94420 + Interview Questions in Chemistry in Class 12

1.	Equation for the half period in first order reaction is
Answer» `t_(1//2)=(0.602)/K` `t_(1//2)=0.693/k` `t_(1//2)=k/0.693` `t_(1//2)=k/0.602` SOLUTION :DERIVE `t_(1//2)` from KT = `2.303 LOG. (a) /(a-x)`

Discussion

2.	Equation for the half life period in first order reaction is :
Answer» `t_1/2=0.602/k` `t_1/2=0.693/K` `t_1/2=K/0.693` `t_1/2=K/0.602` ANSWER :B

Discussion

3.	Equation ._(17)Cl^(37) + ._(1)H^(2) rarr ._(18)Ar^(38) + ._(0)n^(1) is
Answer» NUCLEAR fission Nuclear fusion TRANSFORMATION of CHLORINE Systhesis of argon Solution :It is a transformation of chlorine

Discussion

4.	Equanil is an example of
Answer» ARTIFICIAL SWEETENERS tranquilizers antihistamines antifertility DRUGS . ANSWER :B

Discussion

5.	Equanil is _____
Answer» ARTIFICIAL sweetener tranquillizer antihistamine antifertility drug Answer :B

Discussion

6.	Equanil is
Answer» ARTIFICIAL sweetener tranquilizer antihistamine antifertility drug. Answer :B

Discussion

7.	Equanil is .. .
Answer» Artificial sweetener TRANQUILIZER Antihistamine ANTIFERTILITY DRUG ANSWER :B

Discussion

8.	Equanil is ………
Answer» ARTIFICIAL sweetener TRANQUILIZER antihistamine antifertility DRUG SOLUTION :tranquilizer

Discussion

9.	Equanil is a ............. .
Answer» ARTIFICIAL sweetener tranquilizer anthistamine antifertility DRUG ANSWER :B

Discussion

10.	Equanil is ……………….
Answer» ARTIFICIAL sweetener Tranquilizer Antihistamine Antifertility drug Answer :B

Discussion

11.	Equanil is __________
Answer» artificial SWEETNER tranquilizer antihistamine ANTIFERTILITY drug Answer :B

Discussion

12.	Equanil is :
Answer» Antibiotic TRANQUILLIZER ANTISEPTIC Analgesic. Solution :EQUANIL is tranquillizer.

Discussion

13.	Equal weights of Zn metal and iodine are mixed together and the iodine is completelyconverted to ZnI_(2). What fraction by weightof the original zinc remains unreacted?(zn = 65 , I = 127 )
Answer» Solution :Let x g be the INITIAL WEIGHT of the Znmetaland iodine each. SINCE `I_(2)` is completelyconverted to `ZnL_(2)`, we have, initial no. of moles :`{:((x)/(65)""(x)/(254)""0),(Zn ""+ ""I_(2)""to ""ZnI_(2)):}` No.of moles at the end of the reaction. `((x)/(65)-(x)/(254))""0""(x)/(254)` `:.` fraction of Zn REMAINED unreacted = `(((x)/(65)-(x)/(254)))/((x)/(65))=0.74.`

Discussion

14.	Equanil is "………."
Answer» ARTIFICIAL sweetener TRANQUILIZER antihistamine antifertility drug Solution :EQUAL is a tranquilizer.

Discussion

15.	Equal weights of phosphorusand oxygenare heated in a closed vessel producing P_(2)O_(3) and P_(2)O_(5) in a 1 : 1mole ratio. Ifthe limitingcomponent is exhausted, findwhichcomponentand alsowhatfraction of it is left over.
Answer» Solution :As1 mole each of `P_(2) O_(3) and P_(2) O_(5)` is formed, the mole ratio of thereacting P and O ATOMSIS 4 : 8 or 1 : 2 As the givenweights of phosphorus and oxygen are same, say w g,their MOLES of ATOMS are `(omega)/(31) and (omega)/(16)` respectively. `:.` 2 moles of O combine WITH1 mole of P. `:. (omega)/(16)` moles of O combine with `(omega)/( 32)`moles of P. Sophosphorous is left over and moles of Premaining `= (omega)/( 31) - (omega)/( 32)` `:.`fractionof P left over`=(((omega)/(31)-(omega)/(32)))/((omega)/(31))= (1)/(32)or 3 . 125 %`

