52671 + Interview Questions in Chemistry in Class 11

1.	When a mixture of LiO_(2)CO_(3) and Na_(2)CO_(3) 10H_(2)O is heated strongly, there occurs a loss of mass due to
Answer» it has LOWER m.pt than NA2CO3 and converts metal SALTS to carbonates which decompose to metal oxides it has HIGHER m.pt than K2CO3 and converts metal salts to carbonates, which decompose to metal oxides it has lower melting POINT than both Na2CO3 and K,CO, and converts the metal salts to carbonates, which decompose to metal oxides it has higher melting point than both Na2CO3 and K2CO3 and converts the metal salts to carbonates which decompose to metal oxide Answer :C

Discussion

2.	A mixture of Na_(2)CO_(3) and NaHCO_(3) having a total weight of 100 gm on heating produced 11.2L of CO_(2) under STP conditions. The percentage of Na_(2)CO_(3) in the mixture is
Answer» `55.8 %` `44.2 %` `84 %` `16 %` ANSWER :D

Discussion

3.	A mixture of N_(2) and H_(2) in the ratio 1 : 3 is allowed to attain equilibrium. At equilibrium, the total pressure is 5xx10^(-5)Nm^(-2) and the mixture contains 40% by volume of NH_(3). Calculate K_(p).
Answer» Solution :Total pressure of the mixture = `5xx10^(5) Nm^(-2)` `therefore` Partial pressure of `NH_(3)=(40xx5xx10^(5))/100=2xx10^(5)Nm^(-2)` The SUM of the partial pressure of `N_(2)andH_(2)=5xx10^(5)-2xx10^(5)=3XX10^(5)` Since `N_(2)andH_(2)` are in the RATIO 1 : 3 `P_(N_(2))=1/4xx3xx10^(5)=0.75xx10^(5)Nm^(-2),P_(H_(2))=3/4xx3xx10^(5)=2.25xx10^(5)Nm^(-2)` `K_(p)=(P_(NH_(3))^(2))/(P_(N_(2))xxP_(H_(2))^(3))=((2xx10^(5))^(2))/((0.75xx10^(5))xx(2.25xx10^(5))^(3))=0.468xx10^(-10)(Nm^(-2))^(-2)`

Discussion

4.	A mixture of N_(2) and H_(2) in the molar ratio1:3 attains equilibrium when 50% of the mixture has reacted , If P is the total pressure of the mixture , then partial pressure of NH_(3) formed is
Answer» `P//2` `P//3` `P//4` `P//6` Solution :`{:(,N_(2),+,3 H_(2),hArr,2 NH_(3)),("Intial",1 " mole",,3 "moles",,0),("At eqm." ,1-x,,3-3x,,2 x):}` But out of4 moles of the mixture , 2 moles have reacted and HENCE 2 moles are LEFT . ` :. (1-x) + ( 3-3x) = 2 OR4 x = 2 or 4 x = 2 or x = 0.5` ` :. " Total moles at equilibrium " ` ` = (1-x) + (3-3x )+ 2 x ` ` = 4 - 2 x = 4 - 1 = 3 ` No . of moles of `NH_(3)` at equilibrium = 2 x = 1 ` :. p_(NH_(3)) = 1/3 xx P = P/3 `

Discussion

5.	A mixture of MgO and Mg weighing 10 g is treated with excess of dil HCl. Then 2.24 lit of H_(2) gas was liberated under STP conditions . The mass of MgO present in the sample is
Answer» 2.4 G 7.6 g 8 g 2 g Answer :B

Discussion

6.	A mixture of methyl alochol and acetone can be separated by
Answer» DISTILLATION Fractional distillation STREAM distillation Distillation under REDUCED pressure Answer :A::B::C::D

Discussion

7.	A mixture of methane and ethylene in the volume ratio x : y has a total volume of 30 ml . On complete combustion it gave 40 ml of CO_(2) . If the ratio had been y :x , instead of x : y , what volume of CO_(2) could have been obtained ?
Answer» 50 ML 100 ml 25 ml 75 ml Answer :A

Discussion

8.	A mixture of KOH and Na_(2)CO_(3) solution required 15 mL of N//20 HCl using phenolphthalein as indicator. The same amount of alkali mixture when titrated using methyl orange as indicator required 25 mL of same acid. Calculate amount of KOH and Na_(2)CO_(3) present in solution.
Answer» SOLUTION :N//A

