, each of frequency 2500 s^(-1) , would have approximately a total energy of

Answer»

10 erg
1 JOULE
1 eV
1 MeV

Solution :`upsilon=2500 s^(-1)`
`therefore` Energy of a photon , E=hv
`=(6.626xx10^(-34)xx2500)`J
`because` Energy of one MOLE of photon
`=(6.626xx10^(-34)xx2500xx6.023xx10^23)J`
`=9.977xx10^(-7) J ~~ 10` erg
Discussion
51803.

1 mole of NH_(3) gas at 27^(@) C is expanded under reversible adiabatic conditions to make volume 8 times (gamma=1.33). Final temperature and work done respectively are :

Answer»

150 K, 900 cal
150 K, 400 cal
250 K, 1000 cal
200 K, 800 cal

Answer :A
Discussion
51804.

1 mole of PCl_(5) is placed in a closed vessel at 523K. At equilibrium, if it dissociates to an extent of 35%, calculate K_(p)" for "PCl_(5)hArrPCl_(3)+Cl_(2). Equilibrium pressure is found to be 5xx10^(5) Pa.

Answer»

<P>

Solution :Initial no. of MOLES: `underset(1)(PCl_(5))hArrunderset(0)(PCl_(3))+underset(0)(Cl_(2))`
`{:("No. of moles present"),("at equilibrium"):}}1-x=0.65" "x""x""(x=0.35)`
TOTAL number of moles present at equilibrium = 0.65 + 0.35 + 0.35 = 1.35
We know that partial PRESSURE = Mole fraction = Mole fraction `xx` total pressure
`thereforeP_(PCl_(5))=(0.65/1.35)xx5xx10^(5)PA,P_(PCl_(3))=(0.35/1.35)xx5xx10^(5)Pa,P_(Cl_(2))=(0.35/1.35)xx5xx10^(5)Pa`
`K_(p)=(P_(PCl_(3))xxP_(Cl_(2)))/(P_(PCl_(5)))=((0.35/1.35)xx5xx10^(5)(0.35/1.35)xx5xx10^(5))/((0.65/1.35)xx5xx10^(5))=6.98xx10^(4)Pa`
Discussion
51805.

1 mole of N_(2) and 3 mole of H_(2) are mixed in a closed vessel of 1 dm^(3) capacity. At equilibrium if the vessel contains a total of 2.4 moles, calculate equilibrium constant K_(c)" for "N_(2)+3H_(2)hArr2NH_(3).

Answer»

Solution :Initial no. of moles: `underset(1)(N_(2))+underset(3)(3H_(2))hArrunderset(0)(2NH_(3))`
`{:("No. of moles reacting"),("at equilibrium"):}}x""3x""-`
`{:("No. of moles present"),("at equilibrium"):}}1-x""3-3x""2x`
`therefore` TOTAL no. of moles present at equilibrium = (1 - x) + (3 - 3x) + 2x = 4 - 2x
Given 4 - 2x = 2.4 `""therefore2x=4-2.4orx=1.6/2=0.8`
Since the volume of the vessel is 1 `DM^(3)`,
`[N_(2)]=(1-x)/1=(1-0.8)/1=0.2" mol"//dm^(3),[H_(2)]=(3-3x)/1=(3-2.4)/1=0.6"mol"//dm^(3)`
`[NH_(3)]=(2x)/1=1.6/1=1.6"mol"//dm^(3)`
`K_(c)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))=((1.6)^(2))/((0.2)xx(0.6)^(3))=59.26("mol"//dm^(3))^(-2)`.
Discussion
51806.

1 mole of methane (CH_4) contains

Answer»

`6.02 xx 10^23` atoms of H
4 gram atoms of HYDROGEN
`1.81 xx 10^23` molecules of methane
3.0 G of carbon.

Answer :B
Discussion
51807.

1 mole ofIO_3^(-)ions is heated with excess ofI^(-) ions in the presence of acidic conditions as per the following equationIO_(3)^(-) +I^(-) rarrI_2 .How many moles of acidified hypo solution will be required toreact completley withI_2thus produced ?

