52671 + Interview Questions in Chemistry in Class 11

1.	At 25^(0)C, the hydroxyl ion concentration of a basic solution is 6.75 xx 10^(-3) M.Then the value of K_(w) is
Answer» ` 13.5 xx 10^(-6) M^(2) ` ` 13.5 xx 10 ^(-12)M^(2) ` ` 13.5 xx 10 ^(-8)M^(2) ` `10 ^(-14) M^(2) ` Solution :At ` 25 ^(@) C , K_W = 1xx 10 ^(-14) M^(2) ` It is temperature dependant only.

Discussion

2.	At 25^@ C, the dissociation constant of NH_4 OH is 2 xx 10^(-5). If the molar concentration of NH_2 OH is 0.2 M, calculate the proton concentration of aqueous ammonia solution.
Answer» SOLUTION : `K_b=` Dissociation constant of BASE `= 2 XX 10^(-5) C = `Concentration of base `= 2xx 10^(-1)` `[OH^(-) ]= sqrt(K_b C) =sqrt(2 xx 10^(-5) xx 2 xx 10^(-1) ) = 2 xx 10^(-3) M` `[H^(+) ]=K_(w) // [OH^(-) = 1 xx 10^(-14) // 2 xx 10^(-3) = 5xx 10^(-12) M`

Discussion

3.	At 25^@C, 500 cm^3 flask contains hydrogen at a pressure of 120 mm of Hg and a 250cm^3 flask contains oxgyen at a pressure of 300 mm of Hg at the same temperature. If the two gases are transferred into a one litre flask at the same temperature, find the pressure of the mixture.
Answer» SOLUTION :`P=135 CM`

Discussion

4.	At 227^(@) C & 4 atm , PCl_(5) is dissociated to an extent 50% . At same temperature , extent of dissociation of PCl_(5) is 0.75 at a pressure of [Hint : alpha prop (1)/(sqrtP)]
Answer» 1.77 ATM 2 atm 4 atm 8 atm ANSWER :A

Discussion

5.	At 21^(@)C temperature volume of 212 g O_(2) gas is 34 dm^(3) . If pressure of gas becomes 1.24 bar then how many gram O_(2) gas is left from container ? (R=0.083 dm^(3)" bar " K^(-1)mol^(-1))[Note : Calculate intital moles by pV = nRT]
Answer» ANSWER :156.74 G

Discussion

6.	At 20^(@)C two balloons of equal volume and porosity are filled to a pressure of 2 atm one with 14kgN_(2) and other with 1kg of h_(2) The N_(2) balloon leaks to a pressure of 1//2 atm in 1hr How long will it take for H_(2) balloon to reach a pressure of 1//2atm ? .
Answer» ANSWER :16 MINUTE

Discussion

7.	At 20^@Cthe electrical conductivity of water is 5.7 xx 10^(-8) ohm^(-1)cm^(-1). At the same temperature, the electrical conductivity of heavy water in ohm^(-1) cm^(-1)is
Answer» `5.5 XX 10^(-8)` `5.9 xx 10^(-8)` `5.7 xx 10^(-8)` `6.6 xx 10^(-8)` ANSWER :D

Discussion

8.	At 200^(@)C the velocity of hydrogen molecule is 2.0 xx 10^(5) cm/sec. In this case, the de-Broglie wavelength is about ________Å.
Answer» ANSWER :1

Discussion

9.	At 200^(@) C PCl_(5) dissociates as follow , PCl_(s(g)) hArr PCl_(3(g)) + Cl_(2(g)) . It was found that the equilibrium vapour. are 62 times as heavy as hydrogen. The degree of dissociation of PCl_(5) at 200^(@) C is nearly.
Answer» 0.1 0.42 0.5 0.68 Solution :`ALPHA=(M-m)/(m(n-1))` `m=2 XX 62=124 IMPLIES alpha=(208.5 -124)/(124)=0.68` `1-alpha=68%`

Discussion

10.	At 18^(@)C aniline and acetic acid have dissociation constants 5xx10^(-10) and 1.8xx10^(-5) respectively. An aqueous solution of anilium acetate is hydrolysed to the extent of x% under equilibrium, what is pH of the solution ? (K_(w)=10^(-14))
Answer» ANSWER :`4.7219;`

Discussion

11.	At 17^(@)C temperature volume of gas is 400 mL. Then, at which temperature (i) Volume becomes double (ii) Volume becomes half.
Answer» ANSWER :(i) `580K = 370^(@)C`(ii) `145 K =-128^(@)C`

