52671 + Interview Questions in Chemistry in Class 11

1.	A reaction S_((g))hArr 4S_(2(g))is s carried out by taking 2 moles of S_(8(g))and 0.2 mole of S_(2(g))in a reaction vessel of l lit. Which one is not correct. If K_(c) = 6.3 xx 10^(-6).
Answer» Reaction quotient is `8 xx 1O^(-4)` Reaction proceeds in backward direction Reaction proceeds in forward direction `K_p = 2.55 "atm"^3` SOLUTION :`S_(8) harr 4S_(2)` `Q=((0.2)^(4))/(2)=8 xx 10-^(-4) lt K` `:.` reaction proceds backward `KP = KC (RT)^(Delta n_(g))`, T not given

Discussion

2.	A reaction rate contant doubles between 300 and 310 k. By which of the following factors does the rate constant increases between 400-410 K.
Answer» 3.39 2 1.48 1.1 Solution :For any reaction , the value of RATE CONSTANT DOUBLES between 300 and 310 . For any other temp the rate constant decrease `"log"("log"(k_(2))/(k_(1)))/("log"(k_(2))/(k_(1)))=(10)/(400xx410)*(300xx310)/(10)` `K'_(2)` and `k'_(1)` are rate constant at 410 and 400 K respectively. `k_(2)` and `k_(1)` are rate constant at 310 and 300 K respectively .Here `k_(2).k_(1)=2` `therefore "log"(k_(2)')/(k_(1)')=(3xx31)/(4xx41)xx"log"2` `=(3xx31xx0.3010)/(4xx41)=0.1707` `(k_(2))/(k_(1))`=antilog 0.1707 =1.48

Discussion

3.	(A) : Reaction of NaOH with chlorine is a disproportionation reaction . (R) : All redox reactions are disproportionation reactions .
Answer» Solution :All REDOX reactionsare not disproportianation reaction. So , statement A is TRUE but reason is false.

Discussion

4.	(A) : Reaction of 1-butene with HBr gives 1-bromobutane as major product (R ) : Addition of hydrogen halides to alkenes proceeds according to Markovnikov's rule
Answer» A and R are TRUE, R EXPLAINS A A and R are true, R does not explain A A is true, but R is false A is false, but R is true' Answer :D

Discussion

5.	A reaction occursspontaneously if
Answer» `T Delta S GT Delta H ` and `Delta H` is`+ve` and`Delta S` is -ve `T DELTAS =Delta H ` and both `Delta H` and `DeltaS` are `+ve` `T DeltaS ltDeltaH` and both `DeltaH ` and `DeltaS` are `+ve` `T DeltaS gt Delta H` and both `DeltaH ` and `DeltaS` are `+ve` ANSWER :d

Discussion

6.	A reaction occurs spontaneously if
Answer» `T DeltaS lt DeltaH` and both `DeltaH` and `DeltaS` are +ve `T DeltaS GT DeltaH` and`DeltaH` is +ve and `DeltaS` are -ve `T DeltaS gt DeltaH` and both `DeltaH` and `DeltaS` are +ve `T DeltaS = DeltaH` and both `DeltaH` and `DeltaS` are +ve Solution :For a SPONTANEOUS REACTION, `DELTAG = (-ve), DeltaS = + ve and DeltaH = + ve and T DeltaS gt DeltaH`

Discussion

7.	A reaction N_(2)+ 3H_(2) hArr 2NH_(3) + 92k.j is at equilibrium. If the concentration of N_(2)is increased the temperature of the system
Answer» decreases INCREASES remains constant becomes half Solution :on adding `N_(2)` EQM SHIFTS forward, i.e. more heat released `:.` T of system increases

Discussion

8.	A reaction is carried out aniline as a reactant as well as a solvent. How will you remove unreacted aniline ?
Answer» Solution :The b.p. of ANILINE (457 K) is very high. If aniline (large excess) is DISTILLED froma small amount of the product by SIMPLE distillation, it may cause DECOMPOSITION of the product. Therefore, to avoid decomposition of the product, aniline is REMOVED either by vacuum distillation or by steam distillation.

