52671 + Interview Questions in Chemistry in Class 11

1.	Calculate the wavelength of the light required to break the bond between two chlorine atoms in a chlorine molecule. The Cl - Cl bond energy is 243 kJ mol^(-1) (h = 6.6 xx 10^(-34) Js, c = 3 xx 10^(8) ms^(-1), Avogadro's number = 6.02 xx 10^(23) mol^(-1))
Answer» `8.18 xx 10^(-31) m` `6.26 xx 10^(-21) m` `4.91 xx 10^(-7)m` `4.11 xx 10^(-6) m` Solution :Energy required to break one `Cl - Cl` bond `= ("Bond energy per mole")/("Avogado's No.") = (243 xx 10^(3) J)/(6.02 xx 10^(23))` `= 40.36 xx 10^(-20)J` `E =HV = h (c)/(LAMDA)` `:. lamda = (hc)/(E) = ((6.6 xx 10^(-34) Js) (3 xx 10^(8) ms^(-1)))/(40.36 xx 10^(-20) J)` `= 4.91 xx 10^(-7) m`

Discussion

2.	Calculate the wavelength of the spectral line in Lyman series corresponding to n_(2) = 3
Answer» Solution :For LYMAN series, `n_(1) = 1` `:. BAR(V) = R ((1)/(1^(2)) - (1)/(3^(2))) = 109677 xx (8)/(9) = 97490.7 cm^(-1)` `lamda = (1)/(bar(v)) = (1)/(97490.7 cm^(-1)) = 102.6 xx 10^(7) CMM = 102.6 nm`

Discussion

3.	Calculate the wavelength of the radiation emitted when an electron in a hydrogen atom undergoes a transition from 4th energy level to the 2nd energy level. In which part of the electromagnetic spectrum does this line lie ?
Answer» Solution :For hydrogen atom, `E_(N) - (21.8 xx 10^(-19))/(n^(2)) J "atom"^(-1)` Energy emitted when the electron jumps from `n = 4 " to " n = 2` will be given by `Delta E = E_(4) - E_(2) = 21.8 xx 10^(-19) ((1)/(2^(2)) - (1)/(4^(2))) = 21.8 xx 10^(-19) xx (3)/(16) = 4.0875 xx 10^(-19) J` The wavelength corresponding to this energy can be calculated using the EXPRESSION, `E = hv - h (c)/(lamda) ( :' c = V lamda)` so that `lamda = (hc)/(E) = ((6.626 xx 10^(-34) Js) (3 xx 10^(8) m s^(-1)))/((4.0875 xx 10^(-19) J)) = 4.863 xx 10^(-7) m = 4863 Å` (or 486.3 nm) It lies in the visible region

Discussion

4.	Calculate the wavelength of an electron that has been accelerated in a particle accelerator through a potential difference of 1 kv. [ " Given 1 eV"=1.6 xx 10^(-19)J]
Answer» ANSWER :`3.87xx10^(-7)`

Discussion

5.	Calculate the wavelength of an electron moving with velocity of 2.05xx10^(7) ms^(-1)
Answer» Solution :By de Broglie equation `lambda=h/(mv)=(6.626xx10^(-34) JS)/((9.11xx10^(-31) KG)(2.05xx10^(7) ms^(-1)))=3.55xx10^(-11)m (1 J=1 kg m^(2) s^(-2))`

Discussion

6.	Calculatethe wavelengthof the lightrequired to breakthe bondbetweentwo chlorine atomsin achlorinemolecule . The Cl-Cl andbondenergyis 243 kj mol^(-1) (n=6.6 xx 10^(34)J sC=3 xx 10^(8)ms^(-1) Avogadrons number=6.02xx 10^(23)mol^(-1) )
Answer» `8.18 XX 10^(31)m` `6.26 xx 10^(21) m` `4.91 xx 10^(-7) m` `4.14 xx 10^(6) m` Answer :C

Discussion

7.	Calculate the wavelength of an electron moving with a velocity of 2.05xx10^7ms^(-1)
Answer» Solution :Mass of ELECTRON `=9.1xx10^(-31)`KG `lambda=h/(MV)=(6.626xx10^(-31))/(9.1xx10^(-31)xx2.05xx10^7)=3.55xx10^(11)m`