Answer»

Solution :`10e^(-)+2IO_3^(-) rarrI_2,2I^(-) rarrI_2+2e^(-)]xx5`
`2IO_3^(-)+10I^(-) rarr6I_2`
`2MOL""10mol ""6mol`
ii) `I_2+2S_2O_3^(2-)rarrS_(4)O_6^(2-)+2I^(-)`
1 mol 2 mol
1 mol of `IO_3^(-) =5 ` mol `5I^(-) -= 3 ` mol of `I_2 =6` mol of `S_2O_3^(2-)`
Discussion
51808.

1 mole of H_(2)SO_(4)will exactly neutralise:

Answer»

2 mole of ammonia
1 mole of `BA(OH)_(2)`
0.5 moles of `Ca(OH)_(2)`
2 mole of KOH

Answer :A::B::D
Discussion
51809.

1 mole of each of Ca(OH)_(2) and H_(3) PO_(4) are allowed to react under dilute conditions . The maximum number of moles of Ca_(3) (PO_(4))_(2) formed is

Answer»

1
`1//2`
`1//3`
3

Answer :C
Discussion
51810.

1 mole of chlorine combines with a certain mass of a metal giving 111g of its chloride. The atomic mass of the metal (assuming its valency to be 2) is

Answer»

40
20
80
60

Solution :`("WT of METAL chloride ")/("wt of chlorine ")=(E_("Metal")+E_(CL^(-)))/(E_(Cl^(-)))`
Discussion
51811.

1 mole of an ideal monatomic gas is subjectedtoi the followingreversible charge of stateAunderset("Expansion")overset("Isothermal")toBunderset("Cooling") overset("Isochoric")to (5 atm , 500K)Cunderset("compression")overset("Adiabatic")toA(300K)Then , which of the following are correct?[ Given(0.6)^(2.5) =0.3 , "in"2=0.7](R=0.08 L-atm//"mol"-K=2cal//"mol"-k)

Answer»

Pressure of point B is 2.5 atm
Volume at point C is 16 litre
Magnitude of work INVOLVED in complete process is 100 cal
Change in enthalpy of process of process `C-A` is`+1` kcal .

Answer :a,b,c,d
Discussion
51812.

1 mole of any gas a) Occupies 22.4 lit at STP b) Contains 3.05 xx 10^22 molecules c) Contains 6.023 xx 10^23 molecules d) Contain same number of molecules as in 22 gm of CO_2

Answer»

B, d 
a,C
b, c 
a, d

Answer :B
Discussion
51813.

1 mole of an ideal gas, maintained at 4.1 atm and at a certain temperature, absorbs heat 3710 J and expands to 2 litres. Calculate the entropy change in expansion process.

Answer»

SOLUTION :`n=1"mole"`
`P=4.1atm`
`V=2" LIT"`
`T=?`
`q=3710J`
`DeltaS=(q)/(T)`
`DeltaS=(q)/(((PV)/(NR)))`
`DeltaS=(nRq)/(PV)`
`DeltaS=(1xx0.082" lit atm "K^(-1)xx3710J)/(4.1" atm"xx2" lit")`
`DeltaS=37.10JK^(-1)`.
Discussion
51814.

1 mole of an ideal gas, maintained at 4.1 atm and at a certain temperature , absorbs heat 3710 J and expands to 2 litres . Calculate the entropy changes in expansion process.

Answer»

SOLUTION :PRESSURE of an ideal gas =`P_i` = 4.1 atm.
EXPANSION in volume `DeltaV` = 2 litres
Heat absorbed =Q = 3710 J
Entropy change = `DeltaS` = ?
For an ideal gas PV = RT for one mole .
`T=(PV)/R=(4.1xx2)/0.0830=100^@C`
T=100+273 = 373 K
`DeltaS=q/T_((K))=3710/373`
Entropy change =9.946 `JK^(-1)`
Discussion
51815.

1 mole of an ideal gas A(C_(v,m)=3R) and 2 mole of an ideal gas B are (C_(v,m)=(3)/(2)R) taken in a container and expended reversibly and adiabatically from 1 litre to 4 litre starting from initial temperature of 320 K. DeltaE "or" DeltaU for the process is :

Answer»

`-240 R`
`+240 R`
480 R
`-960 R`

ANSWER :D
Discussion
51816.