Discussion

9.	A reaction having equal energies of activation for toward and reverse reaction has
Answer» `DELTA H =0` `Delta H = Delta G =Delta S =0` `Delta S =0` `Delta G =0` ANSWER :A

Discussion

10.	A reaction has both DeltaH and DeltaS - ve.The rate of reaction
Answer» cannot be PREDICTED for change in temperature increases with INCREASE in temperature increases with DECREASE in temperature remains unaffected by change in temperature. Answer :C

Discussion

11.	A reaction between ammonia and boron trifluoride is given below: NH_3 + BF_3 to H_3N :BF_3 Identify the acid and base in the given reaction. Which theory explains it? What is the hybridisation of B and N in the reactants ?
Answer» Solution :As `BF_3` does not have a PROTON but acts as LEWIS acid as it is an electron deficient molecule. It reacts with `NH_3` by accepting the lone pair of electrons from `NH_3` and COMPLETE its octet. The reaction is `BF+:NH_3 to BF_3 larr NH_3` Lewis theory of acids and bases can EXPLAIN it. BORON in `BF_3` is `sp^2` hybridised where N in `NH_3` is `sp^3` hybridised.

Discussion

12.	A reaction A (g) + B (g) hArr 2 C (g)is in equilibrium at a certain temperature. Can we increase the amount of products by (i) adding catalyst (ii) increasing pressure ?
Answer» Solution :(i) No, because CATALYST does not DISTURB the STATE of equilibrium (ii) No, because `n_(p) = n_(r)`

Discussion

13.	A reaction, A+B to C+D+q is found to have a positive entropy change. The reaction will be
Answer» possible at HIGH TEMPERATURE. possible only at low temperature. not possible at any temperature. possible at any temperature. Solution :`DELTAH= -ve and DELTAS =+ve` Reaction is spontaneous so `DELTAG= -ve` So, Reaction will be spontaneous at any temperature.

Discussion

14.	A reaction, A+B rarrC + D+ q is found to havea positive entropy change. The reaction will be (i) possible at high temperature(ii) possible only at low temperature (iii) not possible at any temperature(iv)possible at any temperature
Answer» Solution :Here, `DeltaH =-ve` and `DeltaS = + ve,Delta G = DeltaH - T DeltaS`. For the reaction to be SPONTANEOUS, `DeltaG` shoulbebe `-ve` which will BESO at any TEMPERATURE, i.e., option (iv) is correct.

Discussion

15.	A reaction has both DH and DS negative. The rate of reaction
Answer» Increase with increase in TEMPERATURE Cannot be PREDICTED for CHANGE in temperature Increase with decrease in temperature Remains unaffected by change in temperature ANSWER :C

Discussion

16.	Rate of diffusion of gases is not influenced by
Answer» Both A and R are CORRECT. R is the correct explanation of A. Both A and R are correct. 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

17.	A rarr B, Graph between long_(10) P and (1)/(T) is a straight line of slope (1)/(4.606). Hence, Delta H is
Answer» 1 cal `-4.18` J 4 cal `-1` cal Solution :`Delta G^(0) = - RT ln K_(P) = - RT ln P= - 2.303RT log P` log P `= (-Delta G^(0))/(2.303R).(1)/(T) = (-Delta H^(0))/(2.303RT) + (Delta S^(0))/(2.303R)` `RARR` SLOPE `= (-Delta H^(0))/(2.303RT) = (1)/(4.606)` `rArr Delta H^(0) = - 1 "cal" = -4.18J`

Discussion

18.	A radioisotope will not emit
Answer» alpha and BETA RAYS simultaneously beta and gamma rays simultaneously gamma and alpha rays simultaneously gamma rays only Solution :A RADIO isotope first EMITS `alpha` or `beta` particles , then it becomes UNSTABLE and emits `gamma`-rays

Discussion

19.	A radioactive substance has a constant actity of 2000 disintegraion/mintue. The material is swparated into two fractions, one of which has an initlial activity of 100 disntegrationper secoundwhile the other fraction decays with t_(1//2) = 24 hour. The total activity in both samples after 48 hour of separation is:
Answer» `1500` `1000` `1250` `2000` SOLUTION :The problem refers that rate is constant.