Discussion

9.	A mixture of K_(2)C_(2)O_(4) and KHC_(2)O_(4) required equal volumes of 0.1 M K_(2)Cr_(2)O_(7) for oxidation and 0.1 M NOH for neutralisation is separate titratiosn. The molar ratio of K_(2)CrO_(4) and KHC_(2)O_(4) in the mixture is
Answer» `1:1` `2:1` `1:2` `3:1` SOLUTION :LET no. of moles of `K_(2)C_(2)O_(4)=a&` no. of moles of `KHC_(2)O_(4)=b` Eqts of `K_(2)C_(2)O_(4)+` Eqts of `KHC_(2)O_(4)` = Eqts of `K_(2)Cr_(2)O_(7)` `(axx2)+(bxx2)=0.1xx6xxv....(1)` Eqts of `KHC_(2)O_(4)` = Eqts of NaOH `(bxx1)=0.1xx1xxv....(1)` On SOLVING (1), (2) a : b = 2 : 1

Discussion

10.	A mixture of hydrocarbon C_(2)H_(2).C_(2)H_(4) & CH_(4) in mole ration of 2:1:2 is burnt completely in the pressence of air containing 80%N_(2) % 20% O_(2) by volume. The mass of air required for the complete combustion of the one gm of mixture is
Answer» `(1728)/(112)` `(1528)/(73)` `(1920)/(120)` `(112)/(1728)` Solution :`{:(,C_(2)H_(2),:,C_(2)H_(4),:,CH_(4)),(,2,:,1,:,2),(,2x,:,x,:,2x):}` `52x+28x+32x=1` `112x=1` `x=(1)/(112)` `{:(C_(2)H_(2)+(5)/(2)O_(2)rarr2CO_(2)+H_(2)O(l)),((2)/(112)""(5)/(112)),(CH_(4)+3O_(2)rarr2CO_(2)+2H_(2)O(l)),((1)/(112)""(4)/(112)),(CH_(4)+2O_(2)rarrCO_(2)+2H_(2)O(l)),((2)/(112)""(4)/(112)):}` Total moles of`O_(2) = (5)/(112)+(3)/(112)+(4)/(112) = (12)/(112)` moles of `O_(2)= (12)/(112)xx32 = (384)/(112)gm` moles of `N_(2) =(12)/(112)xx4 = (48)/(112)` moles of `N_(2) = (48)/(112)xx28=(1344)/(112)gm` mass of air `=(48+1344)/(112) = (1728)/(112) gm`

Discussion

11.	A mixture of hydrogen and helium is prepared such that the number of collisions on the wall per unit time by molecules of each gas is same. Which gas has higher concentration?
Answer» HELIUM hydrogen both have same concentration can't be determined SOLUTION :More diameter `IMPLIES` less conc.

Discussion

12.	A mixture of hydrazine, hydrogen and Cu(II) catalyst is used as rocked fuel. Why?
Answer» Solution :In the presence of `Cu(II)` catalyst, hydrogen peroxide hydraine `(NH_(2)-NH_(2))` to NITROGEN. The reaction is HIGHLY exothermic and in accompanied by large increasein volume. Therefore, the mixture provides an upward and thrust and is used to a rocket FUEL. `NH_(2)-NH_(2)(l)+2H_(2)O_(2)(l) UNDERSET("Heat")overset("Cu(II)")to N_(2)(g)+4H_(2)O(g)+"heat"`

Discussion

13.	A mixture of hydrazine and ……. With a copper (II) catalyst is used as rocket propellant.
Answer» ANSWER :`H_(2)O_(2)`

Discussion

14.	A mixture of hydrazine and H_(2)O_(2) with Cu(II) catalyst is used as a rocket prepellant . Why ?
Answer» Solution :The reaction between hydrazine and `H_(2)O_(2)` is highl EXOTHERMIC and is accoumpanied by a large INCREASE in the volume of the PRODUCTS and hence this MIXTURE is used as a rocket propellant. `NH_(2)NH_(2)(l) + 2H_(2)O_(2)(l) overset( Cu (II))to N_(2)(g)uarr + 4H_(2)O(g) uarr`

Discussion

15.	A mixture of hydrazine and H_(2)O_(2) is
Answer» antiseptic ROCKET FUEL germicide insecticide Solution :A MIXTURE of `N_2H_4` and `H_2O_2` is USED as rocket fuel

