52671 + Interview Questions in Chemistry in Class 11

1.	A sample of an ideal gas with initial pressure 'P' and volume 'V' is taken through an isothermal process during which entropy change is found to be DS. The work done by the gas is
Answer» `(PV Delta S)/(NR)` `n R Delta S` PV `(P Delta S)/(NRV)` Solution :`Delta S = nR l n (V_(2))/(V_(1)), W = nRT l n (V_(2))/(V_(1))` `rArr W = T. Delta S = Delta S xx (PV)/(nR)`

Discussion

2.	A sample of an ideal gas has a volume of 0.5 litres at 27^@C and 750mm pressure. The number of moles of gas are
Answer» 0.02 0.2 2 0.002 ANSWER :A

Discussion

3.	A sample of air contains Nitrogen, Oxygen and saturated with water vapour under a total pressure of 640 mm. If the vapour pressure of water at that temperature is 40 mm and the molecular ratio of N_(2): O_(2) is 3:1, the partial pressure of Nitrogen in the sample is
Answer» <P>480 mm 600 mm 450 mm 160 mm SOLUTION :`D_(N_2) = (D_(N_2))/(N) cdot P_("Total") implies P_(N_2) = 3/4 xx 600 = 450 mm`.

Discussion

4.	A sampleof air consisting of N_(2) and O_(2) was heated to 2500 K until the equilibrium N_(2) (g) + O_(2) (g) hArr 2 NO (g) was established the intial composition of air in mole fraction of N_(2) and O_(2).
Answer» Solution :`{:(,N_(2)(g),+,O_(2),hArr,2NO,),("Intial ",a,,b,,0,),(,,,,,,a+b = 100"" ...(ii)),("At eqm",a-x,,b-x,,2x,):}` ` K_(c)= (2x)^(2) /((a-x)(b-x)) = (4x^(2))/((a-x)(b-x))` In the question , we are given `2x = 18 ("because total moles " =100)` or ` x= 09 and K_(c) = 21 XX 10^(-3) ` ` :. 21 xx 10^(-3) = (18)^(2)/((a-09 )(b-09)) ` ` ab - 09 a- 09 b + 081 = 1620` ` ab - 09 (a+b) + 081 = 1620 ` ` ab- 09 xx 100 + 081 = 1620 ` ` or ab = 1909 * 19 = 1709` ` Now(a-b)^(2) = (a+b)^(2) - 4 ab = (100)^(2)- 4 xx 709 = 3164 ` or ` a-b = sqrt (3164)= 562 "" `...(ii) Solving (i) and (ii), `a= 781 "moles"` Moles fraction of` N_(2) = (781 )/100 = 0 781 ` Mole fraction of ` O_(2) = 1 - 0* 781 = 0* 219 `

Discussion

5.	A sample of AgCl was treated with 5.00 ml of 1.5 M Na_(2)CO_(3) solution to give Ag_(2)CO_(3). The remaining solution contained 0.0026 g of Cl^(-) per litre. Calculate the solubility product of AgCl (K_(sp) "for" Ag_(2)CO_(3)=8.2xx10^(-2))
Answer» SOLUTION :`1.5 M Na_(2)CO_(3) ` gives `[CO_(3)^(2-)]=1.5M` `K_(sp)` for `Ag_(2)CO_(3)=[Ag^(+)]^(2)[CO_(3)^(2-)]` `:. [Ag^(+)]=SQRT((K_(sp) "for" Ag_(2)CO_(3))/([CO_(3)^(2-)]))=sqrt((8.2xx10^(-12))/(1.5))=2.34xx106(-6)M` `K_(sp) "for" AgCl=[Ag^(+)][CL^(-)]=(2.34xx10^(-6))((0.0026)/(35.5))=1.71xx10^(-10)`

