94420 + Interview Questions in Chemistry in Class 12

1.	At 80^(@)C, distilled water has [H_(3)O^(+)] concentration equal to 1 xx 10^(-6) mole/litre. The value of K_(w) at this temperature will be
Answer» `1 xx 10^(-6)` `1 xx 10^(-9)` `1 xx 10^(-12)` `1 xx 10^(-15)` SOLUTION :`K_(w) = [H_(3)O^(+)][OH^(-)]` Concentration of `H_(3)O^(+)` in distilled WATER `= 1 xx 10^(-6)` mol/l. Now `[H_(3)O^(+)] = [OH^(-)]` `K_(w) = [1 xx 10^(-6)] xx [1 xx 10^(-6)] = 1 xx 10^(-12)`.

Discussion

2.	At 720 mm pressure and at 27^@ C , 200 mL of gas weight 1.653g . Calculate its molecular weight using general gas equation.
Answer» ANSWER :214.7

Discussion

3.	At 800^(@)C, the following equilibrium is established as F_(2)(g)hArr2F(g) The composition of equilibrium may be determined by measuring the rate of effusion of the mixture through a pin hole. It is found that at 800^(@)C and 1 atm mixture effuses 1.6 times as fast as SO_(2) effuse under the similar conditions. (At. mass of F =19) what is the value of K_(p) (in atm) ?
Answer» 0.315 0.685 0.46 1.49 Answer :D

Discussion

4.	At 700 K, the equlibrium constant K_(p) for the reaction 2SO_(3(g))hArr2SO_(2(g))+O_(2(g))is 1.80xx10^(-3) and kP_(a) is 14, (R=8.314,Jk^(-1)mol^(-1)). The numerical value in moles per litre of K_(c) for this reaction at the same temperatue will be
Answer» `8.18xx10^(-9)"mol-litre"` `5.07xx10^(-8)"mol-litre"` `8.18xx10^(-9)"mol-litre"` `9.24xx10^(-10)"mol-litre"` Solution :`UNDERSET(2)(2SO_(3))hArr2SO_(2)underset(2)(+)O_(2)` `DELTAN=3-1=+1,K_(p)=1.80xx10^(-3)` `[RT]^(Deltan)=(8.314xx700)^(1)` `K_(c)=(K_(p))/((RT)^(Deltan))=(1.8xx10^(-3))/((8.314xx700)^(1))` `=3.09xx10^(-7)"mole-litre."`

Discussion

5.	At 700 K, CO_2 and H_2 react to form CO and H_2O. For this process K is 0.11. Amixture of 0.45 mole of CO_2 and 0.45 mole of H_2 is heated to 700 K.(i) Find the amount of each gas at equilibrium. (ii) After the equilibrium is reached, another 0.34 mole of CO_2and 0.34 mole of H_2are added to the reaction mixture. Find the composition of the new equilibrium state.
Answer» SOLUTION :(i) 0.34, 0.11 (II) 0.594, 0.196

Discussion

6.	At 627^(@)C and 1 atm pressure, SO_(3) undergoes partial dissociation into SO_(2) and O_(2) SO_(3) hArr SO_(2) + (1)/(2)O_(2) if the observed density of the equilibrium mixture is 0.925 g/L, calculate degree of dissociation of SO_(3).
Answer» Solution :Let the initial no. of moles of `SO_(3)` be 1and its degree of DISSOCIATION x. `{:(1,,0,,0,"Initial no. of moles"),(SO_(3),+,SO_(2),+,(1)/(2)O_(2),),(1-x,,x,,x//2,"Moles at equilibrium"):}` `therefore` total no. of moles at eqb. `= 1- x + x + (x)/(2) = 1 + (x)/(2)` Thus applying PV = nRT `1 xx V = (1+(x)/(2)) xx 0.0821 xx (627 + 273)` `V = (1+(x)/(2)) xx 73.89` litres. Now w.t of 1 mole of `SO_(3) = 80g` and therefore, from the law of conservation of mass, we have, wt. of gases at eqb. = 80 g. `therefore "DENSITY" = ("wt. in g")/("vol. in LITRE") = (80)/((1+(x)/(2)) xx 73.89) = 0.925` (given) or x = 0.34.

