94420 + Interview Questions in Chemistry in Class 12

1.	For ""_(92)^(238)U the binding energy per nucleon is 7.576 MeV. What is the atomic weight of this isotope? Use the mass of neutron and proton
Answer» ANSWER :239.93 AMU

Discussion

2.	For 500ml 22.4V H_(2)O_(2) solution having density 1.2 gm/ml. Identify correct statement(s)
Answer» Molality =1.76m `(W)/(W)%-5.66` `(W)/(v)%=68%` 11.2 litre of `O_(2)` will be evolved at NTP. Solution :`M_(H_(2)O_(2)) "solution"=('V')/(11.2)=(22.4)/(11.2)=2rArr` In 500 ml solution, `n_(H_(2)O_(2))=1, w_(H_(2)O_(2))=34gm` `d_("solution")=1.2gm//ml=("Mass of solution")/(500)rArr` Mass of solution =600gm Mass of solvent =600-34=566gm (A) `m=(n_(H_(2)O_(2))xx1000)/(W_("water")(gm))=(1000)/(566)=1.76` (B) `(W)/(W)%=(34)/(600)xx100=5.66%` (C) `(W)/(v)%=(34)/(500)xx100=6.8%` (D) 1 litre `H_(2)O_(2)` solution produce `O_(3)` (NTP) =22.4litre `500 ml H_(2)O_(2)` solution produce `O_(2)(NTP)=11.2` litre

Discussion

3.	for 3d_(z^2) orbital the value of l and m respectively are
Answer» `2,0` `2, +1` `2, -1` 2, +2 Answer :A

Discussion

4.	For 3AtoxB,(d[B])/(dt) is found to be 2/3rd of (d[A])/(dt),. Then the value of x is
Answer» 1.5 3 `1//2` 2 Answer :D

Discussion

5.	For 2SO_2 + O_2 to 2SO_3, rate of disappearance of SO_2 is 4 xx 10^(-3)M - s^(-1)att=10 sec. Then, the amount of SO_3 formed & amount of O_2consumed at t = 10 sec respectively are
Answer» 0.1g, 0.1g 0.1g, 0.2 g 0.0169, 0.064g 0.32g, 0.064g Answer :D

Discussion

6.	For 2NH_(3)overset("Au")(to)N_(2)+3H_(2), rate w.r.t N_(2) is 2xx10^(-3)"M - min"^(-1), then rate w.r.t N_(2) after 20 minutes will be ("in M - min"^(-1))
Answer» `2XX10^(-3)` `gt2xx10^(-3)` `10^(-4)` `lt2xx10^(-3)` ANSWER :A

Discussion

7.	For 2NH_3(g) underset(Delta)overset(Pt)(to)products follows zero order kinetics. If t_(1//2)at p = 4 atm is 25 sec, t_(1//2)atp = 16 atm will be (in sec)
Answer» 6.25 625 100 `(25)^(1/4)` ANSWER :C

Discussion

8.	For 2H_2O_2 to 2H_2O +O_2. If(-d)/(dt) [H_2O_2] = K_1 [H_2O_2] (+d[H_2O])/(dt) = K_2[H_2O_2](+d[O_2])/(dt) = K_3[H_2O_2]
Answer» `k_1 = k_2 = k_3` `k_1 = k_2 = 2k_3` `2k_1 = 2k_2= k_3` `k_1 = k_2 = 4k_3` ANSWER :B

Discussion

9.	For 2H_2O_2 to 2H_2O + O_2 , t_(0.5) = 0.301 hr. When [H_2O_2]at t =0 is 0.5 M, initial rate is
Answer» `2.303 M.h^(-1)` `1.151 M.h^(-1)` `4.606 M.h^(-1)` `0.301 M.h^(-1)` ANSWER :B

Discussion

10.	For 2A(g) B(g) +3C(g), Pressure of A at t = 0 is 500 mm, then total pressure of A, B & C at t =10 min is( in mm)
Answer» 100 400 540 250 Answer :C