Discussion

6.	A sample of AgCl was treated with 5 mL of 1.5 M Na_2 CO_3 solution to give Ag_2 CO_3. The remaining solution contained 0.0026 g of chloride per litre. If K_(sp) of Ag_2 CO_3is 8.2 xx 10^(-12), what is K_(sp) of AgCl ?
Answer» SOLUTION :`1.7 XX 10^(-10)`

Discussion

7.	A sample of a hydrate of barium chloride weighing 61g was heated until all the water of hydration is removed. The dried sample weighted 52g. The formular of the hydrated salt is: (Atomic mass, Ba = 137 amu, Cl = 35.5 amu)
Answer» `BaCl_(2). 2H_(2)O` `BaCl_(2).4H_(2)O` `BaCl_(2).H_(2)O` `BaCl_(2).3H_(2)O` Answer :A

Discussion

8.	A sample of a gas was heated from 30^(@)C to 60^(@)C at constant pressure. Which of the following statement(s) is true?
Answer» KINETIC ENERGY of the GAS is doubled Boyle's law will appply Volume of the ga will be doubled None of the above Answer :d

Discussion

9.	A sample of a gas occupies 650.0 cm at 100^@C. At what temperature the gas will occupy a volume of 1050.0 cm^3if the pressure is kept constant ?
Answer» Solution :In the present case, `V_1 = 650.0 cm^3 "" V_2 = 1050.0 cm^3` `T_1 = 100^@C = 100 + 273 = 373 K, "" T_2` = ? The PRESSURE remains constant during expansion. Hence, ACCORDING to Charles. law `V_1/T_1 = V_2/T_2` or `T_2 = (V_2 T_1)/(V_1)=(1050.0 xx 373)/(650.0)= 602.5 K` Hence, the required TEMPERATURE is `602.5 - 273 = 329.5^@C`.

Discussion

10.	A sample of a gas has a volume of 8.5dm^(3) at an unknown tempreture, When the sample is submerged in ice water at 0^(@)C, its volume gets reduce to 6.37dm^(3) .What is its intial tempreture?
Answer» Solution :`V_(1)=8.5dm^(3)``V_(2)=6.37dm^(3)` `T_(1)=?``T_(2)=0^(@)C=273K` `(V_(1))/(T_(1))=(V_(2))/(T_(2))` `V_(1)XX((T_(2))/(V_(2)))=T_(1)` `T_(1)=8.5cancel(dm^(3))xx(273 K)/(6.37cancel(dm^(3)))``T_(1)=364.28K`

Discussion

11.	A sample of a gas is found to occupy a volume of 800 cm^(3) "at" 27^(@) celsius. Calculate the temperature at which it will occupy a volume of 400 cm^(3), provided the pressure is kept constant.
Answer» SOLUTION :Here `v_(1)=800 cm^(-2) "" v_(2)= 400 cm^(3)` `T_(1)=(27+273)K=300K "" T_(2)= ?` Applying Charle.s law, `(V_(1))/(T_(1))=(V_(2))/(T_(2))` `T_(2)=(V_(2)xxT_(1))/(V_(1))=(400cm^(3)xx300K)/(800 cm^(3))=150 K` `=150-273=-123^(@)C` `:. T_(2)=-123^(@)C`

Discussion

12.	A sample of .^(14)CO_(2) was mixed with ordinary.^(12)CO_(2) for stuydinga biologicaltracerexperiment. The10mL of ths mixture at STP possess the rate of 10^(4) disintegration per minute. How manymilli-curie of radioactive carbon is needed to prepare 60 litre of such a mixture?
Answer» ANSWER :`0.27 MCI`