Discussion

16.	The pressure of a mixture of equal weight of two gases of mol wt. 4 and 40, is 1.1 atm. The partial pressure of the lighter gas in this mixture is
Answer» ANSWER :1

Discussion

17.	Equal weights of two gases of molecular mass 4 and 40 are mixed. The pressure of the mixture is 1.1 atm. The partial pressure of the lighter gas in this mixture is
Answer» 0.55 atm 0.11 atm 1.0 atm 0.1 atm Solution :Moles of lighter gas `= ( W )/( 4)` Moles of heavier gas `= ( W)/( 40)` Mole fraction of lighter gas `= ( W //4)/( W //4+ W// 40) = 10 //11` PARTIAL pressure of lighter gas = Mole fraction `xx` Total pressure `= ( 10)/( 11) xx 1.1 = 1.0 atm`

Discussion

18.	Equal weights of SO_(2) and O_(2) are mixed in an empty container at 300 K . The total pressure exerted by the gaseous mixture is 1 atm . The partial pressure of SO_2 gas in the mixture is
Answer» 0.33 ATM 0.20 atm 0.67 atm 0.50 atm ANSWER :A

Discussion

19.	Equal weights of methane and hydrogen are mixeed in an empty container at 25^(@)C. The fraction of total pressure exerted by hydrogen is :
Answer» `1//`8 `8 //9` `1//9` `16//17` Solution :The fraction of total pressure exerted by HYDROGEN is equalt o its mole fraction, i.e., `((1)/(2)) /((1)/(2) + ( 1)/( 16) ) = (( 1)/( 2))/((9)/( 16)) = ( 1)/(2) XX ( 16)/( 9) = ( 8)/( 9 )`

Discussion

20.	Equal weights of methane and hydrogen are mixed in an empty container at 25^@ C. The farction of total pressure exerted by hydrogen is
Answer» `1//2 ` 8//9 16//19 1//9 Answer :B

Discussion

21.	Equal weights of ethane and hydrogen and mixed in an empty container at 25^(@)C. The fraction of total pressure exerted by hydrogen is
Answer» <P>`1:2` `1:1` `1:16` `15:16` SOLUTION :`M_(C_(2)H_(6))=30, M_(H_(2))=2` Suppose w is the weight of each of `C_(2)H_(6)` and `H_(2)` taken. `:.n_(C_(2)H_(6))=w/30,n_(H_(2))=w/2` `:.p_(H_(2))=x_(H_(2))xxp=(n_(H_(2))/(n_H_(2))+n_(C_(2)H_(6)))xxp` `impliesp_(H_(2))=((w//2)/(w/2+w/30))p=((1//2)/(1/2+1/30))p` `impliesp_(H_(2))=15/16pimpliesp_(H_(2)):p=15:16`

Discussion

22.	Equal weight of NaCl and KCl are dissolved separately in equal volumes of solutions, then the molarity
Answer» Will be equal for the TWO solutions For NACL solution will be greater than that of KCl solution For KCl solution will be greater than that of NaCl solution For NaCl solution will be HALF of that of KCl solution Answer :B

Discussion

23.	Equal volumes of two monoatomic gases, A and B, at same temperature and pressure are mixed. The ratio of specific heats (C_(p)//C_(v))of the mixture will be
Answer» <P>0.83 1.5 3.3 1.67 Solution :`(C_(P))/(C_(V))=(5//2R)/(3//2R)=(5)/(3)=1.67`.

Discussion

24.	Equal volumes of two solutions of pH 4 and pH 11 are mixed. The pH of the resulting solution will be :
Answer» `7.5` `3.35` 13 `3.04` ANSWER :B

Discussion

25.	Equal volumes of two gases which do not react together are enclosed in separate vessels. Their pressure are 100mm and 400 mm respectively. If the two vessels are joined together, then what will be the pressure of the resulting mixture ( temperature remaining constant )
Answer» 125 mm 500 mm 1000 mm 250 mm Solution :When two VESSELS are joined TOGETHER, volume will be doubled HENCE effective pressure will be halved. `P=(P_(1)+P_(2))/(2)=(100+400)/(2)=250` mmHg

Discussion

26.	Equal volumes of two gases A and B diffuse through a porous pot in 20 and 10 seconds respectively. If molar mass of A is 80 calculate the molar mass of B.
Answer» SOLUTION :Suppose volume of each gas DIFFUSED = V mL Then by Graham.s Law `=(r_(A))/(r_(B))=(((V)/(20))/((V)/(10)))=sqrt((M_(B))/(M_(A))) rArr 1//2 =sqrt((M_(B))/(80))` `therefore M_(B) =20 GMOL^(-1)`