Discussion

16.	A mixture of H_(2)O_(2) and hydrzine with copper (II) is used as a rocket propellant. Why ?
Answer» Solution :The reaction between HYDRAZINE `(NH_(2)NH_(2))andH_(2)O_(2)` in the presence of Cu(II) is highly exothermic and is ACCOMPANIED by a LARGE INCREASE in energy as well as in volume of the PRODUCTS. Therefore, this mixture is used as a rocket propellant. `2H_(2)O_(2)(I)+H_(2)NH_(2)(I)overset(Cu(II))toN_(2)(g)uarr+4H_(2)O(g)uarr`

Discussion

17.	A mixture of H_2,N_2 and NH_3 with molar concentrations 5.0xx10^(-3) mol L^(-1) , 4.0xx10^(-3) mol L^(-1) and 2.0xx10^(-3) mol L^(-1)respectively was prepared and heated to 500 K. The value of K_c for the reaction: 3H_2(g)+N_2(g) hArr 2NH_3(g) at this temperature is 60. Predict whether ammonia tends to form or to decompose at this stage of concentration.
Answer» SOLUTION :Reaction quotient, `Q_c` for the reaction, `N_2(g) +3H_2(g) hArr 2NH_3(g)` `Q_c=([NH_3]^2)/([N_2][H_2]^3)` `[NH_3]=2.0xx10^(-3) MOL L^(-1) , [H_2]=5.0xx10^(-3) mol L^(-1)` `[N_2]=4.0xx10^(-3) mol^(-1)` `Q_c=(2.0xx10^(-3))^2/((4.0xx10^(-3))(5.0xx10^(-3))^(3))=8.0xx10^3 L^2 mol^(-2)` `K_c=60` Since `Q_c > K_c `, reaction will GO in the LEFT direction and ammonia will decompose.

Discussion

18.	A mixture of gases having different molecular weights is seperated by which method ?
Answer» Atomlysis Metathesis Ostwald and WALKER method Reverse osmosis Solution :(a) Diffusion method is USED to SEPARATE a mixture of gases having different MOLECULAR weight as rate of diffusion VARIES inversely with molecular mass, i.e. `rprop(1)/(sqrtM)` This method is called atmolysis.

Discussion

19.	A mixture of gases having different molecular weights is separated by which method ?
Answer» Atmolysis Metathesis Ostwald and WALKER method Reverse osmosis Solution :Diffusion method is used to SEPARATE a MIXTURE of gases having DIFFERENT MOLECULAR weights as rate of diffusion varies inversely with molecular mass i.e., `r prop (1)/(sqrtM)` This method is called atmolysis .

Discussion

20.	A mixture of gases contains 4.76 mole of Ne, 0.74 mole of Ar and 2.5 mole of Xe. Calculate the partial pressure of gases, if the total pressure is 2 atm, at a fixed temperature.
Answer» Solution :`P_(NE)=X_(Ne)^(-)P_("Total")` `X_(Ne)=(n_(Ne))/(n_(Ne)+n_(Ar)+n_(XE))=(4.76)/(4.76+0.74+2.5)=0.595` `X_(Ar)=(n_(Ar))/(n_(Ne)+n_(Ar)+n_(Xe))=(0.74)/(4.76+0.74+2.5)=0.093` `X_(Xe)=(n_(Xe))/(n_(Ne)+n_(Ar)+n_(Xe))=(2.5)/(4.76+0.74+2.5)=0.312` `P_(Ne)=X_(Ne)P_("Total")=0.595xx2=1.19` ATM. `P_(Ar)=X_(Ar)P_(Total)=0.093xx2=0.186atm` `P_(Xe)=X_(Xe)P_(Total)=0.312xx2=0.624atm`.