Discussion

7.	At 550K, the K_(c) for the following reaction is 10^(4)mol^(-1)LX_((g))+Y_((g))hArrX_((g)). At eqilibrium, it was observed that [X]=1/2[Y]=1/2[Z]. What is value of [Z] (in mol L^(-1)) at equlibrium
Answer» `2xx10^(-4)` `10^(-4)` `2xx10^(4)` `10^(4)` Solution :`K_(c)=([Z])/([X][Y])implies1/2[Y]=[X]=1/2[Z]=a(say)` `therefore[Z]=2A,[Y]=2a,[X]=a` `10^(4)=(2a)/(a.2a)impliesa=10^(-4)` `[Z]=2a=2xx10^(-4)`

Discussion

8.	At 540, the equilibrium constant K_(p) for PCl_(5) dissociation equilibrium at 1.0 atm 1.77 atm. Calculate equilibrium constant in molar concentration (K_(c)) at same temperature and pressure.
Answer» ANSWER :`K_(C)=4XX10^(-2)` moles/litres

Discussion

9.	AT 540 K 0.10 moles of PCI_5 are heated in 8 litre flask. The pressure of the equilibrium mixture is found to be 1.0 atm. Calculate K_p and K_c for the reaction.
Answer» SOLUTION :1.77 ATM

Discussion

10.	AT 518^(@)C , the rate of decomposition of a sample of gaseous acetaldehyde , initially at a pressure of 363 Torrr , was 1.00 Torr s^(-1) when 5% had reached and 0.5 Torr s^(-1) when 33% had reacted . The order of the reaction is
Answer» Solution :RATE = K `("PRESSURE")^(N)` `R_(1) = K(P_(1))^(n) , R_(2) = K(P_(2))^(n) implies (R_(1))/(R_(2)) = [(P_(1))/(P_(2))]^(n)` `P_(1) = 363 - (363 xx 5)/(100) = 344. 85 , P_(2) = 363 - (363 xx 33)/(100) = 243.21` `(1)/(0.5) = ((344.85)/(243.21))^(n) implies 2 = (sqrt2)^(n) , (n = 2)`.

Discussion

11.	At 525 K, the equilibrium constant of the reaction PCl_5 iff PCl_3 + Cl_2is 1.78 atm (K_p) . At what pressure should an equimolar mixture of Cl_2and PCl_3 , be taken for the pressure of PCl_5 to be 5 xx10^4 Pa at equilibrium, volume remaining constant?
Answer» SOLUTION :`28.99 XX 10^4 PA `

Discussion

12.	At 518^(@)C, the rate of decomposition of a simple of gaseous acetaldehyde initially at a pressure of 363 to "rr" was 1 to "rr" s^(-1) when 5% had reacted and 0.5 to "rr" s^(-1) when 33% had reached.The order of reaction is:
Answer» 2 3 1 0 Solution :b) Assume that the order of reaction with RESPECT to gaseous acetaldehyde is x. Ist CASE Reaction rate (r ) = `k[363 " torr " xx 0.95]^(x)` `k[344.85]^(x)`â€¦â€¦â€¦..(i) IIND case: `0.5" torr "s^(-1) = k[363 " torr " xx 0.67]^(x)` `=k[243.21]^(x)`............(ii) Divide i) by ii), `(1 " torr " s^(-1))/(0.5 " torr "s^(-1))= [344.85]/[243.21]` or `(2)^(1) = (1.414)^(x)` or `x=2`

Discussion

13.	At 518^(@) C,the rate of decomposition of a sample of gaseous acetaldehyde,initially at a pressure of 363 torr,was 1.00 torr s^(-1) when 5% has reacted and 0.5 torr s^(-1) when 33% had reacted.The order of the reaction is :
Answer» 2 3 1 0 Solution :`(r_(1))/(r_(2))=((95)/(67))^(X)therefore 2=((95)/(67))^(x)` `therefore` x log `((95)/(67))=2` `thererfore` x=second ORDER