Discussion

12.	At 1300K, 0.3 mol of CO ,0.1 mol H_2, 0.02 mol of H_2O and an unknown amount of CH_4 are present at equilibrium in a vessel. K_c for the reaction CO_(g)+H_2(g) leftrightarrow CH_4(g)+H_2O(g) is 3.9. What is the equilibrium concentration CH_4
Answer» SOLUTION :`0.0585 MOL`

Discussion

13.	At 127^(@)C temperature volume of gas is 3 L. If volume of gas becomes half then calculate temperature. Pressure is 1 bar.
Answer» ANSWER :200 K OR `- 73^(@)C`

Discussion

14.	At 127^(@)C, K_(c)" for "2A Leftrightarrow 2B+C" is " 2xx10^(-4). Calculate DeltaG^(@) for the reaction.
Answer» ANSWER :`28.2 KJmol^(-1)`

Discussion

15.	At 127°C and latm pressure, a mixture of a gas contains 0.3 mole of N_(2), 0.2 mole of O_(2) The volume of the mixture is
Answer» 15 LIT 22.4lit 18.2lit 16.4lit ANSWER :D

Discussion

16.	At 127^@C, for helium if time of flight is 0.1 nanosec, the mean free path of helium (in A) is
Answer» 15.8 158 1580 158000 SOLUTION :`lambda = C xx t = SQRT((3RT)/(M)) xx t = 1580 xx 10^(-10) m`.

Discussion

17.	At 1127K and 1 atm pressure, a gaseous mixture of CO and CO_(2) in equilibrium with soild carbon has 90.55% CO by mass C(s)+CO_(2)(g)hArr2CO(g) Calculate K_(c) for this reaction at the above temperature.
Answer» SOLUTION :`0.153 molL^-1`

Discussion

18.	At 1100 K temperature CaCO_3 and CaO_((s)) are in equilibrium pressure of CO_2 is 2.0 xx 10^5 Pa. Find equilibrium constant. CaCO_(3(s)) hArr CaO_((s)) + CO_(2(g))
Answer» SOLUTION :`K_p=p_(CO_2)=2xx10^5` PA = 2 BAR

Discussion

19.	At 1127 K and 1 atm pressure , a gaseous mixture of CO and CO_(2)in equilibrium with solid carbon has 90*55 % CO by mass in the reaction, C(s) + CO_(2) (g) hArr 2 CO (g). Calculate k_(c)for the reaction at the above temperature.
Answer» Solution :If total mass of the mixture of `CO and CO_(2)`is 100 g , then ` :."At eqm., " [BrCl] = ( 330 xx 10^(-3) - 30 xx 10^(-3))= 030 xx 10^(-3) = 30 xx 10^(-4) "mol"L^(-1)` Number of moles of ` :."At eqm., " [BrCl] = ( 330 xx 10^(-3) - 30 xx 10^(-3))= 030 xx 10^(-3) = 30 xx 10^(-4) "mol"L^(-1)` ` :. p_(CO)= (3234)/(3234 + 0215 ) xx 1" atm " = 0938 " atm", p_(CO_(2)) = (0215)/(3234 + 0215 ) xx 1 "atm " = 0062 "atm "` ` k_(p) = (p_(CO)^(2))/p_(CO_(2)) = (0938 )^(2)/(0062 )= 14 19` . ` Delta n_(g)= 2-1 =1 :.K_(p) = K_(c) (RT) or K_(c)= (K_(p))/(RT) = (1419)/(00821 xx 1127)=0153 ` .

Discussion

20.	At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO_2 in equilibrium with solid carbon has 90.55% CO by mass. C_((s)) + CO_(2(g)) hArr 2CO_((g)) Calculate K_c for this reaction at the above temperature.
Answer» SOLUTION :Suppose at equilibrium the gaseous mixture is 100 gm, so 90.55% CO in it. 90.55 g CO and (100-90.55) = 9.45 g `CO_2` `{:("Equilibrium reaction :", C_((s))+CO_(2(g)) hArr, 2CO_((g))),("Equilibrium in mass (gm):", 9.45,90.55),("(Mole at equilibrium = Mass/Molecular mass): ",(9.45)/44,(90.55)/28),("(Mole fraction at equilibrium =Mole/Total mole):", "0.2148"/"3.4448","3.234"/"3.4448"):}` Total mole = 0.2148 + 3.234 = 3.4488 `APPROX` 3.45 Partial pressure = Total pressure X Mole fraction `therefore p_(CO_2)` = 1 atm x 0.0624 =0.0624 atm `p_(CO)`=1 atm x 0.9388 =0.9388 atm `K_p=(p_(CO))^2/p_(CO_2)=(0.9388)^2/(0.0624)=14.12` atm `K_p=K_c(RT)^((Deltan(g))` where, `Deltan_((g))` = (2-1) =1 , T= 1127 K , R= 0.0821 atm L `mol^(-1) K^(-1)` `therefore K_c=K_p/(RT)^(Deltan(g))` `=14.12/(0.0821xx1127)^1`=0.1526