Discussion

11.	For 2A to B,[A] changed from 0.08M to 0.04M in100 seconds. Now (Delta [B])/(Delta t)will be
Answer» `2 XX 10^(-4) Ms^(-1)` `4 xx 10^(-4) Ms^(-1)` `8 xx 10^(-4) Ms^(-1)` `1.2 xx 10 Ms^(-1)` Answer :A

Discussion

12.	For 2A+B+Cto Products, rate law is given by rate =" k [A] [B]"^(2) & rate constant (k) is 2xx10^(-6)M^(-2)-s^(-1). Then rate of the reaction become 2xx10^(-9)"M - s"^(-1) only when
Answer» `{:(" [A]","[B]","[C]"),(0.1M,01.M,0.2M):}` `{:(" [A]","[B]","[C]"),(0.2M,0.1M,0.2M):}` `{:(" [A]","[B]","[C]"),(0.1M,0.1M,"any VALUE"):}` `{:(" [A]","[B]","[C]"),(0.2M,0.2M,0.1M):}` ANSWER :C

Discussion

13.	For 2A B+3C_(3) 2C overset(k_(2))rarr 3D. Which of the following is correct :
Answer» `d[C]//dt=3k_(1)[A]^(2)-3k_(1)[B][C]^(3)-2K_(2)[C]^(2)` `d[B]//dt=k_(1) [A]^(2)` `d[A]//dt =2K_(1) [B][C]^(3)-2K_(1) [B][C]^(3)` None. Answer :A

Discussion

14.	For ""^(24)Na,t_((1)/(2))=14.8 hours. In what period of time will a sample of this substance lose 90% of its radioactive intensity ?
Answer» Solution :Let the initial radioactive intensity be 100 which corresponds to `N^(0)`. The radioactive intensity after a time period, say t HOURS, will be 10 (CORRESPONDING to N) as the SUBSTANCE has lost 90% of its radioactive intensity. `lamda= (0.6932)/(t_((1)/(2))) = (0.6932)/(14.8)` We have, `lamda= (2.303)/(t) "log" (N^(0))/(N)` `(0.6932)/(14.8) =(2.303)/(t) "log" (100)/(10)` t= 49.17 hours.

Discussion

15.	For I_2+2erarr2I^-,standard reduction potential =+0.54 volt. For Br^(- )rarrBr_2 +2e^-, standard oxidation potential=-1.09 volt. For FerarrFe^(2+) +2e^-, standard oxidation potential=+0.44 volt. Which of the following reactions is non-spontaneous :
Answer» `Br_2+2I^-` `rarr2Br^(-) +I_2` `FE+Br_2 RARRFE^(2+)+2Br^-` `Fe +I_2 rarrFe^(2+)+2I^_` `I_2 +2Br^(-) rarr2I^(-)+Br_2` ANSWER :D

Discussion

16.	For 1=3, which value of m is not possible?
Answer» 4 0 `-3` `-1` ANSWER :A

Discussion

17.	For 10 minutes each, at 27^(@)C, from two identical holes nitrogen and an unknown gas are leaked into a common vessel of 3L capacity. The resulting pressure is 4.18bar and the mixture contains 0.4mol of nitrogen. What is the molar mass of the unknown gas?
Answer» Solution :Let unknown gas is ‘a’ Here, `T = 300K, t_(N_(2)) = t_(a) = 10 "MINUTES"`, `V = 3 L = 3dm^(3), P_(T) = 4.18bar, n_(N_(2))= 0.4 "MOL"` From Dalton’s law of PARTIAL pressures, we known that. `P_(T) = R_(N_(2)) + P_(a)` `P_(T) = n_(N_(2)) (RT)/(V)+n_(2)(RT)/(V)=(n_(N_(2))+n_(2))(RT)/(V)` `4.18 = (0.4 + n_(2)) (0.083 xx 300)/(3)` `implies (0.4 + n_(2) ) = (4.18 xx 3)/(0.083 xx 300)= 0.5036 implies n_(2)=0.5036-0.4=0.1036` Now, from Graham’s law of diffusion, we now that. `(n_(N_(2)))/(n_(2))= (t_(N_(2)))/(t_(a))sqrt((M_(a))/(M_(N_(2))))implies (0.4)/(0.1036)= (10)/(10)sqrt((M_(a))/(28))implies 3.861=sqrt((M_(a))/(28))` `M_(a)=(3.861)^(2) xx 28= 417.4` therefore Molar mass of un NOWN gas `= 417.4 G//"mol"`.