Discussion

20.	A radioactive isotope decays as ._(Z)A^(m) rarr ._(Z-2)B^(m-4) rarr ._(Z-1)C^(m-4) The half lives of A and B are 6 months respectively. Assuming that initially only A was present, will it be possibel to achieeveradioactive equilibrium fo B? If so, what would be the ratio ofnuclei A and B? If so, what wouldhappen if the half-livesfor A and B were 10 months adn 6 months respectively?
Answer» ANSWER :`0.6`, No

Discussion

21.	A radioactive isotope decays as ._(Z)A^(m) rarr ._(Z-2)B^(m-4) rarr ._(Z-1)C^(m-4) The half lives of A and B are 6 months respectively. Assuming that initially only A was present, will it be possibel to achieeveradioactive equilibrium fo B? If so, what would be the ratio ot A and B? If so, what wouldhappen if the half-livesfor A and B were 10 months adn 6 months respectively?
Answer» ANSWER :`0.6, 1.66`

Discussion

22.	A radiation of 2000Å falls on the metal whose work function is 4.2 eV. Then the kinetic energy of the fastest photo eletron is
Answer» `6.4xx10^(-10)` J `16xx10^(-10)` J `1.6xx10^(-19)` J `3.2xx10^(-19)` J Solution :Kinetic energy (KE) `=hv-W` `(HC)/lambda-4.2eV(6.62xx10^(-34)xx3xx10^(8))/(2000xx10^(-10))=-4.2xx1.6xx10^(-19)` `=9.9xx10^(-19)j-6.7xx10^(-19)J=3.2xx10^(-19)` J

Discussion

23.	A quantity of KMnO_(4) was boiled with HCl and the gas evolved was led into a solution of KI. When the reaction was complete, the I_(2) liberated was titrated with titrated with a solution of hypo containing 124 g of Na_(2)S_(2)O_(3). 5H_(2)O per litre. It was found that exactly 60 mL were required to decolourise the solution of I_(2). What weight of KMnO_(4) was used ?
Answer» SOLUTION :N//A

Discussion

24.	"A quantity of " PCl_(5) " was heated in a " 10 dm^(3) " vessel at " 250^(@)C : PCl _ (5) (g) hArr PCl_(3) (g) + Cl_(2) (g) . At equilibrium, the vessel contains 01 mole of PCl_(5)and 02mole of Cl_(2). The equilibrium constant of the reaction is
Answer» `004` `0025` `002` `005` Answer :A

Discussion

25.	A quantity of 4.0 moles of an ideal gas at 20^(@)C expands isothermally against a constant pressure of 2.0 atm from 1.0 L to 10.0L. What is the entropy change of the system (in cals)?
Answer» Solution :Expansion against constant pressure means the process is IRREVERSIBLE `Delta S = nR l N (V_(2))/(V_(1)) = 4 xx 2 l n (10)/(1) = 18.424` CAL

Discussion

26.	A quantity of 25.0 mL of solution containing both Fe^(2+) and Fe^(3+) ions is titrated with 25.0 mL of 0.0200 M KMnO_(4) (in dilute H_(2)SO_(4)). As a result, all of the Fe^(2+) ions are oxidised to Fe^(3+) ions. Next 25 mL of the original solution is treated with Zn metal finally, the solution requires 40.0 mL of the same KMnO_(4) solution for oxidation to Fe^(3+). MnO_(4)^(-)+5Fe^(2+)+8H^(+)toMn^(2+)+5Fe^(3+)+4H_(2)O IF 0.02 M K_(2)Cr_(2)O_(7) is used instead of 0.02 M KMnO_(4) its volume required in these titrations are respectively
Answer» 25 mL, 40 mL 25 mL, 15 mL 20.8 mL, 33.3 mL 10.4 mL, 16.7 mL Solution :m.eqts of `Fe^(+2)` = mEq.ts of `KMnO_(4)` implies m. Eq.ts of `Fe^(+2)=25xx0.02xx5=2.5` In PRESSURE of Zn, m.eqts of `Fe^(+3)` = m eqts of `KMnO_(4)` implies m. Eq.ts of `Fe^(+2)=25xx0.02xx5=2.5` In pressure of Zn, me.eqtsof `Fe^(+3)` = m eqts of `KMnO_(4)` m.eqts of `Fe^(+3)` = m eqts of `KMnO_(4)` m.eqts of `Fe^(+3)=40xx0.02xx5=4` Now in the second case `K_(2)Cr_(2)O_(7)` used instead of `KMnO_(4)`, m eq. `K_(2)Cr_(2)O_(7)` =m eq of `Fe^(+2)` `0.02xx6xxv=2.5` `implies V=(2.5)/(0.12)=20.8ml` In pressure of Zn m. eqts of `K_(2)Cr_(2)O_(7)` = m eq of `Fe^(+3)` `0.02xx6xxv=2` `impliesV=(4)/(0.12)=33.3ml`