Discussion

21.	At 125^@C the K_p for the reaction N_2(g)+3H_2(g) leftrightarrow 2NH_3(g) is 2.15 times 10^-6 atm^-2. Calculate K_c for the equilibrium at same temperature.
Answer» SOLUTION :`2.3 TIMES 10^-3 dm^6 mol^-2`

Discussion

22.	At 1065^@ C the heat of dissociation of H_2S is 42.4K cal and the K_p for the decomposition reaction 2H_2S leftrightarrow 2H_2+S_2 is 0.0118 atm. Find the K_p at 1132^@C
Answer» SOLUTION :0.025 ATM

Discussion

23.	At 100 ^(@) C the vapour pressure of a solution containing 6.5 g a solute in 100 g water is 732mm.IfK_b =0.52, the boiling point of this solution will be
Answer» ` 102 ^(@)C` `100^(@)C` `101^(@)C` `100.52 ^(@)C` Solution :`(Delta P )/( P ^(@)) = (n _(2))/( n _(1))` `W _(2) = 65G` `W_(1) = 100g` ` K _(B ) =0.52` `(Delta P )/( P ^(@)) = (W_(2) M _(1))/( M _(2) W _(1))` `(760 - 732)/(760)= (6.5 xx 18)/(m _(2) xx 100)` `M _(2) = 31.75 ` `Delta T _(b ) = K _(b) . m = (0.52 xx 6.5 xx 1000)/( 31. 75 xx 100) = 1. 06` `T _(b) - 100 =1. 06` `T _(b) = 100 + 1.06 = 101. 06 ~~ 101^(@)C`

Discussion

24.	At 100^(@)C, liquid water and water vapour (stea) are present together in equilibrium. Comment on the average kinetic energy of water molecules in the liquid water and steam.
Answer» SOLUTION :As both liquid WATER and steam are at the same TEMPERATURE, they will have same average kinetic energy of their molecules.

Discussion

25.	At 100^(@)C , ionic product of water is 51.3 xx 10^(-14) M^(2) . Then dissociation constant of water at 100^(@) C is
Answer» `1.72 xx 10^(-15)` `1.009 xx 10^(-15)` `9.243 xx 10^(-15)` `5.2 xx 10^(-16)` Answer :C

Discussion

26.	At 1000K, solid carbon, CaOand CaCO_(3) are mixed and allowed to atttain following equilibrium : CaCO_(3)hArrCaO(s),""K_(P)=4xx10^(-2) C(s)+CO_(2)(g)hArr2CO(g),"" K_(P)=2atm What is the pressure of CO at equilibrium (in atm)?
Answer» `0.04atm` `5sqrt2` 50 ATM `sqrt2/5atm` ANSWER :D

Discussion

27.	At 100°C and 1 atm, if the density of liquid water is 1.0 g cm^(-3) andthat of water vapour is 0.0006 gcm-3 , then the volume occupied by water molecules in 1 litre of steam at that temperature is:
Answer» `6 cm^(3)` `60 cm^(3)` `0.6 cm^(3)` `0.06 cm^(3)` Solution :Mass of 1 LITRE of water = 1000 g (`therefore d = 1 g cm^(3)` ) The volume of steam FORMED by 1000 g of water `V=(nRT)/P = 100/18 xx (0.0821 xx 373)/1 = 1701.3 L` Thus, 1701.3 L of steam has a mass of 1000 g. `therefore` Mass of 1 L of steam `=1000/(1701.3) = 0.6 g` Thus, 1L of steam contains only 0.6 g of water. Since, density of water under given conditions is `1 g cm^(-3)`, the volume occupied by water molecules in 1 L of steam would be `0.6 cm^(-3)`.

Discussion

28.	At 1000 K in 0.654 L vessel CaCO_(3(s)) is taken. CaCO_(3(s)) hArrCaO_((s)) + CO_(2(g)), the equilibrium constant is 3.9 xx 10^(-2) bar. Find the weight of CaO at equilibrium. (Ca=40 , C=12, O=16)
Answer» SOLUTION :`3.07xx10^(-4)` MOL and 0.0172 G CAO

Discussion

29.	At 100^(@)C, the P^(H) of pure water is
Answer» <P>7 Greater than 7 Less than 7 Zero Solution :As TEMPERATURE INCREASES, `K_W`also increases and`P^(K_W) `DECREASES , So, pHdecreases` ( because P^(H)=(P^(K_W))/(2))`