Discussion

8.	Calculatethe wavelengthof anelectronmovingwith avelocityof 2.05 xx 10 ^(7)ms^(-1)
Answer» Solution :ACCORDINGTO de - BROGLIE `lambda = (h)/( mv )` `h= 6.626 xx 10^(34)JS` `v= 2.05 xx 10^(7)ms^(-1)` `m=9.1 xx 10^(31) KG` `=(6.626 xx 10^(34) js)/( (9.1xx 30^(-31) kg )(2.05 xx 10^(7)ms^(-1)))` `=0.3552 xx 10^(10) J s^(2) kg^(-1) m^(-1)` `lambda= 3.552 xx 10^(11)`m= 35.52 pm

Discussion

9.	Calculate the wavelength of an electron moving at 3.0 xx 10^(10) cm sec^(-1) (mass of the electron = 9.11 xx 10^(-31) kg, h = 6.6 xx 10^(-34) kg m^(2) sec^(-1))
Answer» SOLUTION :`2.41 XX 10^(-12) m`

Discussion

10.	Calculate the wavelength of a tennis ball of mass 60 gm moving with a velocity of 10 m per second.(h=6.626 xx 10^(-34) kg m^(2) s^(-1))
Answer» ANSWER :`10^(-3) ` METRE

Discussion

11.	Calculate the wavelength of a track star running 150 metre dash in 12.1 sec if its weight is 50 kg.
Answer» `9.11 xx 10^-34 m` `8.92 xx 10^-37 m` `1.12 xx 10^-45 m` NONE of these Solution :(B) `lamda = h //MV = (6.626 xx 10^-24)/(50 xx (150 //12)) = 8.92 xx 10^-37 m`.

Discussion

12.	Calculate the wavelength of a photon in Angstrom units having enery of one electron volt.
Answer» SOLUTION :`E = 1 eV = 1.609 xx 10^(-19) J` `E = hv = h (C)/(lamda)` or `lamda = (hc)/(E) = ((6.62 xx 10^(34) Js) (3 xx 10^(8) ms^(-1)))/(1.602 xx 10^(-19) J) = 12.40 xx 10^(-7) m = 12.40 xx 10^(3) Å`

Discussion

13.	Calculate the wavelength of a photon in Angstrom units having energy of one electron volt
Answer» Solution :Energy (E) = 1EV = `1.602 XX 10^(-19)J` `h = 6.626 xx 10^(-34) Js, c = 3 xx 10^(8) ms^(-1)` `E = hv = (hc)/(LAMDA)` or `lamda = (hc)/(E) = ((6.626 xx 10^(-34) Js) (3 xx 10^(8) ms^(-1)))/(1.602 xx 10^(-19) J) = 12.42 xx 10^(-7) m` `= 12.42 xx 10^(-7) xx 10^(10) Å = 12.42 xx 10^(3) Å`

Discussion

14.	Calculate the wavelength of a CO_(2) molecule moving with a velocity of 440 m sec^(-1)
Answer» ANSWER :`2.06xx10^(-11)`METRE

Discussion

15.	Calculate the wavelength of 1000 kg rocket moving with a velocity of 3000 km/hr. (h=6.626 xx 10^(-34) kg m^(2)s^(-1))
Answer» ANSWER :`7.9512 XX 10^(-40) m`

Discussion

16.	Calculate the wavelength o fan electron moving with a velocity of 2.05xx10^(7)ms^(-1).
Answer» Solution :According to de Broglie.s equation,`gamma=H/(mv)` Mass of electron(m)`=9.1xx10^(-31)kg` velocity F electron(v)`=2.05xx10^(7)ms^(-1)` Planck.s CONSTANT(h)`=6.626xx10^(-34)kgm^(2)s^(-1)` `gamma((6.626xx10^(-34)Kgm^(2)s^(-1)))/((9.1xx10^(-31)Kg)XX(2.05xx10^(7)ms^(-1)))=3.55xx10^(-4)m`