Discussion

14.	At -50^(@)C liquid NH_(3) has ionic product is 10^(-30) .How many amide (NH_(2)^(-)) ions are present per mm.^(3) in pure liqudi NH_(3)? (Take N_(A)=6xx10^(23))
Answer» Solution :`K=[NH_(4)""^(+)][NH_(2)""^(=)]=10^(-30)` `[NH_(2)^(-)]=[NH_(4)^(+)]=10^(-15)M ""( :.2NH_(3)HARR NH_(4)^(+)+NH_(2)""^(-))` No. of `NH_(2)^(-)` IONS `NH_(2)^(-)=((10^(-15)mol e)/(L))((1L)/(10^(6)mm^(3)))((6XX10^(23)ions)/(mol e))=600ions//mm^(3)`

Discussion

15.	At 50^(@)C, the vapour pressure of pure CS_(2) is 854 torr. A solution of 2.0 g of sulphur in 100 gof CS_(2) has vapour pressure of 848.9 torr. Determine the formula of sulphur molecule.
Answer» Solution :`(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1))=(w_(2)//M_(2))/(w_(1)//M_(1)) ""therefore""(854-848.9)/(854)=(2//M_(2))/(100//76)""("Molecular mass of CS"_(2)=12+2xx32=76u)` `"or"(5.1)/(854)=(2)/(M_(2))XX(76)/(100)"or"M_(2)=254.5u` If formula of SULPHUR molecules is `S_(x)`, molecular mass `=x xx 32.` Hence, `32x=254.5 or x = 8.`

Discussion

16.	At 500K, the half period of a gaseous reaction at an initial pressure of 80 kp_(a)is 350 sec. when the pressure is 40 kp_(a) the half life preiod is 175 sec, the order of reaction is
Answer» ONE two three zero Answer :D

Discussion

17.	At 50^(@)C the vapour pressures of pure water and ethyl alcohol are, respectively 92.5mm and 219.9mmHg. If 6g of nonvolatile solute of mol.wt. 120s is dissolved in 150g of each of these solvents, what will be the relative vapour pressure lowerings in the two solvents ?
Answer» ANSWER :`0.006`, `0.015`

Discussion

18.	At 500 K, the half-life period of a gaseous reaction at the initial pressure of 80 kPa is 350 sec. When the pressure is 40 kPa, the half-life period is 175 sec. The order of the reaction is
Answer» ZERO one two three Solution :When initial pressure is halved, the half-life PERIOD is also halved. This SHOWS that `t_(1//2) PROP` initial pressure. This is so for REACTIONS of zero order.

Discussion

19.	At 500K, for the reaction, PCl_5(g)hArrPCl_3(g)+Cl_2(g) the equilibrium constant, KP = 0.52. In a closed container, these three gases are mixed together. If the partial pressure of each of these gases be 1 atm, then in the reaction system-
Answer» The number of moles of `PCl_5` will increase the number of moles of `PCl_3` will increase the REACTION will attain EQUILIBRIUM when 50% of the the reaction gets completed the reaction will attain equilibrium when 75% of the reaction gets completed Answer :A

Discussion

20.	At 500 K, equilibrium constant, K_(c ), for the following reaction is 5.(1)/(2)H_(2)(g)+(1)/(2)I_(2)(g)hArr HI(g)What would be the equilibrium constant K_(c ) for the reaction 2HI(g)hArr H_(2)(g)+I_(2)(g)
Answer» `0.04` `0.4` 25 `2.5` Solution :`(1)/(2)H_(2)(g)+(1)/(2)I_(1)(g)HARR HI(g)K_(a)=5` `HI(g)hArr (1)/(2)H_(2)(g)+(1)/(2)I_(2)(g)K_(a)=(1)/(5)` `2HI(g)hArr H_(2)(g)+I_(2)(g)K_(a)=((1)/(5))^(2)=0.04`

Discussion

21.	At 5 xx 10^(5) bar pressure, density of diamond and graphite are 3g/cc and 2g/cc respectively, at a temperature T. What will be the value of triangleU-triangleH for the conversion of 1 mole grpahite to 1 mole diamond at diamond at temperature T?
Answer» 100 kJ/mole 50kJ/mole `-100" kJ/mole"` 300kJ/mole Answer :A

Discussion

22.	At 490^(@)C, the equilibrium constant for the stnthesis of HI is 50, the value of K for the dissocisation of HI will be
Answer» `20.0` `2.0` `0.2` 0.02` Solution :K for DISSOCIATION of `HI=? H_(2)+I_(2)hArr2HI` `K_(a)=50,K_(b)=1/50=0.02`