Discussion

13.	A sample of ._(131)^(53) I is iodide ion was admistered to a patient in a carrier consisting of 0.10 mg of stable iiodide ion. After 4 day 67.7% of initial radioactivity was detected in the throidgland of the patient. What mass of stable iodide ion hadmigrated to thyroid glad? (t_(1//2) for iodide ion = 8 day)
Answer» Solution :If `t = 4`day, `lambda = (0.693)/(8)`, then since `r_(0) prop N_(0)` and `r prop N` `:. (r_(0))/(r) = (N_(0))/(N)` `:. t = (2.303)/(lambda) log (r_(0))/(r)` `4 = (2.303xx8)/(0.693) log (1)/(0.693) log(r_(0))/(r)` `r = 0.707 r_(0)` Thus iodide ION left is `0.707` part of intially injected sample, HOWEVER the RATE decreases only `67.7%` or `0.688` in 4 days, thus If `0.707` is left then iodidie ion migrated to thryorid `=1` Thus `0.677`is left than iodide ion migrated to thyroid `= (1xx0677)/(0.707) = 0.958` or `95.8%` of the iodide ion is migrated to gland.

Discussion

14.	A sample of 12 M concentrated hydrochloric acid has a density 1.2g L ^(-1)Calculated the molality.
Answer» Solution :Given: Molarity `= 12 MHCl` DENSITY of the solution `=1.2 G L ^(-1)` In 12 M HCL solution, there are 12 moles of HCl in 1 litre of the solution. Molality `= ("No. of moles of solute")/("Mass of solvent (in kg)")` Caculate mass of water (solvent) Mass of 1 litre HCl solution = density `xx` VOLUME `=1.2 GM L ^(-1) xx 100 mL = 1200g` Mass of HCl = No of moles of HCl` xx` molar mass of HCl Mass of water= mass of HCl solution - mass of HCl Mass of water `=1200 - 438 =762 g` Molality `= (12)/(0.762) =15.75m`

Discussion

15.	A sample of 1.0 mol of a monoatomic ideal gas is taken through a cyclic process of expansion and compression as shown in figure. What will be the value of Delta H for the cycle as a whole ?
Answer» Solution :The net ENTHALPY CHANGE, `DELTA H` for a cyclic process is zero as enthalpy change is a state function, i.e., `Delta H` (CYCLE) `= 0`

Discussion

16.	A sample of 0.50 g of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in 50 mL of 0.5 M H_(2)SO_(4). The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound.
Answer» Solution :Step 1. To determine the volume of `H_(2)SO_(4)` USED. Volume of acid taken = 50 mL of 0.5 M `H_(2)SO_(4) = 25 mL` of 1 M `H_(2)SO_(4)` Volume of alkali used for neutralization of excess acid = 60 mL of 0.5 M NaOH = 30 mL of 1 M NaOH. Now 1 mole of `H_(2)SO_(4)` NEUTRALIZES 2 moles of NaOH (i.e. `H_(2)SO_(4) + 2NaOH rarr Na_(2)SO_(4) + 2H_(2)O`) `:.` 30 mL of 1 M `NaOH -= 15 mL` of 1 M `H_(2)SO_(4)` `:.` Volume of acid used by ammonia = 25 - 15 = 10 mL Step 2. To determine percentage of nitrogen. Again 1 mole of `H_(2)SO_(4)` neutralizes 2 moles of `NH_(3) :. 10 mL` of 1 M `H_(2)SO_(4) -= 20 mL` of 1 M `NH_(3)` But 1000 mL of 1 M `NH_(3)` contain nitrogen = 14G `:.` 20 mL of 1 M `NH_(3)` will contain nitrogen `= (14)/(1000) xx 20 g` But this much amount of nitrogen is present in 0.5 g of the organic compound. `:.` Percentage of nitrogen `= (14)/(1000) xx (20)/(0.5) xx 100 = 56.0`. Alternatively, % of N can be determined by applying the following equation, `% N = (1.4 xx "Molarity of the acid" xx "Basicity of the acid" xx "Vol. of the acid used")/("Mass of substance taken")` Substituting the values of all the items in the above equation, we have, `% N = (1.4 xx 1 xx 2 xx 10)/(0.5) = 56.0`

Discussion

17.	A sample of 0.50g of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in 50ml of 0.5M H_(2)SO_(4). The residual acid required 60mL of 0.5M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound
Answer» SOLUTION :Initially taken `H_(2)SO_(4)= 50mL 0.5M` used NaOH = 60 mL 0.5M `:.` unused `H_(2)SO_(4)= 30 mL 0.5M H_(2)SO_(4)` used `H_(2)SO_(4)= (50-30)= 20 mL 0.5 H_(2)SO_(4)` `=40 mL 0.5M NH_(3)` `1000 mL 1M NH_(3)= 17g NH_(3)=14g` nitrogen 40mL 0.5M `NH_(3)=` `=(40 mL XX 0.5M xx 14g)/(1000mL xx 1M)= 0.280` g nitrogen % N `= (0.280 xx 100)/(0.5)= 56` `:.` % of nitrogen = 56%

Discussion

18.	A sample of 0.5g of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in 50 mL of 0.5M H_(2)SO_(4) . The remaining acid after neutralisation by ammonia consumed 80 mL of 0.5 M NaOH, The percentage of nitrogen in the organic compound is ............. .
Answer» 0.14 0.28 0.42 0.56 Answer :B

Discussion

19.	A sample of 0.50 g of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in 50 mL of 0.5 M H_(2)SO_(4). The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. What would be the percentage composition of nitrogen in the compound?
Answer» 50 60 56 44 Solution :Given: Mass of compound taken=0.50g VOL. of `H_(2)SO_(4)=50`mL Molarity of `H_(2)SO_(4)=0.5M` Vol. of NaOH required=60mL Molarity of NaOH requried=0.5M METHOD adopted Kjeldahl's method Formula used: % of N`=(1.4xxMxx2[V-(V_(1))/(2)])/(m)`. .(i) by substituting the vlaues in the formula, we GET `%` of `N=(1.4xx0.5xx2(50-60//2))/(0.5)=56` `therefore` % of N in the given compound=56%

Discussion

20.	A sample of 0.50 g of an organic compound was treated according to Kjeldahl's method. Ammonia evolved was absorbed in 50 mL of 0.5 M H_2SO_4. The residual acid required 60 mL or 0.5 M NaOH solution for neutralisation. Find the percentage composition of nitrogen in the compound.
Answer» ANSWER :`50 mL of 0.5 M H_2SO_4 = 50 mL 1N H_2SO_4` `60 mL 0.5 M NaOH = 60 mL 0.5 N NaoH = 60 XX 0.5 = 30 mL 1N NaOH` `30mL 1N NaOH solution = 30 mL 1N H_2SO_4` VOL. of ACID used up = 50-30 = 20 mL 1N `H_2SO_4` Percentage `N_2 = (14)/1000 xx (20)/0.5 xx 100 = 5.6%`

Discussion

21.	A sample of 0.5 g of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in 50 mL of 0.5 M H_(2)SO_(4). The remaining acid after neutralisation by ammonia consumed 80 mL of 0.5 M NaOH. The percentage of nitrogen in the organic compound is
Answer» 14 28 42 56 Solution :Volume of acid TAKEN = 50 mL of 0.5 `M H_(2)SO_(4)` Let the volume of the acid left unused = v mL of M/10 `H_(2)SO_(4)` APPLYING molarity equation, `n_(a)M_(a)V_(a)` (acid) `= n_(b)M_(b)V_(b)` (base), we have, `2 XX 0.5 xx v = 1 xx 0.5 xx 80` or v = 40 mL `:.` Volume of the acid used = 50 -40 = 10 mL of 0.5 M `H_(2)SO_(4)` Now `%N = (1.4 xx n_(a)M_(a)V_(a))/("wt. of substance taken")` `= (1.4 xx 2 xx 0.5 xx 10)/(0.5) = 28`