Discussion

27.	A quantity of 25.0 mL of solution containing both Fe^(2+) and Fe^(3+) ions is titrated with 25.0 mL of 0.0200 M KMnO_(4) (in dilute H_(2)SO_(4)). As a result, all of the Fe^(2+) ions are oxidised to Fe^(3+) ions. Next 25 mL of the original solution is treated with Zn metal finally, the solution requires 40.0 mL of the same KMnO_(4) solution for oxidation to Fe^(3+). MnO_(4)^(-)+5Fe^(2+)+8H^(+)toMn^(2+)+5Fe^(3+)+4H_(2)O Zinc aded in the second titration wil
Answer» oxidize `Fe^(2+)` to `Fe^(3+)` REDUCE `Fe^(3+)` to `Fe^(2+)` reduce `Fe^(3+)` to Fe reduce `Fe^(2+)` to Fe Solution :In presence of ZN `Fe^(+3)` reduced to `Fe^(+2)`

Discussion

28.	A proton is moving with kinetic energy 5 xx 10^(-27) J. What is the wavelength of the de Broglie wave associated with it ?
Answer» Solution :Mass of proton `= (1.008 XX 10^(-3))/(6.02 xx 10^(23)) KG = 1.67 xx 10^(-27) kg` Further proceed as in Solved PROBLEM 3.

Discussion

29.	A proton is accelerated to a velocity of3xx 10^(7) m s^(-1). If the velocity can be measured with a precision of +- 0.5%, calculate the uncertainty in position of proton [h = 6.6 xx 10^(-34)Js, mass of proton = 1.66 xx 10^(-27)kg]
Answer» Solution :`DELTA v = (0.5)/(100) xx (3 xx 10^(7)) ms^(-1) = 1.5 xx 10^(5) m s^(-1)` `Delta X = (h)/(4pi m Delta v) = (6.6 xx 10^(-34) KG m^(2) s^(-1))/(4 xx 3.14 xx 1.66 xx 10^(-27) kg xx 1.5 xx 10^(5) ms^(-1)) = 2.11 xx 10^(-13) m`

Discussion

30.	A proton is accelerated to 1/10th of the velocity of light. If its velocity can be measured with a precision of +- 0.5%, what must be its uncertanity in position?
Answer» Solution :Velocity of LIGHT `= 3 xx 10^(8) ms^(-1)` `:.` Velocity of proton `= (1)/(10) xx 3 xx 10^(8) = 3 xx 10^(7) ms^(-1)` `Delta v = (0.5)/(100) xx 3 xx 10^(7) = 1.5 xx 10^(5) ms^(-1)` Applying Heisenberg's uncertanity principle. `Delta x. m Delta v = (h)/(4pi)` We get `Delta x = (h)/(4pi xx m Delta v) = (6.626 xx 10^(-34) kg m^(2) s^(-1))/(4 xx 31.43 xx 1.66 xx 10^(-27) kg xx 1.5 xx 10^(5) ms^(-1)) = 2.11 xx 10^(-13) m`

Discussion

31.	A proton is about 1840 times heavier than an electron. When it is accelerated by a potential difference of 1k V, its kinetic energy will be
Answer» 1840 keV 1/1840 ke V 1 keV 920 keV Solution :ENERGY acquired by ELECTRON (as K.E.) on being ACCELERATED by a potential difference of 1kV = 1KEV. As proton has the same CHARGE and as energy = potential difference `xx` charge, hence proton will have the same kinetic energy

Discussion

32.	Electrons are accelerated through a potential difference of 150V. Calculate thede Broglie wavelength.
Answer» `lambda_3 = lambda_p` `lambda_e lt lambda_p` `lambda_e gt lambda_p` `n_3 to n_1` Solution :`(LAMBDA E)/(lambda p)=SQRT((MP)/(me)) = sqrt(1837)`

Discussion

33.	An alpha-particle and a proton are accelerated from rest by the same potential, thenthe ratio of their de-Broglie wavelength is
Answer» `SQRT2` `1/sqrt2` `2SQRT2` `2` Solution :`lambda = h/(sqrt(2qVm)) , lambda_(alpha) = h/(sqrt(2xx2xxV XX m_(alpha)))` `implieslambda_(alpha) = (h)/(sqrt(2xx2xx V xx 4))` `lambda _P =(h)/(sqrt(2xx1xxVxx m_p)) , lambda_p = (h)/(sqrt(2xx1xxV xx 1))` `(lambda_p)/(lambda_(alpha)) = sqrt((16)/(2)) implies(lambda_P)/(lambda_(alpha)) = 2sqrt2`