Discussion

23.	At 444^(@) C, the equilibrium constant K for the reaction 2AB_((g))hArrA_(2(g)) + B_(2(g)) is 1/64 . The degree of dissociation of AB will be -
Answer» 0.1 0.2 0.3 0.5 Answer :B

Discussion

24.	At 40^@C the vapour pressures of pure liquids, benzene and toluene, are 160 mm Hg and 60 mm Hg respectively. At the same temperature, the vapour pressure of an equimolar solution of the two liquids, assuming the ideal solution should be:
Answer» 140 MM Hg 110 mm Hg 220 mm Hg 100 mm Hg Answer :B

Discussion

25.	At 40^@C the vapour pressures in torr, of methyl alcohol ethyl alcohol solutions is represented by the equation. P=119X_A + 135, where X_A is mole-fraction of methyl alcohol, then the value of lim_(X_(Ararr1)) P_A/X_A is:
Answer» 254 torr 135 torr 119 torr 140 torr Answer :A

Discussion

26.	At 40^(@)C, the vapour pressure os water is 55.3 mmHg. Calculate the vapour pressure at the same temperature over 10% aqueous solution of urea [CO(NH_(2))_(2)].
Answer» Solution :Mass of urea in 100 g of SOUTION = 10 = 90 g `"Mass of WATER in the soltion"=100-10=90g` `"No. of moles of urea"=((10g))/((60"g mol"^(-1)))=5.0 mol` `"No. of moles of water" ((90g))/((18"g mol"^(-1)))=5.0 mol` `"Mole fraction of water"=((5.0 mol))/((5.0mol)+(0.17 mol))=(5.0)/(5.17)=0.967` `"Vapoure pressure of water over aquious solution "=P_("water")^(@)xxC_("water")=(55.3mm)xx0.967` =53.48 MM Hg.

Discussion

27.	At 400 K, the root mean square (rms) speed of a gas X (molecular weight = 40) is equal to the most probable speed of gas Y at 60 K. The molecular weight of the gas Y is.
Answer» SOLUTION :`u_(RMS)=alpha_(mp)` `SQRT((3RT)/(M_(X)))= sqrt((2RT)/(M_(Y)))` `= sqrt((3R xx 400)/(40))=sqrt((2R xx 60)/(M^(Y)))` `M_(Y)=4`

Discussion

28.	At 40^@C, the vapour pressures (in torr) of methyl alcohol (A) and ethyl alcohol (B) solution is represented by: P=120X_A + 138, where X_A is mole-fraction of methyl alcohol, then the value of lim X_A rarr 0 P_B^@/X_B and lim X_B rarr 0 P_A^@/X_A are:
Answer» 138, 258 258, 138 120, 138 138, 125 Answer :A

Discussion

29.	At 407 K the rate constant of a chemical reaction is 9.5xx10^(-5)s^(-1) and at 420 K, the rate constant is 1.9xx10^(-4)s^(-1). The frequency factor of the reaction is x xx 10^(5)s^(-1). The value of 'x' is. Report your answer by rounding it up to nearest whole number.
Answer» ANSWER :5

Discussion

30.	At 400 K, the root mean square (rms) speed of a gas X (molecular weight=40) is equal to the most probable speed of gas Y at 60 K. Calculate the molecular weight of the gas Y.
Answer» SOLUTION :`SQRT((3RT_x)/(M_x)) = sqrt((2RT_y)/(M_y))` 4

Discussion

31.	At 375K and at a total pressure of one atmosphere sulphuryl chloride (SO_(2)Cl_(2)) undergoes dissociation according to the equation : SO_(2) Cl_(2) (g) hArr SO_(2)(g) + Cl_(2)(g) to the extent of 90%. Hence the work done in the process at the same temperature:
Answer» `-4.95kJ` `-2.8kJ` `+53.6kJ` `-1.4 KJ` ANSWER :B

Discussion

32.	At 380^@C , the half-life period for the first-order decomposition of H_2O_2is360 minutes. The energy of activation of the reaction is 200 kJ "mol"^(-1). Calculate the time required for 75% decomposition at 450^@C .
Answer» SOLUTION :20.34 MIN

Discussion

33.	At 380^(@)C, the half-life period for the first order decomposition of H_2O_2 is 360 minute. The energy of activation of the reaction is 200 kJ mol^(-1). What will be the time required for 75% decomposition at 450^(@) C?
Answer» 20.39 min 30.03 min 1223.4 s 2000 s Solution : `K_1=(0.693)/(t_(1//2))=(0.693)/(360) = 1.925xx10^(-3) min^(-1)` `log_(10)(k_2)/(k_1)=(E_a)/(2.303R)((T_2-T_1)/(T_1T_2))` `log_10(k_2)/(1.925xx10^(-3))=(200xx10^3)/(2.303xx8.314)[(723-653)/(653xx723)]` `K_2`=0.068 `min^(-1)` `t=(2.303)/(k_2) log_(10)(a)/(a-x) implies (2.303)/(0.068) log_(10) (100)/(25)` `THEREFORE` t=20.39 min =1223.4 s

Discussion

34.	At 380^(@)C , the half-life period for the first order decomposition of H_(2)O_(2) is 360min the energy of activation of the reaction is 200kJ mol^(-1). Calculate the time required for 75% decomosition at 450^(@)C.
Answer» ANSWER :`20.39` MINUTE

Discussion

35.	At 373 K, a gaseous reactionto2B+C is found to be of first order. Starting with pure A, the total pressure at the end of 10 min. was 176 mm and after a long time when A was completely dissociated, it was 270 mm. The pressure of A at the end of 10 minutes was
Answer» 94 mm 47 mm 43 mm 90mm Solution :`A to2B + C` Initialp00 Eq. PP-p2pp TOTAL `= P-p+2p + p=P+2p= 176M m` (complete DISSOCIATION = 2P + P) Total = 3P = 270 m m or P = 90 mm `:.` 2p = 176 - P = 176 - 90 = 86 mm or p = 43 mm Pressure of A after 10 MIN = P -p `= 90 - 43 = 47mm`

Discussion

36.	At 373 K, a gaseous reaction Ato2B+C is observed to be of first order. On starting with pure A, it was found that at the end of 10 minutes, the total pressure of the system was 176 mm of mercury and after a long time, when dissociation of A was complete, it was 270 mm. From these data, calculate (i) the initial pressure of A (ii) the pressure of A at the end of 10 minutes. (iii) the rate constant.
Answer» <P> Solution :Suppose initial of A = P mm. Decrease in the pressure of A after time t = p mm. `{:(,,,A,,to,,2B,,+,,C,,,"TOTAL pressure"),("Initial pressure",,,P,,,,0,,,,0,,,P),("Pressures after time t",,,P-p,,,,2p,,,,p,,,P+2p),("Final pressures",,,0,,,,2P,,,,P,,,3P):}` (i) Final pressure = 270 mm (Given)` :.3P=270" or "P=90" mm"` (ii) Pressure after 10 minutes = 176 mm (Given) `:.P+2p=176" or "90+2p=176" or "p=43" mm"` `:." Pressure of A after 10 MIN"=P-p=90-43=47" mm"` `:.K=(2.303)/(t)log""(a)/(a-x)=(2.303)/(t)log""(P)/(P-p)` or `k=(2.303)/(10" min")log""(90)/(90-43)=(2.303)/(10" min")log""(90)/(47)=6.496xx10^(-2)" min"^(-1).`

Discussion

37.	At 35^(@)C, the vapour pressure of CS_(2) is 512mmHg. And of acetone is 344mmHg. A solution of CS_(2) and acetone in, which the mole fraction of CS_(2) is 0.25, has a total pressure of 600mmHg. Which of the following statements is//are correct ?
Answer» A mixture of `100mL` of acetone and `100mL` of `CS_(2)` has a volume of `200ML` When acetone and `CS_(2)` are mixed at `35^(@)C`, heat must be absorbed in order to produce a solution at `35^(@)C` Process of mixing is exothermic Entropy of mixing is ZERO Solution :`X_(A)P_(A)^(@)+X_(B)P_(B)^(@)=0.25xx512+0.75xx344=386mm` Now `P_(A)+P_(B)=600 mm Hg`(Given) so `P_(A)+P_(B)gtX_(A)P_(A)^(@)+X_(B)P_(B)^(@)` THEREFORE, there is positive DEVITION from Raoult's law `DELTAHGT0` i.e., heat is absorbed.