Discussion

21.	A mixture of FeCI_(2) and SnCI_(2) can exist togethr but that of FeCI_(3) and SnCI_(3) cannot explain why ?
Answer» SOLUTION :SINCE both `FeCI_(2)` and `SnCI_(2)` being reducing agents can STAY together but an OXIDISING agent `(FeCI_(3))` and reducing agent `(SnCI_(2))` cannot stay together

Discussion

22.	A mixture of ethyl alcohol and boric acid burn with green edged flame. The green edged flame contains
Answer» TRIETHYL borate ethyl boride Acetaldehyde diborane Answer :A

Discussion

23.	A mixture of ethanol and water contains 54% water by mass. Calculate the mole fraction of alcohol in this solution.
Answer» SOLUTION :`X_("ETHANOL") = (46//46)/(46//46+54//18)=0.25`

Discussion

24.	A mixture of ethane, ethylene and acetylene gases are passed through a Woulfe's bottle containingammoniacal silver nitrate solution. The gas coming out is
Answer» ETHANE ACETYLENE Both ethane and ethylene Original mixture Solution :acetylene is ABSORBED

Discussion

25.	A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen . What would be the partial pressure of dihydrogen in bar ?
Answer» SOLUTION :Suppose the total MASS of the MIXTURE is 100 g. `:. " Mass of " H_2 = 20 g, " Mass of " O_2` `= 100 - 20 = 80 g` NUMBER of moles of `H_2 = 20/2 = 10` Number of moles of `O_2 =80/32 =2.5` Total number of moles in the mixture = `10 + 2.5 = 12.5` `:. " Partial pressure of " H_2 = ("Moles of " H_2)/(" Total no. of moles")XX " Total pressure "`. `=10/(12.5)= 0.8` bar

Discussion

26.	A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.
Answer» Solution :As the mixture of `H_(2)` and `O_(2)` CONTAINS 20% by weight of dihydrogen, therefore, if `H_(2)`=20 g, then `O_(2)`=80 g `n_(H_(2))=(20)/(2)=10" MOLES", n_(O_(2))=(80)/(32)=2.5" moles"` `p_(H_(2))=(n_(H_(2)))/(n_(H_(2))+n_(O_(2)))xxP_("total")=(10)/(10+2.5)XX1" bar"=0.8" bar"`

Discussion

27.	A mixture of dihydgrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.
Answer» Solution :SUPPOSE, Weight of mixture gas = 100 g `therefore` Weight of `H_(2)=20 g` and Weight of `O_(2)=80 g` `therefore` Mole of `n_(H_(2))=(20)/(2)=10` mol and Mole of `n_(O_(2))=(80)/(32)=2.5` mol Total mole of mixture gas `= (10+2.5)=12.5` mol Mole FRACTION of `H_(2), chi_(H_(2))=(n_(H_(2)))/("Total Moles")=(10)/(12.5)` Mole fraction of `O_(2), chi_(O_(2))=(n_(O_(2)))/("Total moles")=(2.5)/(12.5)` PARTIAL Pressure = Mole fraction `xx` Total pressure `therefore p_(H_(2))=n_(H_(2))xx P=(10xx"1 bar")/(12.5)=(10xx"1 bar")/(12.5)=0.8` bar `therefore p_(O_(2))=n_(O_(2))xxP=(2.5xx"1 bar")/(12.5)=0.2` bar

Discussion

28.	A mixture of CuSO_(4).5H_(2)O and MgSO_(4).7H_(2)O was heated until all the water was driven off If 5.0g if mixture gave 3g of anhydrous salts, what was the percentage by mass of CuSO_(4).5H_(2)O in the original mixture :
Answer» Solution :LET the MIXTURE contain `x gCuSO_(4). 5H_(2)O`. `IMPLIES (x)/(249)xx159+(5-x)/(246)xx120=3impliesx=3.72` implies Mass percentage of `CuSO_(4).5H_(2)O = 74.4`

Discussion

29.	A mixture of CO and CO_(2) is found to have a density of 1.50" g "L^(-1) at 20^(@)C and 740 mm pressure. Calculate the composition of the mixture.
Answer» Solution :Let the mol% of CO in mixture =X mol%of `CO_(2)= (100-x)` AVERAGE molecular mass `=([(x xx28)+(100-x)xx24])/(100)` `:. ([(x xx28)+(100-x)xx24])/(100)=(dRT)/(P)""(" Because ",M=(dRT)/(P))` `=((x xx28)+(100-x)xx24)/(100)=1.50gL^(-1)xx0.0821 L atm K^(-1) mol^(-1)xx (293K)/(100)` mol% of `CO=43.38" and mol % of "CO_(2)=(100-x)` `=100-43.38=56.62`.