Discussion

34.	A proper control of pH is very essential for many industrial as well as biological processes. Solutions with a definite pH can be prepared from single salts or mixtures ofacids/bases and their salts. We also require solutions which resist change in pH and hence have a reserve value. Such solutions are called Buffer solutions. Henderson gave a theoretical equation for preparing acidic buffers of definite pH. The equation is pH=pK_(a) + log. (["Salt"])/(["Acid"]) a similar equation is used for basic buffers. The pH of aqueoussolution of single salts iscalculated by using an expression whose exact form depends upon the nature of the salt. For example, for salts of strong acid and weak base, the expression is pH = 7-(1)/(2) pK_(b)-(1)/(2) log c For weak acids and bases used by a chemist, data are given below: K_(a)=1.8xx10^(-5), K_(b)=1.8xx10^(-5) Also logarithmic values of some numbers are given below : log 1.8 = 0.2553,log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990 Report the correct pH valuein each of the following cases. 100 mL of 0.05 M NH_(4)OH mixed with 100 mL of 0.10 M HCl solution
Answer» 1.6 2.6 3.6 4.6 Solution :100 ML of 0.05 M `NH_(4)OH = 5`mmol of `NH_(4)OH` 100 mL of 0.10 M HCL = 10 mmol of HCl 5 mmol of `NH_(4)OH ` will react with5 mmol of HCl to FORM 5 mmol of `NH_(4)Cl` HCl left = 5 mmol . Volume of solution = 200 mL `:. [HCl]=(5)/(200) = 0.025M`, `[H^(+)] =0.025 M`, `pH = - LOG (0.025) = 1.602`

Discussion

35.	A proper control of pH is very essential for many industrial as well as biological processes. Solutions with a definite pH can be prepared from single salts or mixtures ofacids/bases and their salts. We also require solutions which resist change in pH and hence have a reserve value. Such solutions are called Buffer solutions. Henderson gave a theoretical equation for preparing acidic buffers of definite pH. The equation is pH=pK_(a) + log. (["Salt"])/(["Acid"]) a similar equation is used for basic buffers. The pH of aqueoussolution of single salts iscalculated by using an expression whose exact form depends upon the nature of the salt. For example, for salts of strong acid and weak base, the expression is pH = 7-(1)/(2) pK_(b)-(1)/(2) log c For weak acids and bases used by a chemist, data are given below: K_(a)=1.8xx10^(-5), K_(b)=1.8xx10^(-5) Also logarithmic values of some numbers are given below : log 1.8 = 0.2553,log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990 Report the correct pH valuein each of the following cases. 100 mL of 0.10 M NH_(4)OH mixed with 100 of 0.05M HClsolution
Answer» 6.25 7.25 8.25 9.25 Solution :100 ml of 0.10 M `NH_(4)OH` = 10mmol of `NH_(4)OH` 100 mL of 0.05 M HCl = 5 mmol of HCl 5 mmol of HCl will react with 5 mmol of `NH_(4)OH` to form 5 mmol of `NH_(4)Cl` `NH_(4)OH ` LEFT = 5 mmol , `NH_(4)Cl` FORMED = 5 mmol Thus, it is a basic buffer with `[NH_(4)OH]=[NH_(4)Cl]` `POH = pK_(b) + log .(["SALT"])/(["Base"])` `=-log (1.8xx10^(-5))=4.75` `:. pH = 14 - 4.759.25`