Discussion

18.	For 100 times increase in concentration of H^(+) ions :
Answer» pH INCREASES by 2 units pH DECREASES by 2 units pH remains unaltered pH decreases by 1 UNIT. Answer :B

Discussion

19.	For 10 minute each, at 27^(@)C, from two identical holes nitrogen and an unknown gas are leaked into a common vessel of 3 litre capacity. The resulting pressure is 4.18 bar and the mixture contains 0.4 mole of nitrogen. What is the molar mass of unknown gas ?
Answer» Solution :P=4.18 bar, V=3L, T=300 K `n_(T) =(PV)/(RT) =(4.18xx3)/(0.83xx300)=0.50` Total mole of gases diffused =0.50 Mole of UNKNOWN GAS `n_(g)` diffused `=n_(T)-n_(N_(2))` `=0.50 -0.40=0.10` `(n_(g)//t_(g))/(n_(g)//t_(n_(2)))=sqrt((M_(n_(2))/(M_((g))))` `rArr =(0.1)/(0.4) =sqrt((28)/(M_((g)))` `rArr Mg =448" gm mole "^(-1)`

Discussion

20.	For1 mole of monoatomic gas . Calculate w, DeltaU, DeltaH,q
Answer» SOLUTION :Isochoricprocess `W=0` `q= DU= C_(V) (T_(2)_T_(1)) =3//2xx(400-300)=150 R` `DeltaH=C_(P)DELTAT=(5)/(2)R(400-300)=250R`

Discussion

21.	For 1 mole of an ideal gas, a graph of pressure vs volume is plotted as shown. Which of the following option is correct ?
Answer» AB process is isothermal. Maximum TEMPERATURE of the gas can be `(10)/(0.0821)K`. MINIMUM temperature of the gas can be `(11)/(4 XX 0.0821) K`. None of the above ANSWER :D

Discussion

22.	For 1 mole of an ideal gas, a graph of pressure vs volume is plotted as shown. Which of the following options is correct ?
Answer» AB process is isothermal Maximum temperature of the gas can be `10/0.0821 K` Minimum temperature of the gas can be `11/(4xx0.0821) K` NONE of these Solution :`P-1=(10-1)/(1-10) (V=10)` `P=-V+10` `RT=-V^(2)+11V` `(RDT)/(dV)=-2V+11=0` `V=1/2` `T_(max)=(-121/4+11xx11/2)/R=121/(4xx0.0821)K`

Discussion

23.	For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?
Answer» `[CO(H_(2)O)_(3)Cl_(3)]3H_(2)O` `[Co(H_(2)O)_(6)]Cl_(3)` `[Co(H_(2)O)_(5)Cl]Cl_(2)H_(2)O` `[Co(H_(2)O)_(4)Cl_(2)]Cl2H_(2)O` Solution :The complex giving least number of ions will show highest freezing POINT

Discussion

24.	For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point ?
Answer» `[Co(H_(2)O)_(6)]Cl_(3)` `[Co(H_(2)O)_(5)CL]Cl_(2).H_(2)O` `[Co(H_(2)O)_(4)Cl_(2)]Cl.2H_(2)O` `[Co(H_(2)O)_(3))Cl_(3)].3H_(2)O` SOLUTION :As the value of i increases the freezing point DECREASES. `{:("AQUEOUS solution",,i),([Co(H_(2)O)_(6)]Cl_(3),to,4),([Co(H_(2)O)_(5)Cl]Cl_(2).H_(2)O,to,3),([Co(H_(2)O)_(4)Cl_(2)]Cl.2H_(2)O,to,3),([Co(H_(2)O)_(3)Cl_(3)].3H_(2)O,to,1):}`