1 mole of an ideal gas A (C_(v,m)=3R) and 2 mole of an ideal gas B are (C_(v.m)= (3)/(2)R) taken in a constainer and expanded reversible and adiabatically from 1 litre of 4 litre starting from initial temperature of 320K. DeltaE or DeltaU for the process is (in Cal) (Give your answer after divide with 240)

Answer»

`-240R`
240R
`-1920` Cal
`-960R`

Solution :`C_(V_("avg")) = (1 xx 3R + 2 xx (3)/(2) R)/(1 +2) = 2R rArr gamma = 3//2`
`T_(1) V_(1)^(gamma-1) = T_(2) V_(2)^(gamma-1)`
`320 xx (1)^(1//2) = T_(2) (4)^(1//2) rArr T_(2) = (320)/(2) = 160K`
`Delta U= nC_(v) Delta T= 3 xx 2R xx (160-320)`
`= -960 R = - 1920` cal
Discussion
51817.

1 mole of a diatomic element X_2 contains 34 and 40 moles of electrons and neutrons respectively. The isotopic formula of the element is:

Answer»

`""_(34)^(74)X`
`""_(17)^(37)X`
`""_(34)^(40)X`
`""_(20)^(40)X`

ANSWER :B
Discussion
51818.

1 mole of a compound contains 1 mole of C and 2 moles of O. The molecular weight of the compound is

Answer»

3
12
32
44

Answer :D
Discussion
51819.

Calculate the number of protons, neutrons and electrons respectively in ""_(7)^(14) N^(3-)

Answer»

`7xx6.023xx10^(23)` ELECTRONS
`7xx6.023xx10^(23)` protons
`7xx6.023xx10^(23)` neutrons
`14xx6.023xx10^(23)` protons

Solution :1 MOLE implies 1 mole ions
implies 7 mole protons
7 mole neutrons
10 mole `e^(-)`
Discussion
51820.

1 mole each of H_2(g)and I_2(g) are introduced in a 1L evacuated vessel at 523 K and equilibrium H_2(g)+I_2(g)hArr2HI(g) is established. The concentration of HI(g) at equilibrium:

Answer»

CHANGES on changing pressure.
CHANGE on changing temperature.
is same even if only 2 mol of HI(g) were introduced in the vessel in the BEGINING.
is same even when a platinum gauze is introduced to catalyse the reaction.

ANSWER :A::B::C::D
Discussion
51821.

1 mole each of H_(2)(g) "and" I_(2)(g) are introduced in a 1L evacuated vessel at 523K and equilibriumH_(2)(g)+I_(2)(g)hArr2Hi(g) is established. The concentration of HI(g) at equilibrium:

Answer»

Changes on changing PRESSURE.
Changes on changing temperature.
Changes on changing volume of the vessel
Is same even if only `2` mol of `HI(g)` were introduced in the vessel in the beginning.
Is same even when a platinum gauze is introduced to catalyst the reaction.

Solution :`H_(2)(g)+I_(2)(g)hArr2HI(g)`
(A) For changing pressure volume has to be changed, THOUGH NUMBER of moles of `HI(g)` do not get changed but its concentration will get changed.
(B) Temperature change will change `K_(P)` and hence concentration.
(C) Volume change will change concentration, not the number of moles.
(D) Same equilibrium will be attained from EITHER direction.
Catalyst does not change equilibrium concentration.
Discussion
51822.

1 mole each of FeC_(2)O_4and FeSO_4is oxidised separately by 1M KMnO_4in acid medium. Calculate the volume ratio of KMnO_4used for FeC_(2) O_4 and FeSO_4

Answer»


Solution :`Fe^(2+)+C_(2)O_(4)^(2-) rarrFe^(3+)+2CO_(2)^(4-)+3e`
`Fe^(2+) +SO_(4)^(2-) rarrFe^(3+)+SO_(2)^(4-)e`
`Mn^(7+)+5erarrMn^(2+)`
Meq.of `F_(e)C_(2)O_4` = Meq.of `KMnO_4`
`1xx3xx1000=1xx5xxV_(1).......(1)`
Meq.of `FeSO_(4)` = Meq.of `KMnO_4`
`1xx1 xx1000=1xx5xxV_2""....(2)`
By EQ.(1) and (2) `V_1/V_2=3`
Discussion
51823.