Discussion

36.	A proper control of pH is very essential for many industrial as well as biological processes. Solutions with a definite pH can be prepared from single salts or mixtures ofacids/bases and their salts. We also require solutions which resist change in pH and hence have a reserve value. Such solutions are called Buffer solutions. Henderson gave a theoretical equation for preparing acidic buffers of definite pH. The equation is pH=pK_(a) + log. (["Salt"])/(["Acid"]) a similar equation is used for basic buffers. The pH of aqueoussolution of single salts iscalculated by using an expression whose exact form depends upon the nature of the salt. For example, for salts of strong acid and weak base, the expression is pH = 7-(1)/(2) pK_(b)-(1)/(2) log c For weak acids and bases used by a chemist, data are given below: K_(a)=1.8xx10^(-5), K_(b)=1.8xx10^(-5) Also logarithmic values of some numbers are given below : log 1.8 = 0.2553,log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990 Report the correct pH valuein each of the following cases. 100 mL of 0.10 M NaOH mixed with 100 mL 0.10 M CH_(3)CO OH solution
Answer» 5.72 6.72 7.72 8.72 Solution :100 ml of 0.1 M NaOH = 10 mmol of NaOH 100 ml of 0.10 M `CH_(3)CO OH` = 10 mmol of `CH_(3)CO OH` They will react completely to form 10 mmol of `CH_(3)CO ON a` Volume of solution = 200 mL `:. [CH_(3)CO ON a]=(10)/(200)= 0.05 M` As `CH_(3)CO ON a` is a salt of WEAK acid with strong BASE, `pH= 7 +(1)/(2) pK_(a)+(1)/(2) log c` `= 7 + (1)/(2) (4.74)+(1)/(2) log 0.05` `=7+2.37 + (1)/(2) (-1.301)=8.72`

Discussion

37.	(A) : Propene reacts with HBr to give isopropyl bromide. (R ) : Addition of hydrogen halide to alkenes follows Markownikoff's rule
Answer» A and R are TRUE, R explains A A and R are true, R does not EXPLAIN A A is true, but R is false A is false, but R is true' Answer :A

Discussion

38.	A proper control of pH is very essential for many industrial as well as biological processes. Solutions with a definite pH can be prepared from single salts or mixtures ofacids/bases and their salts. We also require solutions which resist change in pH and hence have a reserve value. Such solutions are called Buffer solutions. Henderson gave a theoretical equation for preparing acidic buffers of definite pH. The equation is pH=pK_(a) + log. (["Salt"])/(["Acid"]) a similar equation is used for basic buffers. The pH of aqueoussolution of single salts iscalculated by using an expression whose exact form depends upon the nature of the salt. For example, for salts of strong acid and weak base, the expression is pH = 7-(1)/(2) pK_(b)-(1)/(2) log c For weak acids and bases used by a chemist, data are given below: K_(a)=1.8xx10^(-5), K_(b)=1.8xx10^(-5) Also logarithmic values of some numbers are given below : log 1.8 = 0.2553,log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990 Report the correct pH valuein each of the following cases. 100 mLof 0.10 MNaOH mixed with 100 ml of 0.05 M CH_(3)C O O H solution
Answer» 10.4 11.4 12.4 13.4 Solution :100 ml of 0.1 M NAOH = 10 mmol of NaOH 100 mL of 0.05 M `CH_(3)CO OH` = 5 mmol of `CH_(3)CO OH` 5 mmol will REACT with 5 mmol of NaOH to form 5 mmol of `CH_(3)CO Ona` NaOH LEFT = 5 mmol. Volume of the solution = 200 ml `:. [NaOH]=(5)/(200) M = 0.025M` or `[OH^(-)]=0.025M` or `[OH^(-)]=0.025M` `:. [H^(+)]=10^(-14)//0.025=4.0xx10^(-13)`, pH= 13- 0.602 = 12.398 `~=` 12.4

Discussion

39.	A proper control of pH is very essential for many industrial as well as biological processes. Solutions with a definite pH can be prepared from single salts or mixtures ofacids/bases and their salts. We also require solutions which resist change in pH and hence have a reserve value. Such solutions are called Buffer solutions. Henderson gave a theoretical equation for preparing acidic buffers of definite pH. The equation is pH=pK_(a) + log. (["Salt"])/(["Acid"]) a similar equation is used for basic buffers. The pH of aqueoussolution of single salts iscalculated by using an expression whose exact form depends upon the nature of the salt. For example, for salts of strong acid and weak base, the expression is pH = 7-(1)/(2) pK_(b)-(1)/(2) log c For weak acids and bases used by a chemist, data are given below: K_(a)=1.8xx10^(-5), K_(b)=1.8xx10^(-5) Also logarithmic values of some numbers are given below : log 1.8 = 0.2553,log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990 Report the correct pH valuein each of the following cases. 100 mL of 0.05 M NaOH mixed with 100 ml of 0.10 M CH_(3)CO OH solution
Answer» 3.75 4.75 5.75 6.75 Solution :100 ml of 0.05 M NAOH = 5 mmol of NaOH 100 mL of 0. 1 M `CH_(3)CO OH= 10 ` mmol of `CH_(3)CO OH` 5 mmol of NaOH will react with 5 mmol of `CH_(3)CO OH` to form 5 mmol of `CH_(3)CO ON a` `CH_(3)CO OH ` LEFT = 5 mmol, `CH_(3)CO ON a` FORMED = 5 mmol Total VOLUME = 200 mL `:. [CH_(3)CO OH]=[CH_(3)CO ON a]` `=(5)/(200) M = 0.025 M` As it is a buffer of weak acid and its salt with strong base, `pH = pK_(a)+log. (["Salt"])/(["Base"])` `=-log (1.8xx10^(-5))+log 1` `=5- (0.255)=4.745`

Discussion

40.	(A) , product (A) is:
Answer» `H_(2)C=CH-CH=CH_(2)` `CH_(3)-C-=C-CH_(3)` `CH_(3)-CH_(2)-C-=CH` `CH_(2)-CH=C=CH_(2)` Solution :Formation of VICINAL di-halide followed by two consective `E_(2)` REACTION. (elimination biomolecular)

Discussion

41.	(A) Product (A) is
Answer» Both 1& 2 None of these Solution :ALDOL PRODUCT

Discussion

42.	A process is taking place at constant temperature and pressure. Then
Answer» `DELTA H = Delta E` `Delta H = T DeltaS` `Delta H = 0` `Delta S = 0` Solution :At const.T & P,PROCESS is in EQUILIBRIUM . `DeltaG=0`.Hence,`DeltaH = T DeltaS`

Discussion

43.	A process is spontaneous at all temperatures when
Answer» `DELTA H =- ve, Delta S=-ve ` `Delta H =+ ve, Delta S=-ve ` `Delta H =-ve, Delta S=+ ve` `Delta H =+ ve, Delta S =+ ve ` ANSWER :C

Discussion

44.	A process in which no heat change takes place is called
Answer» An ISOTHERMAL process An ADIABATIC process An ISOBARIC process An isochoric process Answer :B

Discussion

45.	A process in which heat can flow from system to surroundings or vice-versa is called "…......................" process.
Answer» SOLUTION :ADIABATIC

Discussion

46.	(A): Pressure of the gas increases with the decrease in volume at constant temperature (R): Number of molecules per unit volume increases with decrease in volume consequently number of collisions on the wall increases
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 :B

Discussion

47.	A pressure cooker reduces cooking time for food because.
Answer» Cooking INVOLVES chemical changes helped by a rise in temperature Heat is more evenly distributed in the cooking space BOILING point of water INVOLVED in cooking is increased The higherpressure inside the COOKER crushes the food MATERIAL Answer :C

Discussion

48.	Apre-weighed vessel was filled with oxygen at NTP and weighed. It was thenevacuated, filled with SO_2 at the same temperature andpressure and again weighed. The weight of oxygen will be
Answer» The same as that of `SO_2` 1/2 that of `SO_2` TWICE that of `SO_2` 1/4 that of `SO_2` ANSWER :B

Discussion

49.	A porpous cup is filled with H_(2) gas at the atmospheric pressure and is connected to a thin glass tube a vertical position. The second end of the tube is immersed in water below it. After some time, water rises in the glass tube. Explain giving reasons.
Answer» Solution :Pressure is EXERTED DUE to bombardment of gaseous molecues with the WALL of CONTAINER.

Discussion

50.	A porous tube containing a mixture of H_2 and O_2 is placed in a flask. After the diffusion for 25 seconds into the flask, what would be the composition of the gases in the flask?
Answer» ANSWER :`4:1`

Discussion

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A reaction S_((g))hArr 4S_(2(g))is s carried out by taking 2 moles of S_(8(g))and 0.2 mole of S_(2(g))in a reaction vessel of l lit. Which one is not correct. If K_(c) = 6.3 xx 10^(-6).

A reaction rate contant doubles between 300 and 310 k. By which of the following factors does the rate constant increases between 400-410 K.

(A) : Reaction of NaOH with chlorine is a disproportionation reaction . (R) : All redox reactions are disproportionation reactions .