Discussion

25.	For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point -
Answer» `[Co(H_(2)O)_(6)]Cl_(3)` `[Co(H_(2)O)_(5)Cl]Cl_(2).H_(2)O` `[Co(H_(2)O)_(4)]Cl_(2).2H_(2)O` `[Co(H_(2)O_(3)Cl_(3)].3H_(2)O` Solution :Depression of freezing point will be maximum for that solution which will contain maximum number of SOLUTE particles in it. Among the given SOLUTES, (D) will contain minimum number of solute particles in its solution. Hence depression of reezing point will be minimum for the solution of (D), i.e, solution of (D) will exhibit highest freezing point.

Discussion

26.	For 1 mol ofan ideal gas at a constant temperature T, the plot of (log P) against (log V) is a (P: Pressure, V: Volume)
Answer» STRAIGHT line parallel to x-axis. Straight line with a negative slope. Curve starting at origin. Straight line passing through origin. Solution :ACCORDING to Boyle.s law, PV = constant `thereforelogp+logV` =constant LOGP=-logV+ constant Hence, the plot of logP vs LOG V is straight line with negative slope.

Discussion

27.	For 0.0128 N solution of acetic acid at 25^(@)C equivalent conductance of the solution is 1.4 "mho cm"^(2) eq^(-1) and Lambda_(0) = 391 "mho cm"^(-2)eq^(-1). Calculate dissociation constant (K_(a)) of acetic acid.
Answer» SOLUTION :`1.6 XX 10^(-7)`

Discussion

28.	For 1 g molecule of an ideal gas :
Answer» `(PV)/T` = 2cal `(PV)/T` =3/2 CAL `(PV)/T` = 8.31 cal `(PV)/T` = 0.831 cal Answer :A

Discussion

29.	For 0.01 N KCl, the resistivity 709.22 ohm cm. Calculate the conductivity and equivalent conductance.
Answer» ANSWER :A::B::D

Discussion

30.	For 1/2X_(2)+Y_(2)toXY_(2) relartive rates of species is given as
Answer» ` "RATE" =(-d[x_(2)])/(dt)=(-d[y_(2)])/(dt)=+(d[xy_(2)])/(dt)` ` "Rate" =-2(d[x_(2)])/(dt)=(-d[y_(2)])/(dt)=+(d[xy_(2)])/(dt)` ` "Rate" =(-1)/(2)(d[x_(2)])/(dt)=(-d[y_(2)])/(dt)=+(d[xy_(2)])/(dt)` ` "Rate" =-(1)/(2)(d[x_(2)])/(dt)=(+d[y_(2)])/(dt)=+(d[xy_(2)])/(dt)` Answer :B

Discussion

31.	Fool's gold is
Answer» `Cu_(2)S` `FeS_(2)` `Al_(2)O_(5)` `CuFeS_(2)` Solution :Iron pyrite, `(FeS_(2))` is also known as FOOL's gold.

Discussion

32.	Fool'sgoldis
Answer» ` Cu_ 2 S ` `FeS_2 ` `Al_ 2 O _5` `CUFES _2` Solution :Iron pyrite ,` FeS_2 `is ALSO KNOWN asfool's gold.

Discussion

33.	Fool's gold is:
Answer» `FeS_2` `CuFeS_2` BOTH (A) AND (B) None Answer :C

Discussion

34.	Food preservative in tomato ketchup is
Answer» SODIUM ACETATE sodium benzoate sodium salicylate sodium propionate Answer :A::B::D

Discussion

35.	Food preservatives prevent spoilage of food due to microbial growth. The most commonly used preservatives are
Answer» table SALT, sugar vegetable OILS and SODIUM BENZOATE `C_6H_5COONa` All of the above ANSWER :D

Discussion

36.	Food preservatives prevent spoilage of food due to microbial growth. The most commonly used preservatives are :
Answer» table SALT, sugar vegetable OILS and SODIUM BENZOATE `C_6H_5COONa` all of the above Answer :D

Discussion

37.	Food preservative among the following
Answer» Vanomycine Sodium bisulphate BHT Sodium benzosulphate Solution :Butylated hydroxytoluene (BHT) are phenolic COMPOUNDS that are OFTEN added to FOODS to preserve fat.