Discussion

27.	Equal volumes of three and acid solutions of pH 1,2 and 3 are mixed in a vessel, What will be the H^+ ion concentration in the mixture?
Answer» `3.7xx10^(-2)` `10^-6` `0.111` NONE of these Answer :A

Discussion

28.	Equal volumes of three acid solutions of pH 3, 4and 5 are mixed in a vessel. What will be the H^+ ion concentration in the mixture?
Answer» `3.7xx10^(-3)` M `1.11xx10^(-3)` M `1.11xx10^(-4)` M `3.7xx10^(-4)` M Solution :pH=-log `[H^+]` `[H^+]=10^(- pH)` , `[H^+]` of SOLN. 1= `10^(-3)` `[H^+]` of soln. 2= `10^(-4)` , `[H^+]` of soln. 3= `10^(-5)` Total concentration of `[H^+] = 10^(-3) (1+1xx10^(-1) + 1xx10^(-2))` `10^(-3) (1/1+1/10 +1/100) rArr 10^(-3) ((100+10+1)/100)` `rArr 10^(-3) (111/100) = 1.11xx10^(-3)` M So, `H^+` ION concentration in MIXTURE of equal mixture of these acid solution `= (1.11xx10^(-3))/3 = 3.7xx10^(-4)` M

Discussion

29.	Equal volumes of three acid solutions of pH 3, 4 and 5 are mixed in a vessel. What will be the H^(+) ion concentration in the mixture ?
Answer» `1.11xx10^(-4)M` `3.7xx10^(-4)M` `3.7xx10^(-3)M` `1.11xx10^(-3)M` Solution :`[H_(3)O]^(+)` for a solution having PH=3 is GIVEN by `[H_(3)O]^(+)=1xx10^(-3)` MOLES/litre `[therefore [H_(3)O]=10^(-pH)`] SIMILARLY for solution having pH=4, `[H_(3)O]^(+)=1xx10^(-4)` moles/litre and for pH=5 `[H_(3)O^(+)]=1xx10^(-5)` moles/litre Let the volume of each solution in mixture be 1L, then total volume of mixture solution L=(1+1+1)L=3L Total `[H_(3)O]^(+)` ion PRESENT in mixture solution `=(10^(-3)+10^(-4)+10^(-5)` moles Then `[H_(3)O]^(+)` ion concentration of mixture solution `=(10^(-3)+10^(-4)+10^(-5))/(3)M=(0.00111)/(3)M` `=0.00037 M=3.7xx10^(-4)M`

Discussion

30.	Equal volumes of the following Ca^(2+) and F^(-) solutions are mixed. In which of the solutions will precipitations occur ? [K_(sp) of CaF_(2)=1.7 xx 10^(-10)] 1. 10^(-2) M Ca^(2+)+10^(-5)MF^(-) 2. 10^(-3) M Ca^(2)+10^(-3)MF^(-) 3. 10^(-4) MCa^(2)+10^(-2)MF^(-) 4. 10^(-2) MCa^(2)+10^(-3)MF^(-) Select the correct answer using the codes given below
Answer» In 4 only In 1 and 2 In 3 and 4 In 2,3 and 4 Solution :In case I `[Ca^(2+)]` in reaction mixture `=0.5 xx 10^(-2)M` `[F^(-)]` in reaction mixture `=0.5 xx 10^(-5)M` `:. ` Ionic product `=[Ca^(2+)][F^(-)]^(2)=(0.5 xx 10^(-2))xx (0.5 xx10^(-5))^(2)` `=125 xx 10^(-15)=1.25 xx 10^(-13)xx1.7 xx 10^(-10)` `:. `No. ppt is formed in this case In case 2 `[Ca^(2+)]` in reaction mixture `=0.5 xx 10^(-3)M` `[F^(-)]` in reaction mixture `=0.5 xx 10^(-3)M` `:. `Ionic product `=(0.5 xx 10^(-3))(0.5 xx 10^(-3))^(2)` `=125 xx 10^(-12)=1.25 xx 10^(-10)lt 1.7 xx 10^(-10)` No. ppt. is formed in this case In case 3 `[Ca^(2+)]` in reaction mixture `=0.5 xx 10^(-4)M` `[F^(-)]` in reaction mixture `=0.5 xx 10^(-2)M` `:. ` Ionic product `=[Ca^(2+)]xx[F^(-)]^(2)` `=[0.5 xx 10^(-4)] xx [0.5 xx 10^(-2)]^(2)` `=125 xx 10^(-11)` `=1.25 xx 10^(-9) gt 1.7 xx 10^(-10)` Thus a ppt. of `CaF_(2)` will be formed In case 4 `[Ca^(2+)]` in reaction mixture `=0.5 xx 10^(-2)M` `[F^(-)]`in reaction mixture `=0.5 xx 10^(-3)M` Ionic product `[Ca^(2+)][F^(-)]^(2)` `=[0.5 xx 10^(-2)][0.5 xx 10^(-3)]^(2)=125 xx 10^(-11)` `=1.25 xx 10^(-9) gt 1.7 xx 10^(-10)` Thus a ppt of `CaF_(2)` will be formed.