Discussion

38.	At35^(@)C, the vapour pressure ofisand that of acetone is. A solution of CS_(2)in acetone has a total vapour pressure of. The false statement amongst the following is:
Answer» RAOULT’s law is not OBEYED by this system `CS_(2)`and acetone are less attracted to each other than to themselves Heat must be absorbed in order to produce the solution at`35^(@)C` A MIXTURE of`100 mL CS_(2)` and 100 mLacetone has a volume` LT 200 mL` Solution : So `{:([A…………………A]),([B…………………..B])]gtA…………….B` So, it is non ideal solution showing positive deviation. So, volume should be greater than 200ml

Discussion

39.	At 35^@C , the vapour pressure of CS_2 is 512 mm Hg and that of acetone is 344 mm Hg . A solution of CS_2 in acetone has a total vopour pressure of 600 mmHg. The false statement amongst the following is
Answer» A MIXTURE of 100 ml of acetone and 100 ml of `CS_2` has a total volume of 200 ml. When acetone and `CS_2` are mixed at `35^@C` , HEAT must be absorbed in order to PRODUCE a solution at `35^@C` When acetone and `CS_2` are mixed at `35^@C` , heat is released . Raoult's law is obeyed by both , `CS_2` and acetone for the solution in which the moles fraction of `CS_2` is 0.25 Answer :B

Discussion

40.	At 353 K, the vapoure pressure of pure ethylene bromide and propylene bromide are 22.93 and 16.93 K N m^(-2). Respectively and these compounds forms nearly ideal solution 3. moles of ethylen bromide and 2mole of propylene bromide are equilibrated at 353 K and at a total pressure of20.4 KN m^(-2). (a) What is the composition of the liquids phase (b) How many moles of each compound are present in the vapour phase ?
Answer» Solution :(a) Let ethylene BROMIDE `rarr to A` And propylene bromide `to B` Then from EQUATION, `P_(A)^(@)=22.93 N Km^(-2)` `P_(B)^(0) 18.93 K Nm^(-2)` `n_(A)=3` mole `n_(B)=3` mole `n_(B)=2` mole Total pressure`P_(T)=20.4 K Nm^(-2)` `:.P_(T)=P_(A)^(@)xxP_(B)^(@)X_(B)` `=P_(A)^(@)xxP_(B)^(@)(1-X_(A))=(P_(A)^(@)-P_(B)^(@))X_(A)+P_(B)^(@)` `rArr X_(A)=(P_(T)-P_(B)^(@))/(P_(A)^(@)-P_(B)^(@))=(20.4-16.93)/(22.93-16.93)=0.578` `:.X_(A)=1-0.578=0.422` (b) Let mole fraction in VAPOUR phase `=X_(A)` `X_(A)=(P_(A)^(0))/(P_(T))` `X_(A)=(22.93xx0.598)/(20.4)=0.64""...(1)` Assuming that the no. of moles of A and B that are VAPORIZED are a and b then `X_(A)=(a)/(a+b)=0.64` But composition of A in liquid phase. `X_(A)=(3-a)/((3-a)+(2-b))=0.578` `=(3-a)/(5-(a+b))=0.578 ""...(2)` Solving equation (1) and (2) `a=0.9967` mole `b=0.537` mole

Discussion

41.	At 300 K a gaseous reaction A rarr B +Cwas found to follow first order kinetics . Starting with pure A the total pressure at the end of 20 minutes was 100 mm of Hg . The total pressure after the completion of the reactionn is 180 mm of Hg .The partial pressure of A ( in mm of Hg ) is :
Answer» 100 90 180 80 Solution :`A to B + C` On decomposition of 1 mole of A , 1 mole of B and 1 mole of C are PRODUCED . THUS the total pressure after completion of the reaction corresponds to 2 MOLES and initial pressure to 1 mole . Initial pressure of A , (i.e, `p_(0) = (1)/(2) xx 180 = 90` MM Hg .... (1) `{:(, A to ""B + C) , ("at t = 0" , p_(0) ""0 ""0) , ("after 20 min" , p_(0)-p""p "" p):}` After 20 minutes , the total pressure = 100 mm Hg . `p_(0) - p + p + p = 100` `p + p = 100` from equation (1) `90 + p = 100` `p = (100 - 90) = 10`mm Hg So partial pressure of `A = p_(0) - p` `= 90 - 10 = 80` mm Hg.