Discussion

38.	Food preservation by removal of heat involves_____.
Answer» refrigeration freezing cold storage all of these Answer :D

Discussion

39.	Following two reactions can occurs at cathode in the electrolysis of aqueous sodium chloride. Na^(+)+e^(-)toNa(s),E_(Red)^(@)=-2.71V 2H_(2)O(l)+2e^(-)toH_(2)(g)+2OH^(-)(aq)E_(Red)^(@)=-0.83V Which reaction takes place preferentially and why?
Answer» Solution :As the standard REDUCTION potential of `H_(2)O` is greater than that of `Na^(+)` ION, reduction of WAER takes place preferentially, i.e., `H_(2)` is liberated at CATHODE.

Discussion

40.	Following two graphs are based on the conductometric titration of acid-base reaction. Answer the questions given at the end of it. Select correct statement,
Answer» GRAPH A is for weak monobasic acid while graph B is for monobasic STRONG acid Graph A is for strong monobasic ADD and graph B is for weak dibasic acid Graph A is for weak monobasic ACLD and graph B is for weak dibasic acid Graph A is for strong monobasic acid and graph B is for weak monobasic acid Answer :D

Discussion

41.	Following statements is/are correct about mixture :
Answer» MIXTURE is 3-types of OXIMES mixture is 2-types of oximes all are OPTICALLY ACTIVE ONE is optically active Answer :A

Discussion

42.	Following solutions were prepared by mixing different volumes of NAOH of HCL different concentrations. (i) 60 mL (M)/(10) HCI+40 mL (M)/(10) NaOH (ii) 55 mL (M)/(10) HCI+45 mL (M)/(10) NaOH (iii) 75 mL (M)/(5) HCI+25 mL (M)/(5) NaOH (iv) 100 mL (M)/(10) HCI+100 mL (M)/(10) NaOH pH of which one of them will be equal to I?
Answer» iv (i) (II) (iii) Solution :No of MOLES of HCL `=0.2 xx 75 xx 10^(-3) = 15 xx 10^(-3)` No of mole of `NaOH = 0.2 xx 25 xx 10^(-3) = 5 xx 10^(-3)` No of moles of HCl after mixing `=15 xx 10^(-3) -5 xx 10^(-3)` `:.` Concentration of HCl `= ("No. of moles of HCl")/("Vol in litre") = (10 xx 10^(-3))/(100 xx 10^(-3)) = 0.1M` For (iii) solutiion, pH of 0.1 M HCl = `-log_(10) (0.1) = 1`

Discussion

43.	Following solutions at the same temperature will be isotonic:
Answer» 3.42 g of cane SUGAR in ONE litre WATER and 0.18 g of GLUCOSE in one litre water 3.42 g of cane sugar in one litre water and 0.18 g of glucose in 0.1 litre water 3.42 g of cane sugar in one litre water and 0.585 g of NaCl in one litre water 3.42 g of cane sugar in one litre water and 1.17 g of NaCl in one litre water Answer :B

Discussion

44.	Following solutions at the same temperature will be isotonic
Answer» 3.42 g of cane sugar in one litre water and 0.18 g of glucose in one litre water 3.42 g of cane sugar in one litre water and 0.18 g of glucose in 0.1 litre water 3.42 g of cane sugar in one litre water and 1.17 g of NaCl in one litre water 3.42 g of cane sugar in one litre water and 1.17 g of NACI in one litre water Solution :For ISOTONIC solution, `pi_(1) = pi_(2)` (andfor nonelectrolytes also `C_(1) = C_(2))` `pi_(1) = W/(M xx V) xx RT = 3.42 /(342 xx 1)RT = 0.01 RT` `pi_(2) = 0.18/(180 xx 0.1) RT = 0.01 RT`