Discussion

31.	Equal volumes of three acid solutions of PH 1,2 and 3 are mixed in a vessel. What will be the H^+ ion concentration in the mixture?
Answer» `3.7 times 10^-2` `10^-6` 0.111 none of these Solution :`PH=-log_10[H^+]` `therefore[H^+]=10^(-ph)` LET the VOLUME be x mL `V_1M_1+V_2M_2+V_3M_VM` `thereforexmL of 10^-1 M+xmL of 10^-2 M+xL of 10^-3M` `=3xmL of [H^+]` `therefore[H^+]=(x[0.1+0.01+0.001])/3=(0.1+0.01+0.001)/3=0.111/3` `=0.037=3.7 times 10^-2`

Discussion

32.	Equal volumes of three acid solution of pH 3,4 and 5 are mixed in a vessel. What will be the H^(+) ion concentration in the mixture
Answer» `3.7 xx 10^(-3) M` `1.11 xx 10^(-3) M` `1.11 xx 10^(-4) M` `3.7 xx 10^(-4) M` Solution :`[H^(+)]_(MIX) = (10^(-3)x + 10^(-4)x + 10^(-5)x)/(3X)` `= (1)/(3)(10^(-3) + 10^(-4) + 10^(-5)) = 37 xx 10^(-5) = 3.7 xx 10^(-4)`.

Discussion

33.	Equal volumes of molar hydrochloric acid and sulphuric acid are neutralised by dilute NaOH solution andx kcal and y kcal of heat are liberated respectively. Which of the following is true?
Answer» x=y x=y/2 x=2y None of the above Solution :By equal volumes of molar HCL and `H_2SO_4` the ratio of `H^+`ions furnished = 1:2 ` therefore ` Ratio of heat LIBERATED` = x/y = 1/2 RAR x = y/2`

Discussion

34.	Equal volumes of methanoic acid and sodium hydroxide are mixed. If x is the heat of formation of water, then heat evolved on neutralisation is
Answer» More than x Equal to x Twice of x Less than x Solution :As methanoic ACID is WEAK acid, heat of NEUTRALIZATION `lt x`.

Discussion

35.	Equal volumes of following solutions are mixed, in which case the pH of resulting solution will be average value of pH of two solutions.
Answer» pH=2 HCl and pH=12 NaOH pH=2 HCl and pH=4 HCl pH=2 HCN and pH= 12 NaOH (`k_a` for HCN=`10^(-10)`) pH=5 `CH_3COOH` and pH=9 `NH_(3(AQ.)),K_a(CH_3COOH)=K_b(NH_(3_(aq)))` Solution :(A)`pH=2 HCl` so [HCl]=`10^(-2)` pH=12 NaOH so [NaOH]=`10^(-2)` HENCE pH of mixture =7 (neutral ) (D)pH=5 `CH_3COOH` and pH=9 `NH_3` (aq), both must be of equal concentrations as `pK_a=pK_b`.Hence pH=pOH only if concentration are equal so on mixing equal volume we will get `CH_3COONH_4` salt solution and its pH is given is given by `pH=1/2(pK_w+pK_a-pK_b)=7`

Discussion

36.	Equal volumes of different gases at any definite tempearture and pressure have:
Answer» EQUAL WEIGHTS equal masses equal densities equal NUMBER of MOLES ANSWER :D

Discussion

37.	Equal volumes of different gases at a fixed temperature and pressure :
Answer» have EQUAL WEIGHTS equal MASSES equal densities equal NUMBER of moles Solution : have equal weights