Discussion

22.	A sample contains two radioactive nuclie x and y with half-lives2 hour and 1 hour respectively. The nucleus x-decays to y and y-decays into a stable nucelus z.At t = 0, the activates of the components in the samewere equal. Find the ratio of the number of the active , nuclei of y at t = 4 hoursto the numberat t = 0.
Answer» ANSWER :`0.25`

Discussion

23.	A sample containing 0.496 gm of (NH_(4))_(2) C_(2)O_(4) (MW = 124) and inert material was dissolved in water and made strongly alkaline with KOH which converts NH_(4)^(+) into NH_(3). The liberated NH_(3) was distilled into exactly 50ml of 0.05M H_(2)SO_(4). The excess H_(2)SO_(4) was back titrated with 10ml of 0.1MNaOH. The percentage of (NH_(4))_(2) C_(2)O_(4) with sample is
Answer» `40%` `50%` `60%` `75%` Solution :m. eqts of EXCESS `H_(2)SO_(4) = 10 XX 0.1 = 1` m.eqts of `H_(2)SO_(4)` taken `= 50 xx 0.05 xx 2 = 5` m.eqts of `H_(2)SO_(4)` reacted with `NH_(4) = 4=` m.eqts of `(NH_(4))_(2) C_(2)O_(4)` wt of `(NH_(4))_(2)C_(2)O_(4) = 4 xx 62 xx 10^(-3)` % purity `=(4 xx 0.062)/(0.496) xx 100 = 50%`

Discussion

24.	A sample containing 0.4775 of (NH_(4))_(2)C_(2)O_(4) and inert material was dissolved in water and made strongly alkaline with KOH which converted NH_(4)^(o+) to NH_(3) The liberated NH_(3) was distilled of H_(2)SO_(4) was back titrated with 11.3 " mL of " 0.1214 M NaOH. Calculate (a) % of (NH_(4))_(2)C_(2)O_(4)=124.10 And atomic weight of N=14.0078.
Answer» Solution :`(NH_(4))_(2)C_(2)O+(4)+2KOHtoK_(2)C_(2)O_(4)+2NH_(3)+2H_(2)O` Total `H_(2)SO_(4)` USED `=50xx0.05035xx2NH_(2)SO_(4)` `=5.035m" Eq of "H_(2)SO_(4)` Excess of `H_(2)SO_(4)=11.3xx0.124M NaOH` `=1.372 m" Eq of " NaOH` `=1.372 m" Eq of "H_(2)SO_(4)` `H_(2)SO_(4)` used `=5.035-1.372` `=3.663 m" Eq of "H_(2)SO_(4)` `=3.663 MEQ NH_(3)` `=3.663 m" Eq of "(NH_(4))_(2)C_(2)O_(4)` Ew of `(NH_(4))_(2)C_(2)O_(4)=(124.1)/(2)=62.06g` Weight of `(NH_(4))_(2)C_(2)O_(4)=3.663xx10^(-3)xx62.05=0.2273g` `124.1 g of (NH_(4))_(2)C_(2)O_(4)` CONTAINS `14.0078 g of N`. `0.2273 g of (NH_(4))_(2)C_(2)O_(4)=(14.0078xx0.2273)/(124.1)=0.02565` g `% of N=(0.02565)/(0.4775)xx100=5.373%` `% of (NH_(4))_(2)C_(2)O_(4)=(0.2273)/(0.4775)xx100=5.373%`

Discussion

25.	A samll amount of solution containing Na^(24) radio nuclide with activity A = 2xx10^(3) dps was administeredinto blood of a patientin a hospital. Afer 5 hour a sample of the blooddrawn out form the patientshowed an activity of 16 dpm per cc t_(1//2) for Na^(24) = 15 hr. Find: (a) Volume of the blood in the patient. (b) Activity fo blood sample drawn after a further time fo 5 hr.
Answer» Solution :Let `V mL` blood is present in patient. (a) `r_(0)` pf `Na^(24) = 2xx10^(3) DPS = 2xx10^(3) xx 60 dp m` `= 120xx10^(3) dp m` for `V mL` blood `r` of `Na^(24) = 16 dp m//mL` at `t = 5 HR` `= 16xx V dp m//V mL` `:' (r_(0))/(r) = (N_(0))/(N)` `:. (N_(0))/(N) = (120xx10^(3))/(16V)` `:. t = (2.303)/(lambda) log_(10) ((N_(0)))/(N)` `5 = (2.303xx15)/(0.693) log_(10)((120xx10^(3)))/(16V)` `:. V = 5.95 xx10^(3) mL` (b) Activity of blood sample after`5 hr` more i.e., `t = 10 hrt` `t = (2.303)/(lambda) log_(10) (N_(0))/(N)` `10 = (2.303xx15)/(0.693) log_(10) (120xx10^(3))/(A)` `:. A = 75.6xx10^(3) dp m` per `5.95 xx10^(3)mL` `= (75.6xx10^(3))/(5.95xx10^(3)) = 12.71` dpm per `mL`. `= 0.2118` dpm per `mL`.

Discussion

26.	(A) : Salts of Mg does not impart any colour to the flame (R): Due to small size and high effective nuclear charge, 'Mg' requires a large amount of energy for excitation of electrons.
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. R is False but A is True. ANSWER :A

Discussion

27.	A rrange s, p an d d subshells o f a shell in the in creasin g o rd er of effective n u clear charge (Z_("eff")) experienced by th e electron p resen t in them .
Answer» Solution :s-orbital is spherical in shape, it shields the electrons from the nucleus m ore EFFECTIVELY than p-orbital which in TU RN shields m ore effectively than d-orbital. So, the effective nuclear charge `(Z_(eff))` experienced by electrons PRESENT in them is `d lt p lt s`

Discussion

28.	A room of10 m xx 15m xx4m dimenstion having perfectly insulated walls, ceiling and floor has 60 students seated inside. The air inside the room is at 25^(@)C and 1 atm pressure. If each student loses 200 joules of heat in one second, calculate the rise in temperature noticedin 20 minutes ( Neglect loss of air to the outside as temperature is raised and C_(p) for air = ( 7)/( 2) R).
Answer» Solution :Volume of room `= 10 m XX 15m xx4m = 600m^(3)= 600 xx 10^(3) L =6 xx 10^(5) L` Molesof air in the room at `25^(@)C` at 1 atm pressure `n = ( PV)/( RT) = ( 1 xx 6 xx 10^(5))/( 0.0821 xx 298) = 2.45 xx 10^(4)` Heat producedin1 sec by each student `= 200 J` `:. `Heat produced in 1 sec by 60 students `= 200 xx 60 J = 12000J` Heat produced in 20 minutes `=12000 xx 20 xx60 J = 144 xx 10^(5)J` Change in enthalpy of air, `DELTA H = n C_(p) DeltaT` `:. 144 xx 10^(5) = ( 2.45 xx 10^(4)) = ( ( 7)/(2) xx 8.314)xx Delta T ` or ` Delta T = 20.2 K`

Discussion

29.	A rigid and insulated tank of 3m^(3) volume is divided into two compartments. One second compartment of volume of 2m^(3) contains an ideal gas at 0.8314 Mpa and 400K and while the second compartment of volume 1m^(3) contains the same gas at 8.314 MPa and 500K. If the partition between the two compartments isruptured. the final temperature of the gas is:
Answer» 420K 450K 480K none of these Solution :Number of MOLES in the first compartment = `(PV)/(RT) = (8.314 xx 10^(5) xx 2)/(8.314 xx 400) = 500`. Number of moles in the second compartment `=(8.314 xx 10^(6) xx 1)/(8.314 xx 500) = 2000` HEAT lost by the GAS in second compartment = Heat gained by the gas in first compartment So `2000 xx (500-T) xx C = 500 xx (T - 400)xx C` On solving, T= 480K

Discussion

30.	A reversible reaction has DeltaG^(@)negative for forward reaction ? What will be sign of DeltaG^(@)for backworkreaction ?
Answer» SOLUTION :NEGATIVE.

Discussion

31.	A reversible reaction achieves equilibrium when the rates of forward and backward reactions equal. At equilibrium, the ratio of product of molar concentrations of prodcuts and the prodcut of molar concentration of reactants each raised to the powers equal to their stoichiometric coefficients, becomes constant. In case of gaseous reactios, the partial pressure of gases may be used in place of their molar concentrations. The equilibrium partial pressure of N_(2)O_(4)(g) and NO_(2)(g) are 4 and 2 atmm, respectively. Now, at constant temperature the pressure of system is increased to 60 atm. The new equilibrium partial pressure of N_(2)O_(4)(g) becomes.
Answer» 40atm 46.6atm 20atm 33.4atm Answer :B

Discussion

32.	A reversible reaction achieves equilibrium when the rates of forward and backward reactions equal. At equilibrium, the ratio of product of molar concentrations of prodcuts and the prodcut of molar concentration of reactants each raised to the powers equal to their stoichiometric coefficients, becomes constant. In case of gaseous reactios, the partial pressure of gases may be used in place of their molar concentrations. If 1 mole of PCI_(5)(g) and 1 mole of CI_(2)(g) is taken in a 10L vessel, then the equilibrium concentration of PCI_(3)(g) will be : PCI_(3)(g)+CI_(2)(g)hArrPCI_(5)(g),""K_(c)=(20)/(3)M^(-1)
Answer» `0.05` `0.04` `0.06` `0.025` ANSWER :A

Discussion

33.	A reversible cyclic process involves6 steps. In step -1 and 3 system absorbs500 , 800 J of heat from a heat reservoir at temperature 250 K and 200 K respectively. Step 2,4,6 are adiabaticsuch that the temperature of one reservior changes to that of next. Total work done by the the system in whole cycle is 700 J. Find the temperature during step 5 if it exchanges heat from a reservoir at temperatureT_(5)
Answer» ANSWER :100

Discussion

34.	A reutral moleculeXF_(3)has zerodipole menent. The element X is is most likely
Answer» chlorine boron nitrogen BROMINE SOLUTION :` BF_(3)`is planar triangular

Discussion

35.	(A): Research must be carried in such a manner that there will not be any waste by product in the reactions. (R) : The reaction which gives no by product is an environment friendly reaction.
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 Solution :Both (A) and (R) are true and (R) is the correct explanation.of (A)

Discussion

36.	A relation between vapour pressure and temperture is known as
Answer» IDEAL gas equation Boltzmam equation Clausious equation Clausius - CLAPEYRON equation Solution :`"LOG" (P_2)/(P_1) = (DeltaH_V)/(2.303R) (1/(T_1) - 1/(T_2))`, clausius uapeyron equation.

Discussion

37.	A reflection from (111) planes of a cubic crystal was observed at a glancing angle of 11.2^@ when X-rays of wavelength 154 pm were used. What is the length of the side of the unit cell ? (sin 11.2^@=0.1944)
Answer» Solution :Here, `lambda`=154 PM, n=1 , `THETA=11.2^@` Applying Bragg's equation `2 d SIN theta= n lambda` i.e., `d_111=(nlambda)/(2 sin theta)=(1xx154)/(2xx0.1944) ` pm =396 pm Further , the separation between the PLANES of a CUBIC crystal is given by `d_"hkl"=a/(sqrt(h^2+k^2+l^2))` `therefore d_11=a/(sqrt(1^2+1^2+1^2))`=396 pm (calculated above ) or `a=396xxsqrt3`=686 pm

Discussion

38.	A reducing agent is a substance which can
Answer» ACCEPT ELECTRONS DONATE electrons accept protons donate protons. Answer :B