Discussion

45.	Following solution are prepared by the mixing different volumes of NaOH of HCl different concentrations.
Answer» `60 mL M/10 HCl+40 mL M/10 NaOH` `55 mL M/10 HCl+45 mL M/10 NaOH` `75 mL M/5 HCl+25 mL M/5 NaOH` `100 mL M/10 HCl+100 mL M/10 NaOH` SOLUTION :`75 mL M/5 HCl+25 mL M/5 NaOH` No. of moles of HCl= `0.2 TIMES 75 times 10^-3=15 times 10^-3` No of Moles of NaOH=`0.2 times 25 times 10^-3=5 times 10^-3` No of moles of HCl after mixing `=15 times 10^-3 -5 times 10^-3` `therefore"Concentration"= ("No. of moles of HCl")/("Vol in litre")=(10 times 10^-3)/(100 times 10^-3)=0.1M` for (III) solution pH of 0.1 M HCl=`-log_10(0.1)=1`.

Discussion

46.	Following sidwick.s rule of EAN, Co(CO)_(x) will be:
Answer» `CO(CO)_(4)` `Co(CO)_(3)` `Co(CO)_(6)` `Co(CO)_(10)` Solution :`Co_(2)(CO)_(x)` then x =8 according to EAN RULE

Discussion

47.	Following Sidwick's rule of EAN, Co(CO)_(x) will be :
Answer» `Co_(2)(CO)_(4)` `Co_(2)(CO)_(3)` `Co_(2)(CO)_(8)` `Co_(2)(CO)_(10)` Answer :C

Discussion

48.	Following Sidgwick's rule of EAN, Co(CO)_(x) will be.
Answer» `Co_(2)(CO)_(4)` `Co_(2)(CO)_(3)` `Co_(2)(CO)_(8)` `Co_(2)(CO)_(10)` Solution :`[Co_(2)(CO)_(8)IMPLIES EAN =27+8+1=36`

Discussion

49.	Following reactions occur at cathode during the electrolysis of aqueous silver chloride solution: Ag^(+)(aq)+e^(-)toAg(s),E^(@)=+0.80V,H^(+)(aq)+e^(-)to(1)/(2)H_(2)(g),E^(@)=0.00V On the basis of their standard reduction electrode potential (E^(@)) values, which reaction is feasible at the cathode and why?
Answer» Solution :Higher the standard REDUCTION potential of a species, more easily it is reduced at the CATHODE. As `Ag^(+)(aq)` has GREATER standard reduction potential, therefore, the reaction that will OCCUR at the cathode is `Ag^(+)(aq)+e^(-)TOAG(s)`.

Discussion

50.	Following reactionsoccur spontaneously as written below: 2Fe^(+++)+2l^(-) to 4Fe^(++)+I_(2) O_(2)(g)+H^(+)+4Fe^(++) to 4Fe^(+++)+2H_(2)O Oxidizing power of O_(2),Fe^(+++) and I_(2) will be in the order:
Answer» `O_(2) GT Fe^(3+) gt I_(2)` `O_(2) gt I_(2) gt Fe^(3+)` `I_(2) gt O_(2) gt Fe^(3+)` `I_(2) gt Fe^(3+) gt O_(2)` Solution :`Fe^(3+)` undergoes reduction in presence of `I^(-)` but undergoes OXIDATION in presence of `O_(2)`.

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

For ""_(92)^(238)U the binding energy per nucleon is 7.576 MeV. What is the atomic weight of this isotope? Use the mass of neutron and proton

For 500ml 22.4V H_(2)O_(2) solution having density 1.2 gm/ml. Identify correct statement(s)

for 3d_(z^2) orbital the value of l and m respectively are

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For 2H_2O_2 to 2H_2O + O_2 , t_(0.5) = 0.301 hr. When [H_2O_2]at t =0 is 0.5 M, initial rate is

For 2A(g) B(g) +3C(g), Pressure of A at t = 0 is 500 mm, then total pressure of A, B & C at t =10 min is( in mm)

For 2A to B,[A] changed from 0.08M to 0.04M in100 seconds. Now (Delta [B])/(Delta t)will be

For 2A+B+Cto Products, rate law is given by rate =" k [A] [B]"^(2) & rate constant (k) is 2xx10^(-6)M^(-2)-s^(-1). Then rate of the reaction become 2xx10^(-9)"M - s"^(-1) only when

For 2A B+3C_(3) 2C overset(k_(2))rarr 3D. Which of the following is correct :

For ""^(24)Na,t_((1)/(2))=14.8 hours. In what period of time will a sample of this substance lose 90% of its radioactive intensity ?