Discussion

38.	Equal volumes of all gases under same conditions of temperature and pressure contains equal number of
Answer» Moles Atoms Electrons Radicals Answer :A

Discussion

39.	Equal volumes of all gases under similar conditions of temperataure and pressure contain equal number of molecules. This statement is called………………….. .
Answer» SOLUTION :AVOGADRO's LAW

Discussion

40.	Equal volumes of 30% by mass of H_(2)SO_(4) (density 1.218g mL^(-1)) and of H_(2)SO_(4) (density 1.610 g mL^(-1)) are mixe. If the density of te mixture is 1.452 g mL^(-1) , calculate the molarity oand molality of the solution.
Answer» Solution :Calculation of molarity of the solution Let V mL of each sample of `H_(2)SO_(4)` bemixed for 30% `H_(2)SO_(4)` Mass of `H_(2)SO_(4)`=30g ,Mass of solution=100 g `" Volume of solution"=("Mass of soulution")/("Density")= ((100g))/((1.218gmL^(-1)))=100/(1.218)=mL` `100/1.218` mL of sample contains `H_(2)SO_(4)=30 g` V mL of contains `H_(2)SO_(4)=((30g)xx(V mL))/((100//1.18mL))=(0.3654 V)g` For 70% `H_(2)SO_(4)` Mass of `H_(2)SO_(4)`=70g, Mass of solution=100 g `"Volume of solution"=("Mass of solution")/("Density")=((100g))/((1.610gmL^(-1)))=100/(1.610)mL` `100/(1.610)` mL of sample sontains `H_(2)SO_(4)=70g` V mL of sample contains `H_(2)SO_(4)=((70g)(VML))/((100//1.610mL))=(1.127V)g` On mixing the two sample os `H_(2)SO_(4)` Total mass of `H_(2)SO_(4)=0.3654V+1.127V=(1.127V)g` Total volume of solution=2VmL=(0.002V)L `"Molarity of solution(M)"= ("Total mass of" H_(2)SO_(4)//"Molar mass")/("Voluume of solution in litres") ` `=((1.4924 Vg)//(98gmol^(-1)))/((0.002V)L)=7.614 MOL L^(-1)=7.614 M` Calculation of malality of the solution. Total mass of solution= `d xx" volume "=(1.425 gmL^(-1)xx2VmL)=(2.85v)g` Total mass of ACID =(1.4924 V)g Mass of WATER = `(2.85 V-1.4924 V) g=(1.3576xx10^(-3)V) kg` `"Molality of solutin (m)"=("Mass of" H_(2)SO_(4)//"Molar mass")/("Mass of SOLVENT in kg")=((1.4924V)g//(98gmol^(-1)))/((1.3576xx10^(-3)V)kg)` =11.217 mol `kg ^(-1)`=11.217 m

Discussion

41.	Equal volumes of 1.0M AgNO_(3) and 1.0M KCl are mixed . The depression of freezing point of the resulting solution will be (K_(f)(H_(2)O)=1.86 K kg "mol"^(-1), 1M=1m)
Answer» `3.72K` `1.86 K` `0.93K` NONE of these Answer :B

Discussion

42.	Equal volumes of 1M HCl and 1M H_2SO_4 are neutralised by dilute NaOH solution and x kJ and y kJ of heat are libreated respectively . Which of the following is correct ?
Answer» `x=y` `x=(1)/(2)y` `x=2y` `x=4y` Solution :1M solution of `H_2SO_4` GIVES 2 EQUIVALENTS of `H^(+)` ions

Discussion

43.	Equal volumes of 1.0 M Na_2 CO_3and 1.0 M HCl are mixed. Calculate [CO_3^(2-)]at equilibrium. K_1and K_2 for H_2CO_3 are 4.5 xx 10^(-7) and 4.7 xx 10^(-11)respectively.
Answer» Solution :Solution is 0.50 M `NaHCO_3`(plus 0.5 M NACL). `5 xx 10^(-3)M`

Discussion

44.	Equal volumes of 10% solution (by wt) of the solute A and 15 % solution (by wt) of the solute B are mixed. The mass percent of A and B in the mixture would be respectively
Answer» 5 and 7.5 10 and 25 5 and 10 20 and 30 Solution :As equal volume of two solution are mixed the final volume i.e. the volume of MIXTURE would be DOUBLED. HENCE each % will become half.

Discussion

45.	Equal volumes of 1 M HCl and H_(2)SO_(4) are neutralised by dil NaOH solution and x kJ and y kJ of heat are librated respectively. Which of the following is true ?
Answer» X = y `x = (1)/(2)y` x = 2y None of these. Answer :B

Discussion

46.	Equal volumes of 0.1M AgNO_(3) and 0.2M NaCI are mixed. The concentration of NO_(3)^(-) ions in the mixture will be
Answer» `0.1M` `0.05M` `0.2M` `0.15M` Solution :`UNDERSET(0.1M)(AgNO_(3))+underset(0.2M)(NaCI) rarr AgCI +NaNO_(3)` `0.1M AgNO_(3)` reacts with `0.1M NaCI` to produce `0.1M AgCI` and `0.1M NaNO_(3)` `:. NO_(3)^(-) = (0.1M)/(2) = 0.05M` `[ :'` When EQUAL volumes are mixed dilution occurs]

Discussion

47.	Equal volume of all gases contain equal number of atoms at same temperature and pressure. This statement is
Answer» In ACCORDANCE with Dalton's ATOMIC theory and is KNOWN as BERZELIUS Hypothesis Not in accordance with Dalton's atomic theory and is known as Avogadro's Hypothesis Not in accordance with Dalton's atomic theory and is known as Berzelius Hypothesis In accordance with Dalton's atomic theory and is known as Avogadro's Hypothesis Answer :C

Discussion

48.	Equal volume of 0.02 M NH_(3)OH is mixed with 0.02 M HCI. Calculate [NH_(4)OH] at equivalence point : K_(b)[NH_(4_OH)]=10^(-5)
Answer» `1.414xx10^(-3.5)` `(1.414)/(3.16)xx10^(-5)` `3.16xx10^(-5)` `3.16xx10^(-6)` ANSWER :d

Discussion

49.	Equal volume each of two sols of Agl, one obtained by adding AgNO_(3) to slightly excess of Kl and another obtained by adding Kl to slight excess of AgNO_(3), are mixed together . Then :
Answer» the TWO sols will stablize each other the SOL PARTICLES will acquire more electric charge the sols will coagulate each other mutually a TRUE solution will be obtained Solution :The sols obtained in the two cases will be oppositely charged so coagulate each other.

Discussion

50.	Equal volume each of two sols of Agl one obtained by adding AgNO_3 to slight excess of KI and another obtained by adding KI to slight excess of AgNO_3 are mixed together then
Answer» The sols will COAGULATE each other mutually The two sols will STABILIZE each other The SOL PARTICLES will acquire more electric charge A true solution will be obtained Solution :The sols obtained in the two cases will be oppositely charged so coagulate each other

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Equation for the half period in first order reaction is

Equation for the half life period in first order reaction is :

Equation ._(17)Cl^(37) + ._(1)H^(2) rarr ._(18)Ar^(38) + ._(0)n^(1) is

Equanil is an example of

Equanil is _____

Equanil is

Equanil is .. .

Equanil is ………

Equanil is a ............. .

Equanil is ……………….

Equanil is __________

Equanil is :

Equal weights of Zn metal and iodine are mixed together and the iodine is completelyconverted to ZnI_(2). What fraction by weightof the original zinc remains unreacted?(zn = 65 , I = 127 )

Equanil is "………."

Equal weights of phosphorusand oxygenare heated in a closed vessel producing P_(2)O_(3) and P_(2)O_(5) in a 1 : 1mole ratio. Ifthe limitingcomponent is exhausted, findwhichcomponentand alsowhatfraction of it is left over.

The pressure of a mixture of equal weight of two gases of mol wt. 4 and 40, is 1.1 atm. The partial pressure of the lighter gas in this mixture is

Equal weights of two gases of molecular mass 4 and 40 are mixed. The pressure of the mixture is 1.1 atm. The partial pressure of the lighter gas in this mixture is

Equal weights of SO_(2) and O_(2) are mixed in an empty container at 300 K . The total pressure exerted by the gaseous mixture is 1 atm . The partial pressure of SO_2 gas in the mixture is

Equal weights of methane and hydrogen are mixeed in an empty container at 25^(@)C. The fraction of total pressure exerted by hydrogen is :

Equal weights of methane and hydrogen are mixed in an empty container at 25^@ C. The farction of total pressure exerted by hydrogen is

Equal weights of ethane and hydrogen and mixed in an empty container at 25^(@)C. The fraction of total pressure exerted by hydrogen is

Equal weight of NaCl and KCl are dissolved separately in equal volumes of solutions, then the molarity

Equal volumes of two monoatomic gases, A and B, at same temperature and pressure are mixed. The ratio of specific heats (C_(p)//C_(v))of the mixture will be

Equal volumes of two solutions of pH 4 and pH 11 are mixed. The pH of the resulting solution will be :

Equal volumes of two gases which do not react together are enclosed in separate vessels. Their pressure are 100mm and 400 mm respectively. If the two vessels are joined together, then what will be the pressure of the resulting mixture ( temperature remaining constant )

Equal volumes of two gases A and B diffuse through a porous pot in 20 and 10 seconds respectively. If molar mass of A is 80 calculate the molar mass of B.

Equal volumes of three and acid solutions of pH 1,2 and 3 are mixed in a vessel, What will be the H^+ ion concentration in the mixture?

Equal volumes of three acid solutions of pH 3, 4and 5 are mixed in a vessel. What will be the H^+ ion concentration in the mixture?

Equal volumes of three acid solutions of pH 3, 4 and 5 are mixed in a vessel. What will be the H^(+) ion concentration in the mixture ?

Equal volumes of three acid solutions of PH 1,2 and 3 are mixed in a vessel. What will be the H^+ ion concentration in the mixture?

Equal volumes of three acid solution of pH 3,4 and 5 are mixed in a vessel. What will be the H^(+) ion concentration in the mixture

Equal volumes of molar hydrochloric acid and sulphuric acid are neutralised by dilute NaOH solution andx kcal and y kcal of heat are liberated respectively. Which of the following is true?

Equal volumes of methanoic acid and sodium hydroxide are mixed. If x is the heat of formation of water, then heat evolved on neutralisation is

Equal volumes of following solutions are mixed, in which case the pH of resulting solution will be average value of pH of two solutions.

Equal volumes of different gases at any definite tempearture and pressure have:

Equal volumes of different gases at a fixed temperature and pressure :

Equal volumes of all gases under same conditions of temperature and pressure contains equal number of

Equal volumes of all gases under similar conditions of temperataure and pressure contain equal number of molecules. This statement is called………………….. .

Equal volumes of 30% by mass of H_(2)SO_(4) (density 1.218g mL^(-1)) and of H_(2)SO_(4) (density 1.610 g mL^(-1)) are mixe. If the density of te mixture is 1.452 g mL^(-1) , calculate the molarity oand molality of the solution.

Equal volumes of 1.0M AgNO_(3) and 1.0M KCl are mixed . The depression of freezing point of the resulting solution will be (K_(f)(H_(2)O)=1.86 K kg "mol"^(-1), 1M=1m)

Equal volumes of 1M HCl and 1M H_2SO_4 are neutralised by dilute NaOH solution and x kJ and y kJ of heat are libreated respectively . Which of the following is correct ?

Equal volumes of 1.0 M Na_2 CO_3and 1.0 M HCl are mixed. Calculate [CO_3^(2-)]at equilibrium. K_1and K_2 for H_2CO_3 are 4.5 xx 10^(-7) and 4.7 xx 10^(-11)respectively.

Equal volumes of 10% solution (by wt) of the solute A and 15 % solution (by wt) of the solute B are mixed. The mass percent of A and B in the mixture would be respectively

Equal volumes of 1 M HCl and H_(2)SO_(4) are neutralised by dil NaOH solution and x kJ and y kJ of heat are librated respectively. Which of the following is true ?

Equal volumes of 0.1M AgNO_(3) and 0.2M NaCI are mixed. The concentration of NO_(3)^(-) ions in the mixture will be

Equal volume of all gases contain equal number of atoms at same temperature and pressure. This statement is

Equal volume of 0.02 M NH_(3)OH is mixed with 0.02 M HCI. Calculate [NH_(4)OH] at equivalence point : K_(b)[NH_(4_OH)]=10^(-5)

Equal volume each of two sols of Agl, one obtained by adding AgNO_(3) to slightly excess of Kl and another obtained by adding Kl to slight excess of AgNO_(3), are mixed together . Then :

Equal volume each of two sols of Agl one obtained by adding AgNO_3 to slight excess of KI and another obtained by adding KI to slight excess of AgNO_3 are mixed together then