Discussion

30.	A mixture of CO and CO_(2) is found to have a density of 1.50g litreat 30^(@)C and 730mm What is composition of mixture ? .
Answer» ANSWER :`CO =32.19%, CO_(2) =67.81%`

Discussion

31.	A mixture of CO and CO_2 is found to have a density of 1.5 g/L at 30^(@)C and 740 torr . What is the composition of the mixture .
Answer» `CO = 0.3575 , CO_2 = 0.64225` `CO = 0.64225 , CO_2 = 0.3575` `CO = 0.500 , CO_(2) = 0.500` `CO = 0.2500 , CO_(2) = 0.7500` Solution :Useexpression , ` d = (PM)/(RT)` `implies M = (DRT)/(P) = (1.5 xx 0.0821 xx 303 xx 76)/(74)` `= 38.276` Let one mole of the MIXTURE CONTAIN , N mole of CO and (1 -n) mole of `CO_(2)` `n xx 28 + (1 - n) xx 44 = 1 xx 38.276` [Molecular weights , `CO = 28 CO_(2) = 44]` since n= 0.35775 Number of moles `CO = 0.35775` & number of moles of `CO_(2) = 0.64225`

Discussion

32.	A mixture of camphor and NaCl can be separated by
Answer» Sublimation Evaporation Filtration Decantation Answer :A

Discussion

33.	A mixture of C_(4)H_(8) and C_(2)H_(4) was completely burnt in excess of oxygen yielding equal volumes of CO_(2) and steam. Calculate the percentage (by volume ) of the compounds in the original mixture:
Answer» 25% `C_(4)H_(8)` and 75% `C_(2)H_(4)` 30% `C_(4)H_(8)` and 70% `C_(2)H_(4)` 75% `C_(4)H_(8)` and25% `C_(2)H_(4)` 50% `C_(4)H_(8)` and 50% `C_(2)H_(4)` Answer :A::B::C::D

Discussion

34.	A mixture of an acid anhydride (A) anda mono basic acid (B) on heating produces another mono basic acid (C) of equivalent weight 74 and m an anhydride (D). The acids and anhydrides remain in equilibrium. The anhydride (D) contains two identical fluoro alkyl groups. The acid (B) contains a trifluoro methyl group and has an equivalent weight of 128. Give structures of (A) to (D) with proper reasoning (Atomic weight of fluorine =19).
Answer» Solution :`underset((A))((RCO)_2)O+2R'-underset((B))(COOH)tounderset("MOL.wt.74")underset((C))(2RCOOH)+underset((D))(R'CO)_2O` Mol. Wt. of `R-COOH=74,R+12+32+1=74,R=74-45=29(C_2H_5)` Thus, (C) is `CH_3CH_2COOH` and on the basis of above reaction, acid anhydride (A) is `(C_2H_5CO)_2O`.The acid anhydride (D) contains two identical fluoro alkyl groups. The acid (B) contians a trifluoro methyl group and has an equivalent weight of 128. Hence, mono basic acid (B) is `CF_3CH_2COOH` and so, acid anhydride (D) is `(CF_3CH_2CO)_2O`. Reactions: `underset((A))((CH_3CH_2CO)_2O)+underset("Eq.wt.=128")underset((B))(CF_3CH_2COOH)tounderset("Eq.wt.= 74")underset((C))(CH_3CH_2COOH)+underset((D))((CF_3CH_2CO)_2O)`.

Discussion

35.	A mixture of aluminium and zinc weighing 1.67 g was completely dissolved in acid and 1.69 litres of hydrogen measured at 0°C and 1 atmospheric pressure were evolved. What was the original weight of aluminium in the mixture?
Answer» Solution :Both aluminium and zinc react with sulphuric acid to form hydrogen. Suppose, the mixture contains x g of aluminium. HENCE, amount of zinc in the mixture = `1.67 -x g`. The corresponding equations are: `underset(53.96 g)(2Al) + H_(2)SO_(4) to underset(3 xx 22.4 "L at S.T.P.")(Al_(2)(SO_(4))_(3)) + 3H_(2)` `underset(65.38 g)(Zn) + H_(2)SO_(4) to underset(22.4 L "at S.T.P")(ZnSO_(4)) + H_(2)` `therefore 53.96` g of Al evolve `H_(2) = 3 xx 22.4` L at S.T.P. `therefore x GM` of Al will evolve `H_(2) = (3 xx 22.4)/(53.96) xx x` `=1.24 x` L at S.T.P. SIMILARLY, `therefore 1.24 x + (22.4)/(65.38) xx (1.67 -x) = 1.69` which gives `x = 1.24` Therefore, the WEIGHT of aluminium in the given sample is 1.24 g.