For I_2+2erarr2I^-,standard reduction potential =+0.54 volt. For Br^(- )rarrBr_2 +2e^-, standard oxidation potential=-1.09 volt. For FerarrFe^(2+) +2e^-, standard oxidation potential=+0.44 volt. Which of the following reactions is non-spontaneous :

For 1=3, which value of m is not possible?

For 10 minutes each, at 27^(@)C, from two identical holes nitrogen and an unknown gas are leaked into a common vessel of 3L capacity. The resulting pressure is 4.18bar and the mixture contains 0.4mol of nitrogen. What is the molar mass of the unknown gas?

For 100 times increase in concentration of H^(+) ions :

For 10 minute each, at 27^(@)C, from two identical holes nitrogen and an unknown gas are leaked into a common vessel of 3 litre capacity. The resulting pressure is 4.18 bar and the mixture contains 0.4 mole of nitrogen. What is the molar mass of unknown gas ?

For1 mole of monoatomic gas . Calculate w, DeltaU, DeltaH,q

For 1 mole of an ideal gas, a graph of pressure vs volume is plotted as shown. Which of the following option is correct ?

For 1 mole of an ideal gas, a graph of pressure vs volume is plotted as shown. Which of the following options is correct ?

For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?

For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point ?

For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point -

For 1 mol ofan ideal gas at a constant temperature T, the plot of (log P) against (log V) is a (P: Pressure, V: Volume)

For 0.0128 N solution of acetic acid at 25^(@)C equivalent conductance of the solution is 1.4 "mho cm"^(2) eq^(-1) and Lambda_(0) = 391 "mho cm"^(-2)eq^(-1). Calculate dissociation constant (K_(a)) of acetic acid.

For 1 g molecule of an ideal gas :

For 0.01 N KCl, the resistivity 709.22 ohm cm. Calculate the conductivity and equivalent conductance.

For 1/2X_(2)+Y_(2)toXY_(2) relartive rates of species is given as

Fool's gold is

Fool'sgoldis

Fool's gold is:

Food preservative in tomato ketchup is

Food preservatives prevent spoilage of food due to microbial growth. The most commonly used preservatives are

Food preservatives prevent spoilage of food due to microbial growth. The most commonly used preservatives are :

Food preservative among the following

Food preservation by removal of heat involves_____.

Following two reactions can occurs at cathode in the electrolysis of aqueous sodium chloride. Na^(+)+e^(-)toNa(s),E_(Red)^(@)=-2.71V 2H_(2)O(l)+2e^(-)toH_(2)(g)+2OH^(-)(aq)E_(Red)^(@)=-0.83V Which reaction takes place preferentially and why?

Following two graphs are based on the conductometric titration of acid-base reaction. Answer the questions given at the end of it. Select correct statement,

Following statements is/are correct about mixture :

Following solutions at the same temperature will be isotonic:

Following solutions at the same temperature will be isotonic

Following solution are prepared by the mixing different volumes of NaOH of HCl different concentrations.

Following sidwick.s rule of EAN, Co(CO)_(x) will be:

Following Sidwick's rule of EAN, Co(CO)_(x) will be :

Following Sidgwick's rule of EAN, Co(CO)_(x) will be.

Following reactionsoccur spontaneously as written below: 2Fe^(+++)+2l^(-) to 4Fe^(++)+I_(2) O_(2)(g)+H^(+)+4Fe^(++) to 4Fe^(+++)+2H_(2)O Oxidizing power of O_(2),Fe^(+++) and I_(2) will be in the order: