52671 + Interview Questions in Chemistry in Class 11

1.	A weak acid HX has the dissociation constant 1xx10^(-5)M. It forms a salt NaX on reaction with alkali. The degree of hydrolysis of 0.1 M solution of NaX is
Answer» `0.0001%` `0.01%` `0.1%` `0.15%` Solution :Hydrolysis reaction is `X^(-) + H_(2)O HARR HX+OH^(-)` For a salt of WEAK acid with STRONG base, `K_(h)=(K_(w))/(K_(a))=ch^(2)` `:. (10^(-14))/(10^(-5))=0.1xxh^(2) or h^(2) = 10^(-8) or h = 10^(-4)` % hydrolysis `= 10^(-4) xx 100 = 10^(-2) = 0.01`

Discussion

2.	A weak acid HX has K_(a) =1 xx 10^(-5)M. If forms a salt NaX onreaction with alkali. The perentage degree of dissociation of 0.1 M solution of NaX is
Answer» 1.0E-5 0.01 0.1 0.15 Solution :`underset(("Salt"))(X^(-)) +H_(2)O hArr HX+OH^(-)` For the salt `K_(h) =(K_(w))/(K_(a)) =Calpha^(2)` `:. (10^(-4))/(10^(-5)) =0.1 alpha^(2)` `" or """alpha^(2) =(10^(-14))/(10^(-6)) =10^(-8)` `alpha =(10^(-8))^(1//2) =10^(-4)` `% " degree of DISSOCIATION "underset(=10^(-2) =0.01%)(=10^(-4) XX 100)`

Discussion

3.	A weak acid HX, after mixing with 12 mL of decimolar potash has a pH 5. At the end point of neutralisation 26.6 mL of potash is consumed. What is the dissociation constant of HX ?
Answer» SOLUTION :`8.2 XX 10^(-6)`

Discussion

4.	A weak acid HX has the dissociation constant 1xx 10 ^(-5)M.It forms a salt NaX on reactions with alkali. The percentage hydrolysis of 0.1 M solution of NaX is :
Answer» `0.0001 %` `0.01 %` ` 0.1 %` ` 0.15%` Solution :` h= SQRT ((K_h)/( C) )=sqrt((K_w)/( K_a.C) )=sqrt( 10 ^(_4)/( 10^(-5)xx 10^(-1))), h = 10 ^(_4) ` ` % h= 0.01 %`

Discussion

5.	A water sample has ppm level concentration of the following metals: Fe = 0.2 , Mn = 5.0 , Cu = 3.0, Zn = 5.0. The metal that makes the water sample unsuitable for drinking is....
Answer» Cu Mn Zn Fe Solution :Prescribed LEVEL of Mn is 0.5 PPM. So water SAMPLE containing Mn = 5ppm is water unsuitable for DRINKING.

Discussion

6.	A wave has a frequency of 3 xx 10^(15) sec^(-1). The energy of that photon is
Answer» `1.6xx10^(-12)` erg `3.2xx10^(-11)` erg `2.0xx10^(-11)` erg `3xx10^(15)` erg Solution :`E= H upsilon`

Discussion

7.	A water sample is said to contain permanent hardness if water contains
Answer» SULPHATES and chlorides of CALCIUM and magnesium carbonates of calcium and magnesium bicarbonates of calcium and magnesium sulphates and chlorides of sodium and potassium. Solution :PERMANENT hardness of WATER is due to sulphates and chlorides of calcium and magnesium.

Discussion

8.	A water sample has ppm level concentration of following anions F^(-)=10 , SO_4^(2-)=100 , NO_3^(-)=50 The anion/anions that make/makes the water sample unsuitable for drinking is/are
Answer» only `F^-` only `SO_4^(2-)` only `NO_3^-` both `SO_4^(2-)` and `NO_3^(-)` Solution :ACCEPTABLE LEVEL of `F^-` is upto 1 ppm, `NO_3^-` is upto 50 ppm and `SO_4^(2-)` is upto 500 ppm `F^-` concentration in water sample is 10 ppm which is higher than the permissible level. THEREFORE, only `F^-` MAKES the water sample UNSUITABLE for drinking

Discussion

9.	A water sample has ppm level concentration of following anions. F^(-) = 10, SO_(4)^(2-) = 100, NO_(3)^(-) = 50 The anion/anions that make/makes the water sample unsuitable for drinking is/are……..
Answer» only `NO_(3)^(-)` both `SO_(4)^(2-) and NO_(3)^(-)` only `F^(-)` only `SO_(4)^(2-)` Solution :ACCEPTED amount of negative ion in sample of WATER For `F^(-)` up to 1 ppm `NO_(3)^(-)` up to 50 ppm `SO_(4)^(2-)` up to 500ppm

Discussion

10.	A water sample has ppm level concentration of following anions F^(-)=10,SO_(4)^(2-)=100,NO_(3)^(-)=50 The anion/anions that make/makes the water sample unsuitable for drinking is/are:
Answer» only `F^(-)` only `SO_4^(2-)` only `NO_3^(-)` both `SO_4^(2-) and NO_3^(-)` SOLUTION :only `F^(-)`

Discussion

11.	(A) : Water is a liquid whereas sulphurdi oxide is a gas at room temp (R) : Molecular mass of SO_2 is more than that of H_2 O
Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) Both (A) and (R) are true and (R) is not the correct explanation of (A) (A) is true but (R) is false (A) is false but (R) is true Answer :B

Discussion

12.	(a) W hat do ou mean by catensation? (b) C Cl_(4) cannot be hydrolysed, but SiCl_(4) can be easily hydrolysed, Why?
Answer» Solution : (a) F or answer, CONSULT section 11.6. (B)For answer, Consult section 11.7.

Discussion

13.	(A): Volume to temperature ratio is constant for a fixed amount of gas at constant pressure. (R) : At constant pressure the volume of a given mass of a gas increases or decreases by 1/273 times of its volume at 0^@C, for every 1^@C change in temp.
Answer» Both A and R are CORRECT and R is the correct EXPLANATION of A. Both A and R are correct but R is not the correct explanation of A. A is TRUE but R is false. A is false but R is true Answer :A

Discussion

14.	A vey high cosmic abundance of Fe is due to
Answer» its ferro-magnetic nature its high NUCLEAR BINDING energy its high reactivity towards `O_2` NONE of these. Answer :B

Discussion

15.	A vessel with small opening contained equal volumes of oxygen and an unknown gas. Oxygen effused through the opening 1.8 times faster than the unknown gas. If the atomic mass of oxygen is 16, calculate the molecular mass of the unknown gas.
Answer» Solution :Graham.s LAW of diffusion is also applicable to the phenomenon of EFFUSION. According to this law, `r_1/r_2 = sqrt(M_2/M_1)` SINCE, oxygen EFFUSES 1.8 times FASTER than the unknown gas, we have `r_1/r_2 = 1.8` Suppose the molecular mass of the unknown gas is `M_2`. Hence, `1.8 = sqrt(M_2/32)( :. "Molecular mass of " O_2 = 32)` or`M_2 = 103.68`

Discussion

16.	A vessel of 120 mL capacity contains a certain mass of a gas at 20^(@)C and 750 mm pressure. Thegas was transferred to a vessel whose volume is 180 mL. Calculate the pressure of the gas at 20^(@)C.
Answer» Solution :Since a gas completely fills the vessel in which it is contained, thesefore, we have : `V_(1)=120 mL,V_(2)=180 mL`,`P_(1)=750 mm, P_(2)=?mm` Since the temperature REMAINS CONSTANT, therefore, by applying Boyle's Law, `P_(1)V_(1)=P_(2)V_(2)`

Discussion

17.	A vessel of 25 litre capacity contains 10 moles of steam under pressure of 50.3 atm. Calculate the temperature of steam using van der Waals equation (if for water a=5.46 bar L^2 mol^(-2) and b=0.031 L mol^(-1))
Answer» 1539.5K 153.95K 15.395K 1.5395K Answer :A

Discussion

18.	A vessel of 120 mL capacity contains a certain amount of gas at temperature 35^(@)C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at temperature 35^(@)C, What would be its pressure ?
Answer» Solution :When FIXED mass of any gas is POURED to one container to another container then its pressure, volume and temperature is as under. Container - 1st : Initial volume `(V_(1))=120` ML Initial pressure `(p_(1))=1.2` bar Temperature `(T) = 35^(@)C` Container - 2nd : Final volume `(V_(2))=180` mL Temperature (T) `= 35^(@)C` Final pressure `(p_(2))=(?)` Temperature is constant at `35^(@)C` and mass of gas is constant so according to Boyle.s LAW. `p_(1)V_(1)=p_(2)V_(2)` `therefore p_(2)=(p_(1)V_(1))/(V_(2))=(("1.2 bar")xx(120 mL))/(180 mL)=0.8` bar Pressure of container - 2nd would be 0.8 bar.

Discussion

19.	A vessel of 120 mL capacity contains a certain amount of gas at 35^(@)C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35^(@)C. What would be its pressure ?
Answer» <P> Solution :`V_(1)=120 ML,""P_(1)=1.2 bar, "" T_(1)=35^(@)C` `V_(2)=180 mL, ""P_(2)=?, ""T_(2)=35^(@)C` As temperature remains CONSTANT,APPLYING Boyle's law, `P_(1)V_(1)=P_(2)V_(2)` `(1.2"bar")(120 mL)=P_(2)(180 mL) or P_(2)=0.8 "bar"`

Discussion

20.	A vessel of 120 mL capacity contains a certain amount of gas at 35^(@)C and 1.2 bar pressure. The gas 1p transferred to another vessel of volume 180 mL at 35^(@) C. What would be its pressure ?
Answer» Solution :`P_(1)=12` bar `P_(2) =?, v_(1)=120 ML, v_(2)=180 mL` TEMPERATURE contant. According contant. According to Boyle.s law `P_(1)V_(1)=P_(2)V_(2) `or `P_(2) =(P_(1)V_(1))/(V_(2))=(12 "bar" xx12 mL)/(180mL)=0*8 "bar"`.

Discussion

21.	A vessel of 120 mL capacity contains a certain amount of gas at 35 ^(@)C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35 ^(@)C. What would be its pressure?
Answer» SOLUTION :Since temperature REMAINS constant, we have according to Boyle.s LAW, `P_1 V_1 = P_2 V_2` or `1.2 xx 120 = P_2 xx 180` or `P_2 =(1.2xx120)/180 = 0.8`bar

Discussion

22.	A vessel of 10L was filled with 6 mole of Sb_(2)S_(3) and 6 mole of H_(2) to attain the equilibrium at 440^(@)C as: Sb_(2)S_(3)(s)+3H_(2)(g)hArr2Sb(s)+3H_(2)S(g) After equilibrium the H_(2)S formed was analysed by dissolving it in water and treating with excess of Pb^(2+) to give 708 g "of" PbS as precipitate. What is value of K_(c) of the reaction at 440^(@)C?(At. weight of Pb=206).
Answer» `0.08` `0.8` `0.4` `0.04` Solution :MOLE of `PbS=708//236="mole of" H_(2)S` `Sb_(2)S_(3)(s)+3H_(2)(g)hArr2Sb(s)+3H_(2)S(g)` `{:("Initial",6,6,0,0),("at EQ.",5,3,2,3):}` `K_(c)=((3//10)^(3)xx(2//10)^(2))/((5//10)xx(3//10)^(3))=(4)/(50)=0.08`

Discussion

23.	A vessel is filled with a mixture of oxygen and nitrogen . At what ratio of partial pressures will the mass of the gases be indentical ?
Answer» Solution :`PV=nRT=(W)/(M)RT` `P_(O_(2))V=(w_(O_(2)))/(32)xxRT` `P_(N_(2))V=(w_(N_(2)))/(28)xxRT` `:. "" (P_(O_(2)))/(P_(N_(2)))=(w_(O_(2)))/(32)xx(28)/(w_(N_(2)))=(28)/(32)""(because w_(O_(2))=w_(N_(2)))` `=(7)/(8)=7:8`

Discussion

24.	A vessel of irregular shape has a volume 'V'. It is first evacuated and coupled with a vessel of 4L capacity at 35^@C and 10 atm pressure. If the final pressure in both the vessels is 3atm, calculate the volume V.
Answer» Solution :UNKNOWN vesselKnown vessel `V_(1) = VL ""V_(2) = 4L` `P_(1) = 0 "atm" P_(2) = 10` atm The average pressure (P) of the mixture of the GASES from two vessels ` = P = (P_(1)V_(1)+ P_(2)V_(2))/(V_(1) + V_(2))` ` 3 = (0 XX V + 10 xx 4)/(V + 4) = (40)/(V + 4)` Solving, V = 9.3 L The volume of the given vessel of irregular shape is 9.3 L

Discussion

25.	A vessel has N_2 gas saturated with water vapour at a total pressure of 1 atm. The partial pressure of water vapour is 0.3 atm. The contents of this vessel are completely transferred to another vessel having one third of the capacity of the original volume, at the same temperature. The total pressure of this system in the new vessel is
Answer» 3.0 ATM 1 atm 3.33 atm 2.4 atm Solution :`P_(N_2) = P_("dry GAS") + P_(H_2O)` `1 = P_1 + 0.3 implies P_1 = 0.7 atm` on making volume = `(V_1)/3`, PRESSURE `P_2 = 3P_1 = 3 xx 0.7 = 2.1 atm`. `P_(H_2O)` does not change `:.` NEW `P_(N_2) = 2.1 + 0.3 = 2.4 atm.`

Discussion

26.	A vessel contains N_(2)O_(4) & NO_(2) in 2:3 molar ratio at 10 atm under equilibrium. Now, K_(P) for N_(2)O_(4) harr 2NO_(2) is
Answer» 9 ATM 9 `atm^(-1)` 4.5 `atm^(2)` 10 atm Solution :`N_(2)O_(4) HARR 2NO_(2)` `K_(P)=(P_(NO_(2))^(2))/(P_(N_(2)O_(4)))=(((3)/(5) xx 10)^(2))/((2)/(5) xx 10)=9` atm

Discussion

27.	A vessel contains N_(2) O_(4) at certain temperature & 2 atm . If the degree of dissociation of N_(2) O_(4) at equilibrium is 0.2 , final pressure of system (in atm) ?
Answer» `1.2` `2.4` `3.6` `0.8` ANSWER :B

Discussion

28.	A vessel contains methane and oxygen in the mass ratio 2:1. The fraction of the partial pressure of oxygen in the total pressure is
Answer» `1//3` `2//3` `1//4` `1//2` ANSWER :B

Discussion

29.	A vessel contains helium and methane in 4:1 molar ratio at 20 bar pressure. Due to leakage, the mixture of gases starts effusion. Find the composition of the mixture in the initial stage. (Hint: According to Graham's law of effusion, the rate of effusion of a gas in the mixture r prop Partial pressure sqrt((1)/( "molecular mass"))
Answer» SOLUTION :`He:CH_4 =8:1`

Discussion

30.	A vessel contains He and H_(2) in the molar ratio 1 : 5. The ratio of mean transitional kinetic energies, at the same tempreatures is
Answer» `1 : 1` `1 : 2` `2 : 1` `1 : 5` Solution :(d) `KE=(3)/(2)PV=(3)/(2)nRT` `((KE)He)/((KE)H_(2))=((3)/(2)RT)/((3)/(2)RT)((1)/(5))=1 : 5`

Discussion

31.	A vessel contains helium and methane in 4:1 mole ratio at 20 bar pressure. Find the ratio of initial rates of effusion through a small aperture.
Answer» <P> Solution :Ratio of moles of He and `CH_(4)` = 4 : 1 Ratio of partialpressure of He and `Ch_(4) = 4 :1` Total pressure of the mixture = 20 bar Partial pressure of He ` = 20 xx (4)/(5) = 16` bar partial pressure of `CH_(4) = 20 xx (1)/(5) =4 ` bar Ratio of intial RATES of effusion is ` (r_(1))/(r_(2)) = (P_(1))/(P_(2)) sqrt((M_(2))/(M_(1))) = (16)/(4) sqrt((16)/(4)) = (4)/(1) sqrt((4)/(1))` The ratio of INITIAL rates of effusion of gases, helium and methane ` =4 sqrt(4) : 1 = 8 : 1`

Discussion

32.	A vessel contains equal weights of He and CH_(4)gases at 20 bar pressure. Due to the leakage, the gases in the vessel started effusing out. What is the volume ratio of He and CH_(4)gases coming out initally
Answer» <P> SOLUTION :`n_(He) = x/4 = n_1 , n_(CH_4) = x/16 = n_2 , (n_1)/(n_2) = 4/1`. `(gamma_(He))/(gamma_(CH_4)) = (P_(He))/(P_(CH_4)) SQRT(16/4) = 4/1 sqrt(16/4) = 8:1`.

Discussion

33.	A vessel contains equal volumes of SO_2 and CH_4Through a small hole, the gases effused into vacuum. After 200sec, the total volume is reduced to half. What is the ratio of SO_2and CH_4remaining in the vessel?
Answer» `1 :2` `2 :1` `1 :1` `1 :4` ANSWER :B

Discussion

34.	A vessel contains equal number of moles of helium and methane. Through a small orifice the half of gas effused out. The ratio of the number of mole of helium and methane remaining in the vessel is
Answer» `2 :1` `1:2` `1:4` `4:1` ANSWER :B

Discussion

35.	A vessel contains equal masses of helium and oxygen at 35^@C. What is the ratio of kinetic energies of the components?
Answer» ANSWER :`8:1`

Discussion

36.	A vessel contains a mixture of equal weights of oxygen and SO_2 at a pressure of 600mm of Hg. The partial pressure of oxygen in mm is
Answer» 200 300 400 100 ANSWER :C

Discussion

37.	A vessel contains a mixture of equal weights of methane and sulphur dioxide al pressure of 600 mm. Find the partial pressures of each of these gases in the mixture.
Answer» <P> SOLUTION :`p_(CH_4) = 480mm ,P_(SO_2) = 120 MM`

Discussion

38.	A vessel contains a mixture of equal masses of helium and oxygen at a pressure of 600 torr. Calculate the partial pressures of components in the mixture.
Answer» Solution :Let the mass of each component gas is x. Total number of moles in the mixture = `(XG " of " He)/(4G " of " mol^(-1)) + (xg " of " O_2)/(32 g " of " mol^(-1)) = (9x)/(32 g mol^(-1))` Partial pressure = pressure of the mixture x mole FRACTION Partial pressure of He `= 600 xx (x//4)/(9x//32) = 600 xx 8/9` `= 5333.33"TORR"` Partial pressure of `O_2 = 600 xx (x//32)/(9x//32)` `= 600 xx 1/9 = 66.67` torr (or) Partial pressure of `O_2 = 600 - 533.37 ` = 66.67 torr

Discussion

39.	A vessel contains a mixture of different types of gases. Which of the following satements is correct?
Answer» On the average, the heavier molecules have higher molecules have higher speed The average transaltional energy of DIFFERENT TYPES of molecules is the same The average speed of different molecules is same On an average, the heavier moleules have higher TRANSITIONAL energy Solution :(b) The average TRANSLATIONAL energy of different types of molecules is the same because it depends only on temperature.

Discussion

40.	A vessel contains 1.6 g of dioxygen at STP (273.15 K, 1 atm pressure). The gas is now transferred to another vessel at constant temperature, where pressure becomes half of the original pressure. Calculate (A) volume of the new vessel. (B) number of molecules of dioxygen.
Answer» Solution :(A) `p_(1)=1` atm, `p_(2)=1/2 = 0.5` atm, `T_(1)=273.15, V_(2)=(?)` 32g of dioxygen occupies `=22.4 L` volume at STP `:.1.6g` dioxygen will occupy `=(22.4Lxx1.6g)/(32g)` `=1.12L` `V_(1)=1.12L` From Boyle.s law `p_(1)V_(1)=p_(2)V_(2)` `V_(2)=(p_(1)V_(1))/(p_(2))= (1 "atm"xx1.12L)/(0.5 "atm")=2.24L` Number of moles of dioxygen `=("MASS of dioxygen")/("Molar mass of dioxygen")` `n_(O_(2))=(1.6)/(32)=0.05` mol 1 mol of dioxygen contains `= 6.022xx10^(23)` molecules `:.` 0.5 mol of dioxygen CONTAIN `= 0.05xx6.022xx10^(23)` molecule `=3.011xx10^(22)` molecules

Discussion

41.	A vessel contains 100 litres of liquid X. Heat is suppplied to the liquid in such a fashion that, Heat given = change in ethalpy. The volume of the liquid is incresed by 2 litres. If the external pressure is one atm, and 202.6joules of heat were supplied then, [U= total internal energy] :
Answer» `DeltaU=0,DeltaH=0` `DeltaU=+202.6J,DeltaH=+202.6J` `DeltaU=-202.6J,DeltaH=-202.6J` `DeltaU=0,DeltaH=+202.6J` ANSWER :D

Discussion

42.	A vessel contains 100 litres of a liquid x. Heat is suppllied to the liquid in such a fashion that, Heat given = change in enthalpy. The volume of the liquid increases by 2 lites. If the external pressure is one atm and 202.6 Joules of heat were supplied then,
Answer» `Delta U = 0, Delta H = 0` `Delta U = +202.6J, Delta H = +202.6 J` `Delta U = -202.6J, Delta H = - 202.6J` `Delta U = 0, Delta H = + 202.6J` Solution :from the data `Delta H= Q= 202.6J` from first law of T. D `Delta U = Q + W` `Delta U = 202.6 + (-P Delta V)` `=202.6 + (-1(2) xx 101.3J) = 0`

Discussion

43.	A vessel contains 1 mole PCI_5 (g) at 4 atm and 0.5 mole PCI_3 formed at equilibrium. Now, equilibrium pressure of mixture is (assume ideal behavior)
Answer» 16 ATM 6 atm 2 atm 4.5 atm Solution : given X = 0.5 total moles of equilibrium = 1+ x =1.5 moles `alpha` pressure `1 to 4`atm `1.5 to ? implies` pressure = 6atm

Discussion

44.	A vessel contains 0.25 moles of Helium and0.15 moles of neon at 298K and 2.4 atmosphere pressure. Calculate the partial pressure of each gas in the mixture.
Answer» Solution :TOTAL PRESSURE .P.=2.4atm No. of moles of He=0.25 No. of moles of Ne=0.15 Total moles of n=0.25+0.15=0.4 Partial pressure = Total pressure X Mole-fraction `P_(He) = (n_(He))/(n) xx P = (0,.25)/(0.4) xx 2.4 = 1.5 atm` `P_(N e) = (n_(N e))/(n) xx P = (0.15)/(0.4) xx 2.4 = 0.9 atm`.

Discussion

45.	A vesselat 1000 K containsCO_(2) with a pressure of 0.5 atm . Some of the CO_(2) is converted intoCO on the addition of graphite . If the total pressure at equilibrium is 0*8 atm , the value of K is
Answer» 3 atm `03 `atm `018 atm` `1*8 ` atm Solution :`{:(,CO_(2)(G),+,C(s),hArr,2CO(g)),(" INTIAL ",0.5 atm,,,,),("At eqm.",0.5 -p,,,,2 p):}` `:. (0.5 -p) +2 p = 0.8" atm "or p=0.3 " atm "` `K=(p^(2) CO)/p_(CO_(2)) = (0.6)^(2)/((0.5 -0.3)) = (0.6 xx0.6)/0.2 = 1.8 " atm", `

Discussion

46.	A vessel at 1000 K contains CO_(2) with a pressure of 0.5 atm. Some of CO_(2) is converted into CO on addtion of graphite. The value of K' at equilibrium when total pressure is 0.8 atm will be
Answer» 0.18 atm 1.8 atm 2 atm 1 atm Solution :`underset(0.5)(CO_(2))+C_((s)) harr 2CO` at EQM 0.5-P`""2P` P=0.3 `K=((0.6)^(2))/((0.2))=1.8`atm

Discussion

47.	A vessel at 1000 K contains CO_(2) at 0.4 atm . If certain amount of CO_(2) is converted into CO by the addition of graphite , the following equilibrium is established CO_(2) + C hArr 2CO , at a pressure of 0.6 atm, then equilibrium constant is
Answer» 0.2 ATM 0.05 atm 0.8 atm 0.15 atm ANSWER :C

Discussion

48.	A vessel at 1000 K contains carbon dioxide with a pressure of 0.5 atm. Some of the carbon dioxide is converted to carbon monoxide on addition of graphite. Calculate the value of K it the total pressure at equilibrium is 0*8 atm.
Answer» Solution :`CO_(2) (G) + C (s) hArr 2 CO (g)` Suppose decreasein pressure of `CO_(2)` after reaction = p atm Then increase in pressure due to CO= 2 p Pressure of `CO_(2)` at equilibrium `=(0.5 - p)` atm `:." Final TOTAL pressure " (0.5- p) + 2p = 05 + p = 0.8 " atm"`(Given) `:. p= 03 " atm.Hence we have "` `p_(CO_2) = 05 - 03 = 02 " atm "` `and ""p_(CO) = 2 xx 03 = 06 " atm" ` `:. K = (P_(cO)^(2))/P_(CO_(2)) = (0.6)^(2)/(02) = 1*8`

Discussion

49.	A vessel (A) contains I mole each of N_(2) & O_(2) and another vesser (B) contains 2 mole each of N_(2)& O_(2) . 3oth vessels are heated to same temperature till equilibrium is established in-both cases. Then, correct statement is
Answer» `K_c` for `N_2+O_2 harr 2NO` in A & B are in the ratio 1:2 `K_p` for `N_2+O_2 harr 2NO` in A & B are in ratio of 1:2 `K_c` for `N_2+O_2 harr 2NO` in A & B are equal `K_p` for `N_2+O_2 harr 2NO` in A & B are in ratio of 2:1 Solution :`K_(1)` does not depend on concentrations of REACTANTS

Discussion

50.	A very small amoutn of radioactive isotope of Pb^(213) was mixed with a non-radioactive lead salt containg 0.01g of Pb (atomic mss 207). The whole lead was brought into solution and lead chromate was prcipitated by addition of a soluabolr chromate. Evaporation of 10cm^(3) fo the supernatentliquid gave a residue having a radioactivity (1)/(24000) of that of the original quanity if Pb^(213) caculate the solubility of lead chromate in mol dm^(-3).
Answer» ANSWER :`2.0xx10^(-7)MOL DM^(-3)`

Discussion

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A weak acid HX has the dissociation constant 1xx10^(-5)M. It forms a salt NaX on reaction with alkali. The degree of hydrolysis of 0.1 M solution of NaX is

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A water sample is said to contain permanent hardness if water contains

A water sample has ppm level concentration of following anions F^(-)=10 , SO_4^(2-)=100 , NO_3^(-)=50 The anion/anions that make/makes the water sample unsuitable for drinking is/are

A water sample has ppm level concentration of following anions. F^(-) = 10, SO_(4)^(2-) = 100, NO_(3)^(-) = 50 The anion/anions that make/makes the water sample unsuitable for drinking is/are……..

A water sample has ppm level concentration of following anions F^(-)=10,SO_(4)^(2-)=100,NO_(3)^(-)=50 The anion/anions that make/makes the water sample unsuitable for drinking is/are:

(A) : Water is a liquid whereas sulphurdi oxide is a gas at room temp (R) : Molecular mass of SO_2 is more than that of H_2 O

(a) W hat do ou mean by catensation? (b) C Cl_(4) cannot be hydrolysed, but SiCl_(4) can be easily hydrolysed, Why?

(A): Volume to temperature ratio is constant for a fixed amount of gas at constant pressure. (R) : At constant pressure the volume of a given mass of a gas increases or decreases by 1/273 times of its volume at 0^@C, for every 1^@C change in temp.

A vey high cosmic abundance of Fe is due to

A vessel with small opening contained equal volumes of oxygen and an unknown gas. Oxygen effused through the opening 1.8 times faster than the unknown gas. If the atomic mass of oxygen is 16, calculate the molecular mass of the unknown gas.

A vessel of 120 mL capacity contains a certain mass of a gas at 20^(@)C and 750 mm pressure. Thegas was transferred to a vessel whose volume is 180 mL. Calculate the pressure of the gas at 20^(@)C.

A vessel of 25 litre capacity contains 10 moles of steam under pressure of 50.3 atm. Calculate the temperature of steam using van der Waals equation (if for water a=5.46 bar L^2 mol^(-2) and b=0.031 L mol^(-1))

A vessel of 120 mL capacity contains a certain amount of gas at temperature 35^(@)C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at temperature 35^(@)C, What would be its pressure ?

A vessel of 120 mL capacity contains a certain amount of gas at 35^(@)C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35^(@)C. What would be its pressure ?

A vessel of 120 mL capacity contains a certain amount of gas at 35^(@)C and 1.2 bar pressure. The gas 1p transferred to another vessel of volume 180 mL at 35^(@) C. What would be its pressure ?

A vessel of 120 mL capacity contains a certain amount of gas at 35 ^(@)C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35 ^(@)C. What would be its pressure?

A vessel is filled with a mixture of oxygen and nitrogen . At what ratio of partial pressures will the mass of the gases be indentical ?

A vessel of irregular shape has a volume 'V'. It is first evacuated and coupled with a vessel of 4L capacity at 35^@C and 10 atm pressure. If the final pressure in both the vessels is 3atm, calculate the volume V.

A vessel contains N_(2)O_(4) & NO_(2) in 2:3 molar ratio at 10 atm under equilibrium. Now, K_(P) for N_(2)O_(4) harr 2NO_(2) is

A vessel contains N_(2) O_(4) at certain temperature & 2 atm . If the degree of dissociation of N_(2) O_(4) at equilibrium is 0.2 , final pressure of system (in atm) ?

A vessel contains methane and oxygen in the mass ratio 2:1. The fraction of the partial pressure of oxygen in the total pressure is

A vessel contains He and H_(2) in the molar ratio 1 : 5. The ratio of mean transitional kinetic energies, at the same tempreatures is

A vessel contains helium and methane in 4:1 mole ratio at 20 bar pressure. Find the ratio of initial rates of effusion through a small aperture.

A vessel contains equal weights of He and CH_(4)gases at 20 bar pressure. Due to the leakage, the gases in the vessel started effusing out. What is the volume ratio of He and CH_(4)gases coming out initally

A vessel contains equal volumes of SO_2 and CH_4Through a small hole, the gases effused into vacuum. After 200sec, the total volume is reduced to half. What is the ratio of SO_2and CH_4remaining in the vessel?

A vessel contains equal number of moles of helium and methane. Through a small orifice the half of gas effused out. The ratio of the number of mole of helium and methane remaining in the vessel is

A vessel contains equal masses of helium and oxygen at 35^@C. What is the ratio of kinetic energies of the components?

A vessel contains a mixture of equal weights of oxygen and SO_2 at a pressure of 600mm of Hg. The partial pressure of oxygen in mm is

A vessel contains a mixture of equal weights of methane and sulphur dioxide al pressure of 600 mm. Find the partial pressures of each of these gases in the mixture.

A vessel contains a mixture of equal masses of helium and oxygen at a pressure of 600 torr. Calculate the partial pressures of components in the mixture.

A vessel contains a mixture of different types of gases. Which of the following satements is correct?

A vessel contains 1.6 g of dioxygen at STP (273.15 K, 1 atm pressure). The gas is now transferred to another vessel at constant temperature, where pressure becomes half of the original pressure. Calculate (A) volume of the new vessel. (B) number of molecules of dioxygen.

A vessel contains 100 litres of liquid X. Heat is suppplied to the liquid in such a fashion that, Heat given = change in ethalpy. The volume of the liquid is incresed by 2 litres. If the external pressure is one atm, and 202.6joules of heat were supplied then, [U= total internal energy] :

A vessel contains 100 litres of a liquid x. Heat is suppllied to the liquid in such a fashion that, Heat given = change in enthalpy. The volume of the liquid increases by 2 lites. If the external pressure is one atm and 202.6 Joules of heat were supplied then,

A vessel contains 1 mole PCI_5 (g) at 4 atm and 0.5 mole PCI_3 formed at equilibrium. Now, equilibrium pressure of mixture is (assume ideal behavior)

A vessel contains 0.25 moles of Helium and0.15 moles of neon at 298K and 2.4 atmosphere pressure. Calculate the partial pressure of each gas in the mixture.

A vesselat 1000 K containsCO_(2) with a pressure of 0.5 atm . Some of the CO_(2) is converted intoCO on the addition of graphite . If the total pressure at equilibrium is 0*8 atm , the value of K is

A vessel at 1000 K contains CO_(2) with a pressure of 0.5 atm. Some of CO_(2) is converted into CO on addtion of graphite. The value of K' at equilibrium when total pressure is 0.8 atm will be

A vessel at 1000 K contains CO_(2) at 0.4 atm . If certain amount of CO_(2) is converted into CO by the addition of graphite , the following equilibrium is established CO_(2) + C hArr 2CO , at a pressure of 0.6 atm, then equilibrium constant is

A vessel at 1000 K contains carbon dioxide with a pressure of 0.5 atm. Some of the carbon dioxide is converted to carbon monoxide on addition of graphite. Calculate the value of K it the total pressure at equilibrium is 0*8 atm.

A vessel (A) contains I mole each of N_(2) & O_(2) and another vesser (B) contains 2 mole each of N_(2)& O_(2) . 3oth vessels are heated to same temperature till equilibrium is established in-both cases. Then, correct statement is