Discussion

36.	A mixture of acidified K_(2)Cr_(2)O_(7) and 10% Kl is titrated againstagainst Na_(2)S_(2)O_(3) (sodium thiosulphate) solution using starch indicator. The colour of the reaction m ixture at the end point is:
Answer» Yellow Blue Green Colourless Answer :C

Discussion

37.	A mixture of acetone and methanol can be separated by
Answer» vacuum distillation Steam distillation FRACTIONAL distillation none of these Solution :ACETONE (b.p. `56^(@)C`) and methano (b.p. `67^(@)C`) DIFFER very less in their boiling points so they are separated by fractional distillation.

Discussion

38.	A mixture of 7g of nitrogen and 8g of oxygen at STP occupies a volume of
Answer» 11,200 mL 22, 400 mL 2240 mL 5600 mL Solution :Total moles `=(7)/(28)+(8)/(32)=(1)/(2)` `IMPLIES V_(STP)=22.4xx(1)/(2)=11.2` LIT

Discussion

39.	A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H_(2)SO_(4). The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be............. .
Answer» SOLUTION :`underset(underset(46 g)("Formic acid"))(HCOOH) overset(conc.H_(2)SO_(4))rarrunderset(28g)(CO)+ H_(2)O` Now 46 g of COOH evolve CO = 28 g `:. 2.3 g` of HCOOH will evolve `CO = (28)/(46) XX 2.3` = 1.4 g `{:(""COOH),("\|" overset(conc.H_(2)SO_(4))rarr CO+CO_(2)+H_(2)O),(COOH""28 g),("Oxalic acid"),(""90g):}` When the GASEOUS mixture of `(CO+CO_(2)` is passed through KOH pellets, `CO_(2)` is ABSORBED while CO phase out `2KOH + CO_(2) rarr K_(2)CO_(3) + H_(2)O` Now, 90g of oxalic acid evolve `CO = 28 g` `:. 4.5 g` of oxalic acid will evolve CO `= (28)/(90) xx 4.5 = 1.4 g` Total amount of CO EVOLVED = 1.4 + 1.4 = 2.8 g

Discussion

40.	A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H_2SO_4 . The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be
Answer» 4.4 1.4 2.8 3 Solution :2.3 gm formic ACID (HCOOH) ` = (2.3)/(46) = 1/20 ` mol HCOOH 4.5 gram oxalic acid `(H_2 C_2O_4) ` = (4.5)/(90) = 1/20 ` mol `H_2C_2O_4` (i) `UNDERSET(1/20 "mol")(HCOOH) overset(CONC. H_2SO_4)(to) underset(1/20 "mol")(CO + H_2O)` (ii) `H_2C_2 O_4 overset(conc. H_2SO_4)(to) CO + CO_2 + H_2O` `CO_2`will be absorbed in KOH anc CO isadditional product. Total MOLES of additional `CO_2 = 1/20 + 1/20 = 1/10`mol ` = 1/10 "mol" = 1/10 xx 28 = 2.8 g`

Discussion

41.	A mixture of 2 moles of N_(2) and 8 moles of H_(2) are heated in a 2 lit vessel, till equilibrium is established. At equilibrium, 04 moles of N_(2) was present. The equilibrium concentration of H_(2) will be
Answer» 2 mole/lit 4 mole/lit 1.6 mole/lit 1 mole/lit Solution : at equlibrium no. of moles of `N_(2)`. 2-X=0.4, x=1.6 MOLAR con. Of `H_(2)` at equlibrium = `(8-3x)/(2)=1.6` at eqm `underset(1-x)underset(1)(SO_(2)Cl_(2)) HARR underset(x)underset()(SO_(2))+underset(1+x)underset(1)(Cl_(2))`

Discussion

42.	A mixture of157 " mol of " N_(2) , 192 " mol of " H_(2) and813 " mol of " NH_(3) " is introduced into a20 L reaction vessel at 500 K. At this temperature , the equilibrium constant ," K_(c) for the reaction , N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) " is " 17 xx 10^(2) Is the reaction mixture at equilibrium ? If not, what is the direction of the net reaction ?
Answer» SOLUTION :The reaction is : ` N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g) ` ` Q_(c) = ([NH_(3)]^(2))/([N_(2) ][H_(2)]^(3)) = ((813)/20"mol"L^(-1))^(2)/(((157)/2"mol" L^(-1))((192 )/20 "mol" L^(-1))^(3))=2 38 xx 10^(3)` As ` Q_(c) != K_(c),` the reaction mixture is not in equilibrium . As `Q_(c) gtK_(c)`, the et reaction will be in the backward direction .