1 mole each of CaC_(2), Mg_(2)C_(3) reacts with excess water in separate open flasks work done by the gas during the dissolution shows the order:

Answer»

`CaC_(2) = Mg_(2)C_(3) lt Al_(4)C_(3)`
`CaC_(2) = Mg_(2)C_(3) = Al_(4)C_(3)`
`Mg_(2)C_(3) lt CaC_(-2) lt Al_(40C_(3)`
`Mg_(2)C_(3) lt Al_(4)C_(3) lt CaC_(2)`

Solution :`CaC_(2) RARR C_(2)H_(2)`
`Mg_(2)C_(3) rarr C_(3)H_(4)`
`Al_(4)C_(3) rarr CH_(4)`
Discussion
51824.

1 mole CH_(4) contain....

Answer»

`6.022xx10^(23)` hydrogen GAS
`6.022xx10^(23)` atoms of hydrogen
8g MOLECULES of hydrogen
4 g atoms of hydrogen

Answer :A::D
Discussion
51825.

1mole CaCO_(3(s)) isheatedin11.2litvesselsothatequilibriumisestablishedat819 K.IfK_pfor CaCO_3 harr CaO + CO_2 at this temperature is 2 atm, equilibrium concentration of CO_2 (in mol-lit^(-1))

Answer»

`1//3`
`1/(11.2)`
`1/(33.6)`
`1/(22.4)`

Solution :`CaCO_(3(s)) harr CaO_((s))+CO_(2(g))`
`K_(P)=PCO_(2)=2`
we know that `K_(P)=K_(C)(RT)^(Deltan)`
`K_(c)=(K_(P))/((RT)^(DELTA N))=(2)/((0.0821 xx 819)^(1))=(1)/(33.6)`
Discussion
51826.

1 mole Ba(OH)_(2) will exactly neutralize

Answer»

0.5 MOLE HCl
1 mole `H_(2)SO_(4)`
1 mole of `H_(3)PO_(3)`
2 mole of `H_(3)PO_(2)`

Solution :1 Mole `BA(OH)_(2)` = 2 equivalents `Ba(OH)_(2)`
1 mole HCl = 1 EQUIVALENT HCl
1 mole `H_(3)PO_(3)` (di-basic acid)
= 2 equivalent `H_(3)PO_(3)`
1mole `H_(3)PO_(2)` (monobasic acid)
= 1 equivalent `H_(3)PO_(2)`
Discussion
51827.

1 molar solution of a non-volatile and non-electrolyte compound will produce an osmotic pressure ….. At 0^(@)C

Answer»

1 atm
44.8 atm
0.5 atm
0.75 atm

Answer :D
Discussion
51828.

1 mol PCl_(3) and 1 mol Cl_2 are taken in a 10 lit vessel and heated . Now , if 0.8 mol PCl_(5) is present at equilibrium , then

Answer»

`[PCl_(3)] = 0.8` M
`[Cl_(2)] = 0.02` M
`[PCl_(5)] = 0.02` M
`[PCl_(3)] = [Cl_(2)]= 0.2 `M

Answer :B
Discussion
51829.

1 mol of PCl_(5) , kept in a closed container of volume of 1 dm^(3) and was allowed to attain equilibrium at 423 K . Calculate the equilibrium composition of reaction mixture . (The K_(C) value for PCl_(5) dissociation at 423 K is 2 )

Answer»

Solution :`PCl_(5) hArr PCl_(3) + Cl_(2)`
Given that `[PCl_(5)]_("initial") =1` mol , `V = 1 dm^(3) , K_(C) = 2`

`K_(C) = ([PCl_(3)][Cl_(2)])/([PCl_(5)])`
`2 = ( x xx x)/((1 - x))`
`2 - 2x = x^(2)`
`x^(2) + 2x - 2 = 0`
Solution for a quadratic equation
`ax^(2) + bx + c = 0` are `x = (- B pm sqrt(b^(2) - 4ac))/(2a)`
`a = 1 . b = 2 . c = -2`
`x = ( - 2 pm sqrt(4 - 4XX 1 xx (-2)))/(2 xx 1) = (-2 pm sqrt(12))/(2) = (-2 pm sqrt(4 xx 3))/(2)`
`x = (-2 pm 2 sqrt3)/(2) = (- 2 + 2 sqrt3)/(2) , ( - 2 - 2 sqrt3)/(3)`
`x = -1+ sqrt3 , - 1 - sqrt3` (SINCE x is +ve ) , `-1 - sqrt3` not possible
`= - 1 + 1.732 = 0.732`
`therefore` Equilibrium concentration of `[PCl_(5)]_(EQ) = (1- x)/(1) = 1 - 0.732 = 0. 268` M
`[PCl_(3)]_(eq) = (x)/(1) = (0.732)/(1) = 0.732`
`[Cl_(2)]_(eq) = (x)/(1) = (0.732)/(1) = 0.732`
Discussion
51830.

1 mol of NH_(3) gas( gamma = 1.33) at 27^(@)Cis allowed to expand adiabatically so that the final volume becomes 8 times. Work done will be

Answer»

450 cal
1800 cal
900 cal
300 cal

Solution :Work done in ADIABATIC expansion` = - nC_(v) (T_(2) -T_(1))`. To calculate FINAL tem. `T_(2)`, proceeding as in the above case, `T_(2) = 150 K`.
`C_(v)` for `NH_(3)= 3R`
`( NH_(3)` is a non-linear molecule with`N=4` .HENCE, translational degrees`=3`, Rotational degrees`=3`. Total `U = ( 3)/(2) RT + (3)/(2)RT =3RT`)
`:. w=-1 xx3 xx2(150-300) = 900cal`
Discussion
51831.

1 mol of equimolar mixture of ferric oxalate and ferrous oxalate will require x mol of KMnO_(4) in acidic medium for complete oxidation. X is

Answer»

0.8 mol
0.9 mol
1.6 mol
1.8 mol

SOLUTION :Equimolar implies 0.5 MOLES each
eq. `FeC_(2)O_(4)+eq. Fe_(2)(C_(2)O_(4))_(3)=eq. KMnO_(4)`
`0.5xx3+0.5xx6=x xx5impliesx=0.9`
Discussion
51832.

1 mol of CH_4, 1 mole of CS_2 and 2 mole of H_2S are 2 mol of H_2 are mixed in a 500 mL flask. The equilibrium constant for the reaction K_C = 4 xx 10^(-2) mol^2 lit^(-2). In which direaction will the reaction proceed to reach equilibrium ?

Answer»

Solution :`CH_4 (g)+ 2H_2S (g) hArr CS_2(g) + 4H_2(g)`
`K_C =4 xx 10^(-2) mol^2 L^(-2)`
VOLUME`= 500 mL= 1/2 L`
`[CH_4]= (1 mol)/(1/2L) = 2mol L^(-1)`
`[CS_2] = (1 mol)/(1/2L) = 2mol L^(-1)`
`[H_2S] = (2 mol)/(1/2L) = 4mol L^(-1)`
`[H_2]= (2mol)/(1/2L) = 4 mol L^(-1)`
`Q = ([CS_2][H_2]^4)/([CH_4][H_2S]^2)`
`:. Q = (2 xx (4)^4)/(2 xx (4)^2) = 16`
`Q gt K_C`
`:.` The reaction will PROCEED in the reverse direction to REACH the equilibrium.
Discussion
51833.

1 mol of CH_(4), 1 mole of CS_(2) and 2 mol of H_(2)S are 2 mol of H_(2) are mixed in a 500 ml flask The equilibrium constant for the reaction K_(C)=4xx10^(-2)"mol"^(2)" lit"^(-2). In which direcition will the reaction proceed to reach equilibrium ?

Answer»

Solution :`CH_(4)(g)+2H_(2)S(g)hArrCS_(2)(g)+4H_(2)(g)`
`K_(C)=4xx10^(-2)" MOL LIT"^(-2)`
`"Volume = 500 ML"=1/2L`
`{:([CH_(4)]_("in")=("1 mol")/(1)""[CS_(2)]_("in")=("1 mol")/(1/2L)),("= 2 mol L"^(-1)"= 2 mol L"^(-1)),([H_(2)S]_("in")=("2 mol")/(1/2L)""[H_(2)]=("2 mol")/(1/2L)),("= 4 mol L"^(-1)"= 4 mol L"^(-1)):}`
`Q=([CS_(2)][H_(2)]^(4))/([CH_(4)][H_(2)S]^(2))`
`:.Q=(2xx(4)^(4))/((2)xx(2)^(2))=64`
`QgtKC`
The reaction will proceed in the reverse direation to reach the equilibrium.
Discussion
51834.

1 mol N_2 and 3 mol H_2 taken in 4L definite temperature of closed vessel. Reaction N_(2(g)) + 3H_(2(g)) hArr2NH_(3(g)) according to 0.25% N_2 convert into ammonia. Calculate K_c and how much reaction of K_c? 1/2N_(2(g)) + 3/2 H_(2(g)) hArr NH_(3(g))

Answer»

SOLUTION :`K_c=1.49xx10^(-5) ("mol L"^(-1) )^(-2)` and `K_c=3.86xx10^(-3) ("mol L"^(-1))`
Discussion
51835.

1 mol N_2 and 3 mol H_2 heated at 473 K and 100 atm pressure. At equilibrium moles of NH_3 is 0.5 mol. Than calculate the equilibrium constant of the given reaction N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g))

Answer»

SOLUTION :`K_p=7.5xx10^(-6)`
Discussion
51836.

1 mol A_((g)) is heated in1 lit closed vessel and equilibrium is reached at 300^(@) C in A_((g)) hArr B_((g)) . If K_(C) =4 , concentration of B_((g)) at equilibrium is (in mol/lit)

Answer»

`0.2`
`0.6`
`0.8`
`0.1`

ANSWER :C
Discussion
51837.

1 ml of 13.6 M HCl is diluted with water to give 1 litre of the solution. Calculate pH of the resulting solution.

Answer»


ANSWER :`1.67`
Discussion
51838.

1 ml of 1 M solution is mixed with 999 ml of pure water.

Answer»

`10^(-3)` M solution is formed
The mass of SOLUTE per ml DECREASES by 1000 times
The quantity of solute DECREASE in the solution
10 ml of resultant solutoin contains `10^(-5)` moles of solute

Solution :`M_(1)V_(1)=M_(2)V_(2)`
`1xx1=M_(2)xx1000=M_(2)=10^(-3)M`
10 ml contains `(10)/(1000)xx10^(-3)` moles `=10^(-5)` moles Quantity of solute in solution remains same but mass of solute/ml decreases by `10^(3)` times
Discussion
51839.

1-Methylcyclopentene can be converted into the given compound by the use of which of the followign reagetns?

Answer»

`BD_(3)` FOLLOWED by `HCOOH`
`BH_(3)` followed HCOOD
`BD_(3)` followed by HCOOD
`BH_(3)` followed by `CH_(3)-overset(O)overset(||)(C)-O-D`

Answer :B::D
Discussion
51840.

1 litres of oxygen is allowed to react with three litres of carbon monoxide at N.T.P. Calculate the volume of each gas found after the reaction.

Answer»


Solution :The reaction takes place ACCORDING to the equation :
`underset(2L)(2CO)+underset(1L)(O_(2))rarrunderset(2L)(2CO_(2))`
1L of `O_(2)` REACTS with CO = 2L
`:.` VOLUME of CO left unreacted `= 3-2 = 1L`
Volume of `CO_(2)` formed = 2L.
Discussion
51841.

1 litre of N_(2) and 7/8 litre of O_(2) at the same temperature and pressure were mixed together. What is the relationship between the masses of the two gases in the mixture ?

Answer»

Solution :`PV=nRT=(w)/(M)RT`
For`N_(2),"" Pxx1=(w_(N_(2)))/(28)xxRT ""`……(i)
For`O_(2), "" Pxx(7)/(8)=(w_(N_(2)))/(32)xxRT ""`......(II)
Dividing eqn.(ii) by eqn. (i)
`(7)/(8)=(w_(O_(2)))/(32)XX(28)/(w_(N_(2)))`. This give `w_(N_(2))=w_(O_(2))`
Thus, the MIXTURE contains equal masses of `N_(2)` and `O_(2)`.
Discussion
51842.

1 lit of oxygen and 3 lit of SO_(2) at STP are reacted to produce sulphur trioxide . Then the ratio of between volume of sulphur trioxide and that of sulphur dioxide after reaction and weight of SO_3 formed (in grams) respectively are

Answer»

`1:2 , 7:14 gms`
`2:1 , 14.28 gms`
`2:1 , 7.14 gms`
`1:1 , 14.28 gms`

Solution :`2SO_(2)+O_(2)rarr2SO_(3)`
2 lit `LARR` 1 lit `larr` 2 lit
IMPLIES 1 lit `SO_(2)` left
`therefore (SO_(3))/(SO_(2))=(2)/(1)`
WEIGHT of `SO_(3)=(2)/(22.4)xx80=7.14g`
Discussion
51843.

1 lit of M/10 Ba(MnO_(4))_(2), in acidic medium can be oxidised completely with 1/6 lit of x M ferricoxalate. The value of x is

Answer»


Solution :(a) `10E^(-) +2MnO_4^(-) rarr2Mn^(2+)(n=5)`
Eq of `MnO_4^(-) -= Eq ` of `Fe_(2)(C_2O_4)_3`
(n=5) `""` (n=6)
`1LxxM/(10)xx10 -=1/6 L xx6 xx "x" , x = 1M `
Discussion
51844.

1 kg of a sample of water contained 222 mg of CaCl_2and 219 mg of Mg(HCO_3)_2 . The permanent and temporary hardness are .... ppm and .... ppm

Answer»

200, 200
200, 150
200, 300
150, 220

Answer :B
Discussion
51845.

1 kg of 2m urea solution is mixed with 2 kg of 4m urea solution. The molality of the resulting solution is

Answer»

`3.33m`
10M
1.67m
5m

Solution :`(m_(1)w_(1)+m_(2)w_(2))/(w_(1)+w_(2))=m` (APPROX)
Discussion
51846.

1-Hydroxypropane can be obtained from propene by treating with

Answer»

`B_(2)H_(6)+NaOH`
`B_(2)H_(6)+H_(2)O //OH^(-)`
`LiAlH_(4) +H_(2)O //H^(+)`
`HCl+H_(2)O //H^(+)`.

SOLUTION :`CH_(3)CH=CH_(2) underset((ii)H_(2)O_(2)//OH^(-))OVERSET((i)B_(2)H_(6))rarrCH_(3)CH_(2)CH_(2)OH`
Discussion
51847.

1 grams of a carbonate (M_(2)CO_(3))on tratment with excess HCl produces 0.01186 mole of CO_(2). The molar mass of M_(2)CO_(3)in gmol^(-1) is .....

Answer»

1186
84.3
118.6
11.86

Solution :`M_(2)CO_(3) + 2HCl rarr MCl_(2) + H_(2)O + CO_(2)`
`(1)/(M_(0))` MOLE `rarr ""0.01186 ` mole
where , `M_(0) = ` MOLECULAR mass of `M_(2)CO_(3)`
`(1)/(M_(0))= 0.01186`
`:.M_(0)=84.3` gram/mole
Discussion
51848.

1 gram of hydrogen contains6xx10^(23) atoms. Then 4 grams of He contains

Answer»

`6XX10^(23)` ATOMS
`12XX10^(23)` atoms
`24xx10^(23)` atoms
`1.5xx10^(23)` atoms

Solution :1 GM atom `H=N_(0)` atoms
1 gm atom `He=N_(0)` atoms
Discussion
51849.

1 gram hydrogen is present in 0.0833 mole carbohydrate. The empirical formula of carbohydrate is CH_(2)O. What will be the molecular formula ?

Answer»

`C_(5)H_(10)O_(5)`
`C_(3)H_(4)O_(3)`
`C_(12)H_(22)O_(11)`
`C_(6)H_(12)O_(6)`

Solution :In 0.0833 mole carbohydrate1 gm hydrogen is present .
`:.` In 1 mole `=(1xx1)/(0.0833)=12` gm hydrogen
The number of H atoms `= 12/1 = 12`
MOLECULAR formula `=C_(6)H_(12)O_(6)`
Discussion
51850.

1 gram of a carbonate (M_(2)CO_(3)) o treatment with excess HCl produces 0.01186 mole of CO_(2). The molar mass of M_(2)CO_(3) in g mol^(-1) is:-

Answer»

118.6
11.86
1186
84.3

Answer :D
Discussion