Discussion

30.	At100^(@)C, the value of P^(k_w) is
Answer» <P>`14` ` LT 14` ` gt 14` ` P^(H) -P^(OH) ` SOLUTION : ` as T uparrow ,K w uparrow therefore pKw uparrow `

Discussion

31.	At 1 bar pressure and 310 K temperature 25%, N_2O_4 decompose. Reaction : N_2O_(4(g)) hArr 2NO_(2(g))(i) Find K_p (ii) At 0 bar pressure and 310 K how much percentage of N_2O_4 is decompose ?
Answer» SOLUTION :`K_p`=0.267 BAR, 63.2% DISSOCIATION

Discussion

32.	At 0^(@)C, the density of a gaseous oxide at 2 bar is same as that of nitrogen at 5 bar. What is the molecular mass of the oxide ?
Answer» Solution :Using the expression, `d=(MP)/(RT)`, at the same temperature and for same DENSITY, `underset(("GASEOUS oxide"))M_(1)P_(1) ""= "" underset((N_(2))M_(2)P_(2)` (as R is constant) `:."" M_(1)xx2=28xx5 ("Molecular mass of "N_(2)=28 U)` or `"" M_(1)=70 u`

Discussion

33.	At 1 bar pressure a gas having volume 0.6 lit. If this gas gained 122 Joule heat at 1 bar pressure its volume become 2 lit. Then calculate the its internal heat change. [1 lit. bar= 101.32 Joule]
Answer» SOLUTION :`DELTAU = - 19.85` J

Discussion

34.	At 0^(@)C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?
Answer» <P> SOLUTION :For nitrogen, `d=(MP)/(RT)` `(28xx5)/(Rxx273) ( :." mol. mass of " N_2 = 28` ) For the gaseous OXIDE, `d= (MP)/(RT) = (Mxx2)/(Rxx273)` Since, the two densities are EQUAL `:. "" (28xx5)/(Rxx273)= (Rxx2)/(Rxx273)` or `M = (28xx5)/2 = 70g mol^(-1)`

Discussion

35.	At 0^@C, the density of a certain oxide of a gas at 2 bar is the same as that of N_2 at 5 bar. What is the molecular mass of the oxide?
Answer» Solution :Density d is GIVEN by the relation d = PM/RT If density and temperature are the same for two GASES, we can write `P_1 M_1 = P_2 M_2` MOLECULAR MASS of `N_2 (M_2)` = 28 therefore `2xxM_1` = `5xx28` therefore `M_1` (Molecular mass of the oxide) = `(5xx28)/2` = `70mol^-1`

Discussion

36.	At 0^(@)C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide ?
Answer» Solution :Relation between DENSITY pressure and molecular mass according to eq. `M=(dRT)/(p)` `d=(pM)/(RT)` `THEREFORE M=(dRT)/(p)` where, T = 0.2 R = GAS constant d = density Thus, dRT = Constant No. (K) `therefore M =(K)/(p)` and `M PROP (1)/(p)` Thus, `M_(1)p_(1)=M_(2)p_(2)` where, `M_(1)=` molecular mass of OXIDE = (?) `M_(2)=` molecular mass of `N_(2)=28 g mol^(-1)` `p_(1)=` pressure of oxide = 2 bar `p_(2)=` pressure of `N_(2)=5` bar Thus, `M_(1)xx2` bar `= 28xx5` `therefore M_(1)=(28xx5)/(2)=70 g mol^(-1)`

Discussion

37.	At 0^(@)C, ice and water in equilibrium and Delta H =6.00 kJ mol^(-1) for the processH_(2)O (s) rarr H_(2)O(l). What will be Delta S and Delta G for the conversino of ice to liquid water ?
Answer» Solution :Since the given PROCESS inequilibrium , `Delta G = 0` Putting this value in the relationship , `Delta G = DELTAH - T DELTAS ` weget `0= Delta H-T Delta S`or `T DeltaS = DeltaH ` or `Delta S = (DeltaH)/(T)` We are given `DeltaH = 6.0 kJ mol^(-1) = 6000 J mol^(-1) `and `T = 0^(@) C = 273 K` `:. Delta S = ( 6000 J mol^(-1))/( 273K ) = 21.98 J K^(-1) mol^(-1)`

Discussion

38.	At 0^(@)C , ice and water are in equilibrium and DeltaS and DeltaG for the conversion of ice to liquid water is
Answer» `0,21.98JK^(-1)MOL^(-1)` `-ve` , ZERO `+ve,21.98JKk^(-1)mol^(-1)` All of these Solution :Since the given process is in EQUILIBRIUM, i.e., `DeltaG=0` `DeltaG=DELTAH-TDeltaS` or, `0=DeltaH-TDeltaS` or `DeltaH=TDeltaS` or, `Delta=(DeltaH)/(T)=(6)/(273)xx10^(3)` `JK^(-1)mol^(-1)=21.98JK^(-1)mol^(-1)`

Discussion

39.	At 0^(@)C, ice and water are present in equilibrium. What will happen on increasing the pressure ?
Answer» Solution :On INCREASING the pressure, ICE MELTS to form water (because water has lesser VOLUME than ice).