(A) : Reaction of 1-butene with HBr gives 1-bromobutane as major product (R ) : Addition of hydrogen halides to alkenes proceeds according to Markovnikov's rule

A reaction occursspontaneously if

A reaction occurs spontaneously if

A reaction N_(2)+ 3H_(2) hArr 2NH_(3) + 92k.j is at equilibrium. If the concentration of N_(2)is increased the temperature of the system

A reaction is carried out aniline as a reactant as well as a solvent. How will you remove unreacted aniline ?

A reaction having equal energies of activation for toward and reverse reaction has

A reaction has both DeltaH and DeltaS - ve.The rate of reaction

A reaction between ammonia and boron trifluoride is given below: NH_3 + BF_3 to H_3N :BF_3 Identify the acid and base in the given reaction. Which theory explains it? What is the hybridisation of B and N in the reactants ?

A reaction A (g) + B (g) hArr 2 C (g)is in equilibrium at a certain temperature. Can we increase the amount of products by (i) adding catalyst (ii) increasing pressure ?

A reaction, A+B to C+D+q is found to have a positive entropy change. The reaction will be

A reaction, A+B rarrC + D+ q is found to havea positive entropy change. The reaction will be (i) possible at high temperature(ii) possible only at low temperature (iii) not possible at any temperature(iv)possible at any temperature

A reaction has both DH and DS negative. The rate of reaction

Rate of diffusion of gases is not influenced by

A rarr B, Graph between long_(10) P and (1)/(T) is a straight line of slope (1)/(4.606). Hence, Delta H is

A radioisotope will not emit

A radiation of 2000Å falls on the metal whose work function is 4.2 eV. Then the kinetic energy of the fastest photo eletron is

"A quantity of " PCl_(5) " was heated in a " 10 dm^(3) " vessel at " 250^(@)C : PCl _ (5) (g) hArr PCl_(3) (g) + Cl_(2) (g) . At equilibrium, the vessel contains 0*1 mole of PCl_(5)and 0*2mole of Cl_(2). The equilibrium constant of the reaction is

A quantity of 4.0 moles of an ideal gas at 20^(@)C expands isothermally against a constant pressure of 2.0 atm from 1.0 L to 10.0L. What is the entropy change of the system (in cals)?

A proton is moving with kinetic energy 5 xx 10^(-27) J. What is the wavelength of the de Broglie wave associated with it ?

A proton is accelerated to a velocity of3xx 10^(7) m s^(-1). If the velocity can be measured with a precision of +- 0.5%, calculate the uncertainty in position of proton [h = 6.6 xx 10^(-34)Js, mass of proton = 1.66 xx 10^(-27)kg]

A proton is accelerated to 1/10th of the velocity of light. If its velocity can be measured with a precision of +- 0.5%, what must be its uncertanity in position?

A proton is about 1840 times heavier than an electron. When it is accelerated by a potential difference of 1k V, its kinetic energy will be

Electrons are accelerated through a potential difference of 150V. Calculate thede Broglie wavelength.

An alpha-particle and a proton are accelerated from rest by the same potential, thenthe ratio of their de-Broglie wavelength is

(A) : Propene reacts with HBr to give isopropyl bromide. (R ) : Addition of hydrogen halide to alkenes follows Markownikoff's rule

(A) , product (A) is:

(A) Product (A) is

A process is taking place at constant temperature and pressure. Then

A process is spontaneous at all temperatures when

A process in which no heat change takes place is called

A process in which heat can flow from system to surroundings or vice-versa is called "…......................" process.

(A): Pressure of the gas increases with the decrease in volume at constant temperature (R): Number of molecules per unit volume increases with decrease in volume consequently number of collisions on the wall increases

A pressure cooker reduces cooking time for food because.

Apre-weighed vessel was filled with oxygen at NTP and weighed. It was thenevacuated, filled with SO_2 at the same temperature andpressure and again weighed. The weight of oxygen will be

A porpous cup is filled with H_(2) gas at the atmospheric pressure and is connected to a thin glass tube a vertical position. The second end of the tube is immersed in water below it. After some time, water rises in the glass tube. Explain giving reasons.

A porous tube containing a mixture of H_2 and O_2 is placed in a flask. After the diffusion for 25 seconds into the flask, what would be the composition of the gases in the flask?

"A quantity of " PCl_(5) " was heated in a " 10 dm^(3) " vessel at " 250^(@)C : PCl _ (5) (g) hArr PCl_(3) (g) + Cl_(2) (g) . At equilibrium, the vessel contains 01 mole of PCl_(5)and 02mole of Cl_(2). The equilibrium constant of the reaction is