Discussion

39.	A reducing agent is
Answer» SNO `SnO_(2)` `SnCl_(2)` `SnCl_(4)` Solution :`SnCl_(2)` is a poweful reducing AGENT

Discussion

40.	A redox reaction is always a // an "_____________".
Answer» PROTON TRANSFER reaction ion COMBINATION reaction reaction in solution electron transfer reaction Answer :D

Discussion

41.	A redox reaction involves oxidation of reductant liberating electrons, which are then consumed by an oxidant. The sum of two half reactions give rise to net redox change. In half reaction charge and atoms are always conserved. In the reaction: As_(2) S_(3) + HNO_(3) rarrH_(2)AsO_(4) +H_(2)SO_4, +NOthe element oxidised is:
Answer» `FE^(+8//3) to Fe^(+3) + 1/3 E` `Fe^(+8//3) to Fe^(+2) - 2/3 e ` `(Fe^(+8//3))_3 to 3Fe^(+2)+ 2e` `2(Fe^(+8//3))_3 to 3 (Fe^(+3))_2 + 2e` SOLUTION :Electrons and atoms are CONSERVED in half reaction .

Discussion

42.	A red solide is insoluble in water. However it becomes soluble if some KI is added to water. On heating the red solid in a test tube, there is liberation of some violet coloured fumes and droplets of a metal appear on the cooler parts of the test tube. The red solid is :
Answer» `HgO` `Pb_(3)O_(4)` `(NH_(4))_(2) Cr_(2)O_(7)` `Hgl_(2)` Solution :The red solide is `Hgl_(2)` `underset(("Red solide"))(Hgl_(2)) + 2 KL rarr underset(("soluble"))(K_(2) [Hgl_(4)])` `Hgl_(2) overset("HEAT")rarr underset(("Droplets"))(Hg(l)) + underset(("VIOLET fumes"))(I_(2)(g))`

Discussion

43.	A real gas obeying van der Walls's equation will resemble ideal gas, if the:
Answer» CONSTANTS a and B are small a is LARGE and b is small a is SMAL and b is large constant a and b are large Answer :a

Discussion

44.	A real gas most closely approaches the behaviour of an ideal gas at:
Answer» 15 ATM and 200K 1 atm and 273 K 0.5 atm and 500K 15 atm and 500K Answer :C

Discussion

45.	A real gas most closely approaches the behaviour of an ideal gas at ,
Answer» LOW PRESSURE `&` low temperature HIGH pressure `&` high temperature low pressure `&` high temperature high pressure `&` low temperature Answer :3

Discussion

46.	A real gas is subjected to an adibatic process causing in a change of state from (3 bar, 50 L, 500 K) to (5 bar, 40 L, 600 K) against a constant pressure of 4 bar. The magnitude of enthalpy change for the process is :
Answer» 4000 J 5000 J 9000 J 1000 J Answer :C

Discussion

47.	A real gas deviates most from ideal behaviour at
Answer» <P>High TEMPERATURE and Low pressure High pressure and Low temperature High pressure and High temperature Low pressure and Low temperature Solution :At high P, low T, VOLUEM of less `IMPLIES` more IMF.

Discussion

48.	A real gas can be liquefied:
Answer» under adiabatic expansion above critical TEMPERATURE when COOLED below critical temperature under applied PRESSURE at temperature LOWER than critical temperature and pressure higher than critical pressure Answer :A::C::D

Discussion

49.	A real gas acts as an ideal gas under which condition ?
Answer» HIGH TEMPERATURE, LOW PRESSURE Low temperature, high pressure High temperature, high pressure Low temperature, low pressure ANSWER :A

Discussion

50.	A reactionwas nono-spontaneous at high temperature but became spontaneousat low temperature . The reaction is "…............"
Answer» SOLUTION :EXOTHERMIC

Discussion

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