Discussion

40.	At 0^(@)C ice and water are in equilibrium and Delta H= 6.0KJ " then " Delta S will be
Answer» `22 JK^(-1) MOL^(-1)` `35 JK^(-1) mol^(-1)` `48 JK^(-1) mol^(-1)` `100 JK^(-1) mol^(-1)` Solution :`DELTA S = (Delta H)/(T) = (6000)/(273) = 22`

Discussion

41.	Assuming the water vapour is an ideal gas, the internal energy change (Delta U) when 1 mole of water is vapourised at 1 bar pressure and 100^(@)C(given molar enthalpy of vapourisation of watyer at 1 bar and 373 K = 41 kJ mol^(-1) and R = 8.3 J mol^(-1)K^(-1)) will be
Answer» `3.7904 kJ"mol"^(-1)` `37.904 kJ"mol"^(-1)` `41.00 kJ"mol"^(-1)` `4.100 kJ"mol"^(-1)` Solution :`H_(2)O(l) to H_(2)O(g)` `DELTAG = 41 kJ mol^(-1)` `DELTAH = DeltaU + Deltan_(g)RT` `Deltan_(g) = 1` `41 = DeltaU + 1 xx 8.3 xx 10^(-3) xx 373` `41 = DeltaU + 3.096` `DeltaU = 41 - 3.096 = 37.904 kJ mol^(-1)`.

Discussion

42.	Assuming the same pressure in each, case, calculate the mass of hydrogen required to inflate a balloon to a certain volume at 100^(@)C if3.5 g He is required to inflate the balloon to half the volume at 25^(@)C. (At masses H=1, He=4).
Answer» Solution :Volume of 3.5 g He at `25^(@)C` and pressure P, `V=(nRT)/(P)=(w)/(M)(RT)/(P)=(3.5)/(4)xx(Rxx298)/(P)` To FILL the BALLOON with `H_(2)` to double this volume means volume of `H_(2)`=V, T=373 K. Hence, `2V=(w)/(M)(RT)/(P)=(w)/(2)xx(Rxx373)/(P)` DIVIDING (ii) by (i) we get `2=(wxx373)/(2)xx(4)/(3.5xx298) "or" w=2.796%" g"`

Discussion

43.	Assuming that water vapour in an ideal gas, the internal energy change (DeltaU) when 1 mol of water is vaporised at 1 bar of pressure and 100^(@) C, (Given : Molar enthalpy of vapourization of water at 1 bar and 373 K = 41 kJ mol^(-1) and R=8.3 J mol^(-1)K^(-1)) will be :
Answer» 37.904 KJ `MOL^(-1)` 41.00 kJ `mol^(-1)` 4.100 kJ `mol^(-1)` 3.7904 kJ `mol^(-1)` ANSWER :A

Discussion

44.	Assuming that, water vapour is an ideal gas the internal energy (Delta U) when 1 mol of water is vapourised at 1 bar pressure and 100^(@)C (given : molar enthalpy of vaporisation of water at 1 bar and 373 K = 41 kJ mol^(-1) and R = 8.3 J K^(-1) mol^(-1))) will be
Answer» `41.00kJ mol^(-1)` `4.100kJ mol^(-1)` `3, 790J mol^(-1)` `37.9 KJ mol^(-1)` SOLUTION :Reaction for the given condition is `H_(2)O_((L)) rarr H_(2)O_((G))` `Delta H = Delta u + Delta n_((g)) RT` `41 xx 10^(3) = Delta u + (1) xx 8.314 xx 373` `Delta u = 37898.878 KJ = 37900KJ`

Discussion

45.	Assuming that the density of water to be 1g//cm^(3), calculate the volume occupied by one molecule of water.
Answer» `2.989 XX 10^(-23)`mL `6.023 xx 10^(23)` `CM^(3)` `0.288 xx 10^(-3)` `1.66 xx 10^9-2)` SOLUTION :a) `THEREFORE` Density of WATER `=1 g//cm^(3)` `therefore

Discussion

46.	Assuming that petrol is octane (C_(2)H_(18)) and has a density of 0.8 gm/m.l, 1.425 litres of pertrol on complete combustion will consume
Answer» 50 moles of `O_(2)` 100 moles of `O_(2)` 125 mole of `O_(2)` 200 moles of `O_(2)` ANSWER :C

Discussion

47.	Assuming that Hund's rule is viovated , the bond order and magnetic nature of diatomic molecule B_(2) is
Answer» 1 and diagagnetic 0 and diamagnetic 1 and paramagnetic 0 and paramagnetic Solution :If Hund's rule is violated, molecular orbital electronic configuration of `B_(2)` will be `sigma_(1S)^(2)sigma_(1s)^(2)sigma_(2s)^(2)sigma_(2s)^(2)sigma_(2p_(Z))^(2)pi_(2p_(x))^(2)pi_(2p_(y))^(2)pi_(2p_(x))^(2)pi_(2p_(y))^(2)` It will be diamgnetic with bond order `(1)/(2) (6-4) = 1`

Discussion

48.	Assuming that dry air contains 79% N_(2) and 21% O_(2) by volume, calculate the density of dry air at if it has a relative humidity of 40%. The vapour pressure of water at 25^(@)C is 23.76 mm.
Answer» Solution :Calculation of density of dry air. Density of air means the mass of air per litre. first let us convert 1 L `(1000 cm^(3))` of air at `25^(@)C` 1 atm pressure to the volume at S.T.P. `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` `:. "" (1 atmxx1000" cm"^(3))/(298" K")=(1 atm xxV_(2))/(273 K) "or" V_(2)=916.1" cm"^(3)` As air contains 79%`N_(2)` and 21% `O_(2)`, `:. ""` Volume of `N_(2)` at S.T.P. in the air `=(79)/(100)xx916.1cm^(3)=723.72" cm"^(3)` Volume of `O_(2)` at S.T.P. in the air `=(21)/(100)xx916.1" cm"^(3)=193.38" cm"^(3)` Mass of `723.72" cm"^(3)`of `N_(2)` at S.T.P. `=(28)/(22400xx723.72=0.905" g"` Mass of `193.38" cm"^(3)` of `O_(2)` at S.T.P. `=(32)/(22400)xx923.38=0.276" g"` `:. ` Total mass of 1 L of air at `25^(@)C` and 1 atm `=0.905+0.276" g"=1.181" g"` Hence,density of dry air `=1.181" g "L^(-1)` Calculation of density of moist air % Relative humidity`=("Partial pressure of "H_(2)O "vapours")/("VAPOUR pressure of "H_(2)O "at the same temp")xx100` `:. ""` Partial pressure of `H_(2)O` vapours`=(40xx23.76)/(100) =9.5" mm"` As total pressure =1 atm=760 mm, `:. ""` Partial pressure of `N_(2)=(79)/(100)xx750.5" mm "=592.9" mm "` and Partial pressure of `O_(2)=750.5-529.9=157.6" mm"` Applying gas equation PV=nRT, i.e., `N=(PV)/(RT)` No. of moles of `H_(2)O` in the air `(n_(H_(2)O))=((9.5//760)"atm "xx1L))/(0.0821" L atm "K^(-1)mol^(-1)xx298" K ")=5.1xx10^(-4)` No. of moles of `N_(2)` in the air `(n_(N_(2)))=((592.9//760)xx1)/(0.0821xx298)=3.19xx10^(-2)` No. of moles of `O_(2)` in the air `(n_(O_(2))=((157.6//760)xx1)/(0.0821xx298)=8.48xx10^(-2)` Total mass of 1 L of air `=5.1xx10^(-4)xx18+3.19xx10^(-2)xx28+8.48xx10^(-3)xx32` `=0.00918+0.8932+0.02714=1.174" g"` Hence, density of moist air `=1.174" g "L^(-1)` Note. Density of moist dry air can also becalculated by this method.

Discussion

49.	Assuming that each has icosahedral strcucture, determine how many isomers are possible for the B_(10) C_(2) H_(12) molecule ?
Answer» SOLUTION :THREE isomers are possible. The carbon atoms may be at adjacent possible, may have one boron ATOM between them or may ne on opposite side of the icosahedan. There are only three isomers possible because the `12` positions of the icosahedran are all inherently equivalent.

Discussion

50.	Assuming that air at STP contained 80% by volume of nitrogen, the volumeof air at STP that contains 4.8xx10^(23) molecules of notrogen is
Answer» 18 L 44.8 L 22.4 L 11.2 L Solution :22.4 litres AIR - STP - `6.023xx10^(23)` molecules `therefore` Nitrogen molecules `=(80)/(100)xx6.023xx10^(23)` `~~4.8xx10^(23)` molecules

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

At 25^(0)C, the hydroxyl ion concentration of a basic solution is 6.75 xx 10^(-3) M.Then the value of K_(w) is

At 25^@ C, the dissociation constant of NH_4 OH is 2 xx 10^(-5). If the molar concentration of NH_2 OH is 0.2 M, calculate the proton concentration of aqueous ammonia solution.