Discussion

42.	At 35^(@)C, the value of K_(p) for the equilibrium reaction N_(2)O_(4)hArr2NO_(2) is 0.3174, Calculate the degree of dissociation when P is 0.2382 atm
Answer» ANSWER :`x=0.5768`

Discussion

43.	At 300 K 50% of molecule collide with energy greater than or equal to E_(a). At what temperature 25% molecule will have energy greater than or equal to E_(a)?
Answer» ANSWER :150

Discussion

44.	At 35 ""^(@)C the vapour pressure of CS_(2) is 512 mm of Hg and that of acetone is 344 mm of Hg. A solution of CS_(2) in acetone has a total vapour pressure of 600 mm of Hg. The false statement among the following is :
Answer» `CS_(2)` and acetone are less attracted to each other than themselves. Heat must be absorbed in order to PRODUCE the SOLUTION at `35^(@)C`. RAOULT's LAW is not obeyed by this system A mixture of 100 mL `CS_(2)` and 100 mL acetone has a volume less than 200 mL. Solution :`p_("total")=p_(A)^(@)x_(A)+p_(B)^(@)x_(B)` Maximum value of `X_(A)` is 1, so `p_("total")` has maximum value of 512 mm, which is less value than observed 600 mm value, so positive deviation would be observed. In which interaction between A-A and B-B is more than A-B interaction. So, for the system which does not follow Raoult.s law and shows positive deviation, `Delta V_(mix)gt 0, Delta H_(mix)gt 0`

Discussion

45.	At 353 K, the vapour pressure of pure liquid A and B are 520 mm and 1000 mm respectively. If a mixture of solutions of A and B boils at 353 K and 1 bar pressure, the mole percent of A in mixture is …… (1 bar = 760 mm)
Answer» `52%` `34%` `48%` `50%` ANSWER :D

Discussion

46.	At 300 K, 36 g of glucose present per litre of the solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bar at the same temperature, what wolud be its concentrain ?
Answer» Solution :`"According to van't Hoff equation", piV=n_(B)VT` `"Fro a solution," n_(B)"and T are all constants"` `thereforepi_(1)V_(1)=pi_(2)V_(2)orV_(2)=(pi_(1)V_(1))/(pi_(2))=((4.98"BAR")xx(1L))/((1.52"bar"))=3.28 L` `"No. of moles of solute"(n_(B))=(36g)//(180" g mol"^(-1))=0.2 mol` `"Molar concentration ( C) in second case"=n_(B)/V_(2)=((0.2 mol))/((3.28 L))=0.061" mol L"^(-1)=0.061 M`

Discussion

47.	At 337 K the vapour pressure of ethanol is 0.526 atm and the vapour pressureof water is 0.236 atm. A solution is prepared from equimolar amounts of water and ethanol at this temperature. The vapour above the solution is removed and condensed. The condensed solution is heated to 337 K and the vapour above the solution is removed and condensed. Determine the mole fraction of the condensed solution
Answer» SOLUTION :0.91, 0.09

Discussion

48.	At 323K , P=120X_(A)+140, The value ofunderset(X_(A)rarr1)(lim)(P_(A))/(X_(A)) is 32.5x Find x.
Answer» ANSWER :8

Discussion

49.	At 300 K, 36 g of glucose present per litre in its solution has an osmotic pressure of 4.98 bar. If osmotic pressure of the solution is 1.52 bar at the same temperature, what would be its concentration?
Answer» Solution :`pi_(1)=C_(1)RT,""pi_(2)=C_(2)RT""THEREFORE""pi_(1)//pi_(2)=C_(1)//C_(2)` `"4.98bar = 1.52 bar"=(36//"180 mol L"^(-1))//C_(2)"or"C_(2)=0.061" mol L"^(-1) = 0.061xx180gL^(-1)=10.98gL^(-1)`

Discussion

50.	At 320 K. a gas A_2is 20 % dissociated to A(g)IF [A_2] to = 1M,find K_c for the reactions.
Answer» ANSWER :2

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