Discussion

17.	Calculate the wavelength (in nanometer) associated with a proton moving at 1.0 xx 10^(3) ms^(-1)
Answer» 0.032 nm 0.40 nm 2.5 nm 14.0 nm Solution :`lamda = (h)/(MV) = (6.62 XX 10^(-34) kg m^(2) s^(-1))/((1.67 xx 10^(-27) kg) (10^(3) MS^(-1)))` `= 3.96 xx 10^(-10) m = 0.40 nm`

Discussion

18.	calculatethe wavelength(innanometer )associated with aelectronmovingat 1.0 xx 10^(3)ms^(-1) (m= 1.656 xx 10^(27))
Answer» 0.40 NM 14. nm 2.5 nm 0.32 nm ANSWER :A

Discussion

19.	Calculate the wavelength from the Balmer formula when n_2 = 3
Answer» SOLUTION :`BAR(V) = R ((1)/(2^(2)) - (1)/(n_(2)^(2))) = 109677 ((1)/(2^(2)) - (1)/(3^(2))) cm^(-1) = 15232.9 cm^(-1)`

Discussion

20.	Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0 xx 10^(-10)s
Answer» Solution :Frequency `(v) =(1)/("Period") = (1)/(2.0xx 10^(-10)s) = 5 xx 10^(9) s^(-1)` WAVELENGTH, `lamda = (c)/(v) = (3.0 xx 10^(8) MS^(-1))/(5 xx 10^(9) s^(-1)) = 6.0 xx 10^(-2) m` WAVE number, `bar(v) = (1)/(lamda) = (1)/(6 xx 10^(-2)m) = 16.66 m^(-1)`

Discussion

21.	Calculatethe wavelengthfrequencyand wavenumberof a lightwavewhoseperiodis 2.0 xx 10^(10)s
Answer» SOLUTION :Frequency(V ) = `(1)/( "Periodic ") = (1)/(2.0xx 10^(10)s)` wavelength`(lambda)= (c )/(v ) = (3.0 XX 10^(8)ms^(-1))/(5.0 xx 10^(9) s^(-1))` Wavenumber`(v ) = (1)/(lambda) = (1)/( 6.0 xx 10^(2) m) ` `=16 .67m^(-1)`

Discussion

22.	Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.
Answer» Solution :Radius of nth orbit of H-like particles `= (0.529 n^(2))/(Z) Å = (52.9 n^(2))/(Z)` pm `r_(1) = 1.3225 nm = 1322.5 " pm" = (52.9 n_(1)^(2))/(Z)` `r_(2) = 211.6 " pm" = (52.9 n_(2)^(2))/(Z) :. (r_(1))/(R_(2)) = (1322.5)/(211.6) = (n_(1)^(2))/(n_(2)^(2)) or (n_(1)^(2))/(n_(2)^(2)) = 6.25 or (n_(1))/(n_(2)) = 2.5` `:.` If `n_(1) = 2, n_(1) = 5`. Thus, the transition is from 5th orbit to 2nd orbit. it belongs to BALMER series. `bar(v) = 1.097 xx 10^(7) m^(-1) ((1)/(2^(2)) - (1)/(5^(2))) = 1.097 xx (21)/(100) xx 10^(7) m^(-1)` or `lamda = (1)/(v) = (100)/(1.097 xx 21 xx 10^(7)) m = 434 xx 10^(-9) m = 434 nm` It lies in the visible region

Discussion

23.	Calculate the wavelength associated with an electron (mass 9.1 xx 10^(-31)kg) moving with a velocity of 10^(3) m sec^(-1) (h = 6.6 xx 10^(-34) kg m^(2) sec^(-1))
Answer» Solution :Here, we are given: `m = 9.1 xx 10^(-31) kg, V = 10^(3) m sec^(-1), h = 6.6 xx 10^(-34) kg m^(2) s^(-1)` `lamda = (h)/(mv) = (6.6 xx 10^(-34) kg m^(2) s^(-1))/((9.1 xx 10^(-31) kg) xx (10^(3) MS^(-1))) = 7.25 xx 10^(-7) m`

Discussion

24.	Calculate the wavelength and wave numbers of the first and second lines in tbe Balmer series of hydrogen spectrum. Given R = 1.096xx10^(7) m^(-1)
Answer» Solution :For Balmer series `n_(1) = 2`, for FIRST LINE=`n_(2)= 3`, for second line`= n_(2) = 4` i. Wave number `barv` of the first line is GIVEN by `bar(v_(3to2))=R[1/(n_(1)^(2))-1/(n_(2)^(2))]=1.096xx10^(7)xx(1/(2^(2))-1/(3^(2)))` `=1.096xx10^(7)xx(1/4-1/9)=1.096xx10^(7)xx5/36=1.5236xx106m^(-1)` `lamda=36/(1.0967xx10^(7)xx5)m=656.3xx10^(-9)m=656.3xx10^(-9)m=656.3nm` Wavelength of `H_(alpha)` line = 656.3 nm For `H_(beta)` line `n_(2//=4).R=10.97xx10^(6)(1/(2^(2))-1/(4^(2)))m^(-1)` `=(10.97xx10^(6)xx3)/16m^(-1)=2.0568xx10^(6)m` `lamda=16/(10.97xx10^(6)xx3)m`

Discussion

25.	Calculate the wave number of the spectral line of shortest wave length appearing in Given R=1.09xx10^(7)m^(-1)
Answer» SOLUTION :`barv=R(1/(n_(1)^(2))-1/(n_(2)^(2)))m^(-1)` `barv=1.09xx10^(7)m^(-1)(1/(2^(2))-1/0)` `barv=1.09xx10^(7)(1/4)` `barv=0.2725xx10^(7) barv=2.725xx10^(6)m^(-1)`

Discussion

26.	Calculate the wave number of radiations having a frequency of 4 xx 10^(14) Hz
Answer» Solution :`v = 4 xx 10^(14) HZ` (i.e. `s^(-1)`) WAVE number `bar(v) = (1)/(lamda) = (v)/(c) = (4 xx 10^(14) s^(-1))/(3 xx 10^(8) MS^(-1)) = 1.33 xx 10^(6) m^(-1)" or " = (4 xx 10^(14) s^(-1))/(3 xx 10^(10) cm s^(-1)) = 1.33 xx 10^(4) cm^(-1)`

Discussion

27.	Calculate the wave number of the first spectral line in the Lyman series of He^(+) spectrum
Answer» SOLUTION :For first LINE in Lyman series, `n_(1) =" lower energy "= 1` `n_(2)= "higher energy "= 2` Wave number `(barv)` is given by the equation `barv=z^(2)R_(H)((1)/(1^(2))-(1)/(2^(2)))=4xx109677xx(3)/(4)=329031cm^(-1)` Wave number of first line in Lyman series of `He^(+)` is `3.29xx10^(5)cm^(-1)`

Discussion

28.	Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.
Answer» Solution :For Balmer SERIES, `n_(1)=2`. Hence, `BAR(v)=R(1/2^(2)-1/n_(2)^(2))` `bar(v)=1/lambda`. For `lambda` to be longest (maximum), `bar(v)` should be minimum. This can be so when `n_(2)` is minimum, i.e., `n_(2)=3`. Hence, `bar(v)=(1.097xx10^(7) m^(-1)) (1/2^(2)-1/3^(2))=1.097xx10^(7)xx5/36 m^(-1) =1.523xx10^(6) m^(-1)`

Discussion

29.	Calculate the wave number for thelongest wavelength transition in the Balmer series of atomic hydrogen.
Answer» Solution :For BALMER series, `bar(v) = 109, 677 ((1)/(2^(2)) - (1)/(n^(2))) cm` `lamda` will be maximum or `bar(v)` is minimum `(bar(v) = (1)/(lamda)) " for " n = 3`. The value will be `bar(v) = 109, 677 cm^(-1) ((1)/(2^(2)) - (1)/(3^(2))) = 109, 677 XX (5)/(36) cm^(-1) = 15232.9 cm^(-1) = 1.523 xx 10^(6) m^(-1)`

Discussion

30.	Calculate the wave number for the longest wavelength transition inthe Balmer series of atomic hydrogen.
Answer» ANSWER :`1.523 XX 10^(6) m^(-1)`

Discussion

31.	Calculate thewavenumberfor thelongestwavelengthtransitionin the Balmerseries of atomichydrogen
Answer» SOLUTION :ForBalmerseries `n_(i)= 2 ` and ` n_(2)= 3 ,4,5 ` forlongestwavelengthtransitionthe isleast Accordingto balmer `VEC( v) =109677 ((1)/(n_(1)^(2) )-(1)/(n_(2)^(2))) cm^(-1)` `=10967 ((9-4)/(36))` `=109677 ((5)/(36))` `=15232.9 cm^(-1)` `=1.52329 xx 10^(4) cm^(-1)` `=1.52329 xx 10^(6)m^(-1)= 1.523 xx 10^(6) m^(-1)`

Discussion

32.	Calculate the wave number, wavelength and frequency of the first line in the Baliner series.
Answer» Solution :`barv_(5to2)=R[1/(n_(1)^(2))-1/(n_(2)^(2))]m^(-1)=10.97xx10^(6)(1/(1^(2))-1/(2^(2)))m^(-1)=(10.97xx10^(6)xx3)/4m^(-1)` `8.2275xx10^(6)m^(-1)` Wave NUMBER `BARV=8.2275xx10^(6)m^(-1)` Wavelength `lamda=1/(8.2272xx10^(6)m)=121.5xx10^(-9)m=121.5nm`.

Discussion

33.	Calculate the wave number and wavelength of the first spectral line of Lyman series of hydrogen spectrum. Rydberg constant R' = 10.97xx10^(6) m^(-1).
Answer» SOLUTION :`BAR(v_(5to2))=R[1/(n_(1)^(2))-1/(n_(2)^(2))]m^(-1)=10.97xx10^(6)(1/(1^(2))-1/(2^(2)))m^(-1)=(10.97xx10^(6)xx3)/4m^(-1)` `8.2275xx10^(6)m^(-1)` WAVE number `barv=8.2275xx10^(6)m^(-1)`. WAVELENGTH `lamda=1/(8.2275xx10^(6))m=121.5xx10^(-9)m=121.5nm`

Discussion

34.	Calculate the wave number and wave length of H_(alpha) line in the Balmer series of hydroge emission spectrum.
Answer» Solution :For `H_(a)` line in the Balmer series `n_(1)=2 and n_(2)=3` `BARV=109677((1)/(2^(2))-(1)/(3^(2)))=15,232.9cm^(-1)` `lambda=(1)/(uspilon)=(1)/(15,232.9)=6.564xx10^(-5)cm` `lambda=(1)/(barv)=(1)/(15,232.9)=6.564xx10^(-5)cm` `"Wave number of "H_(ALPHA)" line" = 1.5239 XX 10^(4)cm^(-1)` `"Wave length of "H_(alpha)" line "= 6564 Å`

Discussion

35.	Calculate the wave length of an electron of mass 9.1 xx 10^(-31) kg, moving with a velocity of 2.05 xx 10^(7)ms^(-1).
Answer» SOLUTION :Mass of electron, `m = 9.1 xx 10^(-31) kg` v = velocity of electron `= 2.05xx 10^(7) ms^(-1)` h = Plancks constant `= 6.6256 xx 10^(-34) Js.` `lambda =` wave length = ? According to de BROGLIE equation `lambda=(h)/(MV)=(6.625xx10^(-34)kgm^(2)s^(-1))/(9.1xx10^(-31)kgxx2.05xx10^(7)ms^(-1))` The wave length of electron is `3.55 xx 10^(-11) m`

Discussion

36.	Calculate the volumes of gases after the reaction and also the weight of CO_(2) formed when 1 litre of oxygen reacts with 3 times of carbon monoxide measured at STP.
Answer» ANSWER :3.92gm. `CO_(2)` 1 LITRE CO

Discussion

37.	Calculate the volume strength of H_(2)O_(2) solution, which normality is 3.57
Answer» 10V 20V 30V 40V Answer :B

Discussion

38.	Calculate the volume strength of a 3% solution of H_(2)O_(2)
Answer» Solution :Step 1. To calculate the amount of `H_(2)O_(2)` present in one litre of 3% solution. 100 mL of `H_(2)O_(2)` solution contain `H_(2)O_(2)` =3g `THEREFORE ` 1000 mL of `H_(2)O_(2)` solution will contain `H_(2)O_(2)=(3)/(100)xx1000=30g` Step 2. To calculate the volume strength Consider the chemical equation, `underset(2xx34=68 g)(2H_(2)O_(2))to 2H_(2)O + underset(22.4 "LITRES at N.T.P.")(O_(2))` Now 68 g of `H_(2)O_(2)` give `O_(2)` at N.T.P. =22.4 litres `therefore `30 g of `H_(2)O_(2)` will give `O_(2)` at N.T.P. `=(22.4)/(68)xx30=9.88` litres=9880 mL. But 30 g of `H_(2)O_(2)` are present in 1000 mL of `H_(2)O_(2)`. Hence, 1000 mL of `H_(2)O_(2)` solution gives `O_(2)` at N.T.P. 9880 mL `therefore ` 1 mL of `H_(2)O_(2)` solution will give `O_(2)` at N.T.P`=(9880)/(1000)=9.88` mL. Hence the volume strength of 3% `H_(2)O_(2)` solution =9.88

Discussion

39.	Calculate the volume strength of 1.7% H_(2)O_(2) solution.
Answer» ANSWER :5.6Vol

Discussion

40.	Calculate the volume strength of 13.6% solution of H_(2)O_(2)
Answer» SOLUTION :According to problem 100 ml of`H_(2) O_(2) ` solutioncontains`13.6 g H_(2)O_(2) `THEREFORE `1000 ml (1 lit)of `H_(2)O_(2)` solution contains 136g `H_(2)O_(2)` We knowthat `68g H_(2) O_(2)`given22.4 lit of`O_(2)`at NTP `136g H_(2) O_(2)` GIVEN `(22.4)/(68)xx136` litre iof`O_(2)` `=44.8 liter of `O_(2)`lt brgt Thus 1 liter of the solutiongiven44.8 lit of O_(2) at NTP `therefore ` Volumestrenght of the`13.6%` solution of `H_(2)O_(2)`=44.8.

Discussion

41.	Calculate the volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution. (K_(sp)of PbCl_(2) = 3.2 xx 10^(-8), atomic mass of Pb = 207 u).
Answer» Solution :Suppose the solubility of `PbCl_(2)` in water is S MOL `L^(-1) `. Then `{:(PbCl_(2)(s)hArrPb^(2+)(aq)+2Cl^(-)(aq)),(""S"" 2S):}` `K_(sp)=(S)(2S)^(2)=4S^(3) = 3.2 xx 10^(-8) ` (Given) or `S^(3) = 0.8 xx 10^(-8) = 8xx10^(-9) or S = 2 xx 10^(-3) "mol" L^(-1)` Molar MASS of`PbCl_(2) = 207 + 71 = 278` `:. ` Solubility of `PbCl_(2)` dissolve in 1 L of water to makea saturated solution. HENCE, water required to dissolve 0.1 g of `PbCl_(2) = (1L)/(0.556g) xx 0.1 g = 0.1798 L = 179.8 mL `.

Discussion

42.	Calculate the volume of oxygen required for the complete combustion of 60mL acetylene. What voume of carbondioxide is obtained in the combustion?
Answer» ANSWER :150ML `O_(2)`; 120ML `CO_(2)`

Discussion

43.	Calculate the volume of oxygen gas required at STP conditions for the complete combustion of 10 cc of methane gas at 20^(@)C and 770mm pressure.
Answer» ANSWER :9.44cc

Discussion

44.	Calculate the volume of oxygen at N.T.P that would be required to convert 5 litres of carbon monoxide into carbon dioxide.
Answer» Solution :Chemicalm EQUATION for the REACTION is : `UNDERSET(2L)(2CO(g))+underset(1L)(O_(2)(g))rarr2CO_(2)(g)` 2 of CO(g) require `O_(2)(g)` at N.T.P = 1L 5L of CO(g) require `O_(2)(g)` at N.T.P `= (5)/(2)=2.5 L`.

Discussion

45.	Calculate the volume of O_(2) gas, when temperature increases 25^(@)C to 50^(@)C ?
Answer» HALF DOUBLE More than double LESS than double ANSWER :D

Discussion

46.	Calculate the volume of O_(2) at STP required to burn completely 70 ml of acetylene.
Answer» Solution :Combustion of acetylene is GIVEN as `2C_(2)H_(2)+5O_(2) to 4CO_(2)+2H_(2)O` 2 moles of `C_(2)H_(5)-="5 moles of "O_(2)` 2 VOLUME of `C_(2)H_(2)="5 volumes of "O_(2)` 2mL of `C_(2)H_(5)="5mL of "O_(2)` Volume of oxygen required to BURN 70mL of acetylene `=70 XX 5/2=175mL`

Discussion

47.	Calculate the volume of H_(2) liberated at 27^(@)C and 760 mm of Hg pressure by action by 0.6 g magnesium with excess of dil HCl.
Answer» ANSWER :0.615L

Discussion

48.	Calculate the volume of CO_(2) obtained at STP when 2.12 g of Na_(2)CO_(3) reacts with excess of dil HCI.
Answer» SOLUTION :0.448 LIT

Discussion

49.	Calculate the volume of carbon dioxide at N.T.P evolved by strong heating of 20 g of calcium carbonate.
Answer» Solution :The balanced chemical equation is : `UNDERSET(underset(100G)("1 mol"))(CaCO_(3))rarrunderset(underset("22.4 L at N.T.P")("1 mol"))(CaO+CO_(2))` 100 g of `CaCO_(3)` EVOLVE carbon dioxide = 22.4 L 20 g of `CaCO_(3)` evolve carbon dioxide `= ((22.4 L))/((100g))xx(20G)=4.48L`.

Discussion

50.	Calculate the volume of air measured at NTP used for the roasting of 3kg of iron pyrites FeS_(2).Assume that air contains 21% O_(2)by volume.
Answer» ANSWER :7.33L

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Calculate the wavelength of the light required to break the bond between two chlorine atoms in a chlorine molecule. The Cl - Cl bond energy is 243 kJ mol^(-1) (h = 6.6 xx 10^(-34) Js, c = 3 xx 10^(8) ms^(-1), Avogadro's number = 6.02 xx 10^(23) mol^(-1))

Calculate the wavelength of the spectral line in Lyman series corresponding to n_(2) = 3

Calculate the wavelength of the radiation emitted when an electron in a hydrogen atom undergoes a transition from 4th energy level to the 2nd energy level. In which part of the electromagnetic spectrum does this line lie ?

Calculate the wavelength of an electron that has been accelerated in a particle accelerator through a potential difference of 1 kv. [ " Given 1 eV"=1.6 xx 10^(-19)J]

Calculate the wavelength of an electron moving with velocity of 2.05xx10^(7) ms^(-1)

Calculatethe wavelengthof the lightrequired to breakthe bondbetweentwo chlorine atomsin achlorinemolecule . The Cl-Cl andbondenergyis 243 kj mol^(-1) (n=6.6 xx 10^(34)J sC=3 xx 10^(8)ms^(-1) Avogadrons number=6.02xx 10^(23)mol^(-1) )

Calculate the wavelength of an electron moving with a velocity of 2.05xx10^7ms^(-1)

Calculatethe wavelengthof anelectronmovingwith avelocityof 2.05 xx 10 ^(7)ms^(-1)

Calculate the wavelength of an electron moving at 3.0 xx 10^(10) cm sec^(-1) (mass of the electron = 9.11 xx 10^(-31) kg, h = 6.6 xx 10^(-34) kg m^(2) sec^(-1))

Calculate the wavelength of a tennis ball of mass 60 gm moving with a velocity of 10 m per second.(h=6.626 xx 10^(-34) kg m^(2) s^(-1))

Calculate the wavelength of a track star running 150 metre dash in 12.1 sec if its weight is 50 kg.

Calculate the wavelength of a photon in Angstrom units having enery of one electron volt.

Calculate the wavelength of a photon in Angstrom units having energy of one electron volt

Calculate the wavelength of a CO_(2) molecule moving with a velocity of 440 m sec^(-1)

Calculate the wavelength of 1000 kg rocket moving with a velocity of 3000 km/hr. (h=6.626 xx 10^(-34) kg m^(2)s^(-1))

Calculate the wavelength o fan electron moving with a velocity of 2.05xx10^(7)ms^(-1).

Calculate the wavelength (in nanometer) associated with a proton moving at 1.0 xx 10^(3) ms^(-1)

calculatethe wavelength(innanometer )associated with aelectronmovingat 1.0 xx 10^(3)ms^(-1) (m= 1.656 xx 10^(27))

Calculate the wavelength from the Balmer formula when n_2 = 3

Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0 xx 10^(-10)s

Calculatethe wavelengthfrequencyand wavenumberof a lightwavewhoseperiodis 2.0 xx 10^(10)s

Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.

Calculate the wavelength associated with an electron (mass 9.1 xx 10^(-31)kg) moving with a velocity of 10^(3) m sec^(-1) (h = 6.6 xx 10^(-34) kg m^(2) sec^(-1))

Calculate the wavelength and wave numbers of the first and second lines in tbe Balmer series of hydrogen spectrum. Given R = 1.096xx10^(7) m^(-1)

Calculate the wave number of the spectral line of shortest wave length appearing in Given R=1.09xx10^(7)m^(-1)

Calculate the wave number of radiations having a frequency of 4 xx 10^(14) Hz

Calculate the wave number of the first spectral line in the Lyman series of He^(+) spectrum

Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

Calculate the wave number for thelongest wavelength transition in the Balmer series of atomic hydrogen.

Calculate the wave number for the longest wavelength transition inthe Balmer series of atomic hydrogen.

Calculate thewavenumberfor thelongestwavelengthtransitionin the Balmerseries of atomichydrogen

Calculate the wave number, wavelength and frequency of the first line in the Baliner series.

Calculate the wave number and wavelength of the first spectral line of Lyman series of hydrogen spectrum. Rydberg constant R' = 10.97xx10^(6) m^(-1).

Calculate the wave number and wave length of H_(alpha) line in the Balmer series of hydroge emission spectrum.

Calculate the wave length of an electron of mass 9.1 xx 10^(-31) kg, moving with a velocity of 2.05 xx 10^(7)ms^(-1).

Calculate the volumes of gases after the reaction and also the weight of CO_(2) formed when 1 litre of oxygen reacts with 3 times of carbon monoxide measured at STP.

Calculate the volume strength of H_(2)O_(2) solution, which normality is 3.57

Calculate the volume strength of a 3% solution of H_(2)O_(2)

Calculate the volume strength of 1.7% H_(2)O_(2) solution.

Calculate the volume strength of 13.6% solution of H_(2)O_(2)

Calculate the volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution. (K_(sp)of PbCl_(2) = 3.2 xx 10^(-8), atomic mass of Pb = 207 u).

Calculate the volume of oxygen required for the complete combustion of 60mL acetylene. What voume of carbondioxide is obtained in the combustion?

Calculate the volume of oxygen gas required at STP conditions for the complete combustion of 10 cc of methane gas at 20^(@)C and 770mm pressure.

Calculate the volume of oxygen at N.T.P that would be required to convert 5 litres of carbon monoxide into carbon dioxide.

Calculate the volume of O_(2) gas, when temperature increases 25^(@)C to 50^(@)C ?

Calculate the volume of O_(2) at STP required to burn completely 70 ml of acetylene.

Calculate the volume of H_(2) liberated at 27^(@)C and 760 mm of Hg pressure by action by 0.6 g magnesium with excess of dil HCl.

Calculate the volume of CO_(2) obtained at STP when 2.12 g of Na_(2)CO_(3) reacts with excess of dil HCI.

Calculate the volume of carbon dioxide at N.T.P evolved by strong heating of 20 g of calcium carbonate.

Calculate the volume of air measured at NTP used for the roasting of 3kg of iron pyrites FeS_(2).Assume that air contains 21% O_(2)by volume.