At 25^@C, 500 cm^3 flask contains hydrogen at a pressure of 120 mm of Hg and a 250cm^3 flask contains oxgyen at a pressure of 300 mm of Hg at the same temperature. If the two gases are transferred into a one litre flask at the same temperature, find the pressure of the mixture.

At 227^(@) C & 4 atm , PCl_(5) is dissociated to an extent 50% . At same temperature , extent of dissociation of PCl_(5) is 0.75 at a pressure of [Hint : alpha prop (1)/(sqrtP)]

At 21^(@)C temperature volume of 212 g O_(2) gas is 34 dm^(3) . If pressure of gas becomes 1.24 bar then how many gram O_(2) gas is left from container ? (R=0.083 dm^(3)" bar " K^(-1)mol^(-1))[Note : Calculate intital moles by pV = nRT]

At 20^(@)C two balloons of equal volume and porosity are filled to a pressure of 2 atm one with 14kgN_(2) and other with 1kg of h_(2) The N_(2) balloon leaks to a pressure of 1//2 atm in 1hr How long will it take for H_(2) balloon to reach a pressure of 1//2atm ? .

At 20^@Cthe electrical conductivity of water is 5.7 xx 10^(-8) ohm^(-1)cm^(-1). At the same temperature, the electrical conductivity of heavy water in ohm^(-1) cm^(-1)is

At 200^(@)C the velocity of hydrogen molecule is 2.0 xx 10^(5) cm/sec. In this case, the de-Broglie wavelength is about ________Å.

At 200^(@) C PCl_(5) dissociates as follow , PCl_(s(g)) hArr PCl_(3(g)) + Cl_(2(g)) . It was found that the equilibrium vapour. are 62 times as heavy as hydrogen. The degree of dissociation of PCl_(5) at 200^(@) C is nearly.

At 18^(@)C aniline and acetic acid have dissociation constants 5xx10^(-10) and 1.8xx10^(-5) respectively. An aqueous solution of anilium acetate is hydrolysed to the extent of x% under equilibrium, what is pH of the solution ? (K_(w)=10^(-14))

At 17^(@)C temperature volume of gas is 400 mL. Then, at which temperature (i) Volume becomes double (ii) Volume becomes half.

At 1300K, 0.3 mol of CO ,0.1 mol H_2, 0.02 mol of H_2O and an unknown amount of CH_4 are present at equilibrium in a vessel. K_c for the reaction CO_(g)+H_2(g) leftrightarrow CH_4(g)+H_2O(g) is 3.9. What is the equilibrium concentration CH_4

At 127^(@)C temperature volume of gas is 3 L. If volume of gas becomes half then calculate temperature. Pressure is 1 bar.

At 127^(@)C, K_(c)" for "2A Leftrightarrow 2B+C" is " 2xx10^(-4). Calculate DeltaG^(@) for the reaction.

At 127°C and latm pressure, a mixture of a gas contains 0.3 mole of N_(2), 0.2 mole of O_(2) The volume of the mixture is

At 127^@C, for helium if time of flight is 0.1 nanosec, the mean free path of helium (in A) is

At 1127K and 1 atm pressure, a gaseous mixture of CO and CO_(2) in equilibrium with soild carbon has 90.55% CO by mass C(s)+CO_(2)(g)hArr2CO(g) Calculate K_(c) for this reaction at the above temperature.

At 1100 K temperature CaCO_3 and CaO_((s)) are in equilibrium pressure of CO_2 is 2.0 xx 10^5 Pa. Find equilibrium constant. CaCO_(3(s)) hArr CaO_((s)) + CO_(2(g))

At 1127 K and 1 atm pressure , a gaseous mixture of CO and CO_(2)in equilibrium with solid carbon has 90*55 % CO by mass in the reaction, C(s) + CO_(2) (g) hArr 2 CO (g). Calculate k_(c)for the reaction at the above temperature.

At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO_2 in equilibrium with solid carbon has 90.55% CO by mass. C_((s)) + CO_(2(g)) hArr 2CO_((g)) Calculate K_c for this reaction at the above temperature.

At 125^@C the K_p for the reaction N_2(g)+3H_2(g) leftrightarrow 2NH_3(g) is 2.15 times 10^-6 atm^-2. Calculate K_c for the equilibrium at same temperature.

At 1065^@ C the heat of dissociation of H_2S is 42.4K cal and the K_p for the decomposition reaction 2H_2S leftrightarrow 2H_2+S_2 is 0.0118 atm. Find the K_p at 1132^@C

At 100 ^(@) C the vapour pressure of a solution containing 6.5 g a solute in 100 g water is 732mm.IfK_b =0.52, the boiling point of this solution will be

At 100^(@)C, liquid water and water vapour (stea) are present together in equilibrium. Comment on the average kinetic energy of water molecules in the liquid water and steam.

At 100^(@)C , ionic product of water is 51.3 xx 10^(-14) M^(2) . Then dissociation constant of water at 100^(@) C is

At 1000K, solid carbon, CaOand CaCO_(3) are mixed and allowed to atttain following equilibrium : CaCO_(3)hArrCaO(s),""K_(P)=4xx10^(-2) C(s)+CO_(2)(g)hArr2CO(g),"" K_(P)=2atm What is the pressure of CO at equilibrium (in atm)?

At 100°C and 1 atm, if the density of liquid water is 1.0 g cm^(-3) andthat of water vapour is 0.0006 gcm-3 , then the volume occupied by water molecules in 1 litre of steam at that temperature is:

At 1000 K in 0.654 L vessel CaCO_(3(s)) is taken. CaCO_(3(s)) hArrCaO_((s)) + CO_(2(g)), the equilibrium constant is 3.9 xx 10^(-2) bar. Find the weight of CaO at equilibrium. (Ca=40 , C=12, O=16)

At 100^(@)C, the P^(H) of pure water is

At100^(@)C, the value of P^(k_w) is

At 1 bar pressure and 310 K temperature 25%, N_2O_4 decompose. Reaction : N_2O_(4(g)) hArr 2NO_(2(g))(i) Find K_p (ii) At 0 bar pressure and 310 K how much percentage of N_2O_4 is decompose ?

At 0^(@)C, the density of a gaseous oxide at 2 bar is same as that of nitrogen at 5 bar. What is the molecular mass of the oxide ?

At 1 bar pressure a gas having volume 0.6 lit. If this gas gained 122 Joule heat at 1 bar pressure its volume become 2 lit. Then calculate the its internal heat change. [1 lit. bar= 101.32 Joule]

At 0^(@)C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?

At 0^@C, the density of a certain oxide of a gas at 2 bar is the same as that of N_2 at 5 bar. What is the molecular mass of the oxide?

At 0^(@)C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide ?

At 0^(@)C, ice and water in equilibrium and Delta H =6.00 kJ mol^(-1) for the processH_(2)O (s) rarr H_(2)O(l). What will be Delta S and Delta G for the conversino of ice to liquid water ?

At 0^(@)C , ice and water are in equilibrium and DeltaS and DeltaG for the conversion of ice to liquid water is

At 0^(@)C, ice and water are present in equilibrium. What will happen on increasing the pressure ?

At 0^(@)C ice and water are in equilibrium and Delta H= 6.0KJ " then " Delta S will be

Assuming the water vapour is an ideal gas, the internal energy change (Delta U) when 1 mole of water is vapourised at 1 bar pressure and 100^(@)C(given molar enthalpy of vapourisation of watyer at 1 bar and 373 K = 41 kJ mol^(-1) and R = 8.3 J mol^(-1)K^(-1)) will be

Assuming the same pressure in each, case, calculate the mass of hydrogen required to inflate a balloon to a certain volume at 100^(@)C if3.5 g He is required to inflate the balloon to half the volume at 25^(@)C. (At masses H=1, He=4).

Assuming that water vapour in an ideal gas, the internal energy change (DeltaU) when 1 mol of water is vaporised at 1 bar of pressure and 100^(@) C, (Given : Molar enthalpy of vapourization of water at 1 bar and 373 K = 41 kJ mol^(-1) and R=8.3 J mol^(-1)K^(-1)) will be :

Assuming that, water vapour is an ideal gas the internal energy (Delta U) when 1 mol of water is vapourised at 1 bar pressure and 100^(@)C (given : molar enthalpy of vaporisation of water at 1 bar and 373 K = 41 kJ mol^(-1) and R = 8.3 J K^(-1) mol^(-1))) will be

Assuming that the density of water to be 1g//cm^(3), calculate the volume occupied by one molecule of water.

Assuming that petrol is octane (C_(2)H_(18)) and has a density of 0.8 gm/m.l, 1.425 litres of pertrol on complete combustion will consume

Assuming that Hund's rule is viovated , the bond order and magnetic nature of diatomic molecule B_(2) is

Assuming that dry air contains 79% N_(2) and 21% O_(2) by volume, calculate the density of dry air at if it has a relative humidity of 40%. The vapour pressure of water at 25^(@)C is 23.76 mm.

Assuming that each has icosahedral strcucture, determine how many isomers are possible for the B_(10) C_(2) H_(12) molecule ?

Assuming that air at STP contained 80% by volume of nitrogen, the volumeof air at STP that contains 4.8xx10^(23) molecules of notrogen is