Discussion

43.	A mixture of 1.57 mol of N_2 1.92 mol of H_2 and 8.13 mol of NH_3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, K_c for the reaction N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g)) is 1.7 xx 10^2. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction ?
Answer» Solution :`{:("EQUILIBRIUM reaction :",N_(2(g)) +, 3H_(2(g)) HARR , 2NH_(3(g))),("Mol of mixture :",1.57,1.92,8.13),("mol" L^(-1) "of equili. mixture :", 1.57/20,1.92/20,8.13/20),(,0.0785,0.096,0.4065):}` The equilibrium reaction mixture quotient, `Q_c` `therefore Q_c=[NH_3]^2/([N_2][H_2]^3)=(0.4065)^2/((0.0785)(0.096)^3)` Here, `(Q_c=2.3792) lt (K_c=1.7xx10^2)` . The VALUE of `Q_c` is less . So, the reaction will proceed in the direction of the products (FORWARD reaction).

Discussion

44.	A mixture of 1.57 mol of N_2 1.92 mol of H_2 and 8.13 mol of NH_3 is introduced into a 20 L reaction vessel at 500 K . At this temperature, the equilibrium constant K_c for the reactin. N_(2)(g)+3H_(2) (g)j hArr 2NH_(3)(g) is 1.7 xx 10^(-2) Is this reaction at equilibriuim ? if not , what is the direction of net rection ?
Answer» SOLUTION :The reaction is `N_(2)(g) + 3H_(2)(g) hArr 2NH_(3)(g)` Conecentration quotient `(Q_c)= ([NH_3]^)/([N_2][H_2]^3)` `=((8.13/20 " mol "L^(-1))^1)/((1.57/20 "mol " L^(-1))xx(1.92/20 " mol "L^(-1))^3)=2.38 xx 10^3` The equilibrium contant `(K_c)` for the reaction `=1.7 xx 10^(-2)` As `Q_(C) NE K_(C)` , this means that the reaction is not in a state of equilibrium.

Discussion

45.	A mixture of 1 mole of N_(2) and 3 moles of H_(2) is allowed to react at a constant pressure of 100 bar. At equilibrium, 0.6 mole of ammonia is formed. Calculate the equilibrium constant for the reaction N_(2)+3H_(2)hArr2NH_(3).
Answer» Solution :INITIAL no. of MOLES: `underset(1)(N_(2))+underset(3)(3H_(2))hArrunderset(0)(2NH_(3))` `{:("No. of moles reacting"),("at equilibrium"):}}x""3X""2x=0.6thereforex=0.3` `{:("No. of moles present"),("at equilibrium"):}}{:(1-0.3,3-0.9),(=0.7,=2.1):}" "0.6` `therefore` Total no. of moles present at equilibrium = 0.7 + 2.1 + 0.6 = 3.4 Partial pressure = MOLE fraction `xx` total pressure Since the total pressure = 10 bar. `P_(N_(2))=0.7/3.4xx100`bar = 20.59 bar `P_(H_(2))=2.1/3.4xx100` bar = 61.76 bar `P_(NH_(3))=0.6/3.4xx100` bar = 17.65 bar `K_(p)=(P_(NH_(3))^(2))/(P_(NH_(2))xxP_(H_(2))^(3))=((17.65)^(2))/(20.59xx(61*76)^(3))=6.420xx10^(-5)`

Discussion

46.	A mixture of 1-iodoethane and 1-iodopropane is treated with sodium metal and dry ether to carry out Wurtz reaction. Which of the following hydrocarbons will be formed?
Answer» Propane+HEXANE Ethane+Propane Butane+Propane Butane+PENTANE+Hexane Solution :`CH_(3)CH_(2)I+2Na+ICH_(2)CH_(2)CH_(3) OVERSET(ether)to underset("Butane")(CH_(3)CH_(2)CH_(2)CH_(3))+ underset("Pentane")(CH_(3)CH_(2)CH_(2)CH_(2)CH_(3))+underset("Hexane")(CH_(3)CH_(2)CH_(2)CH_(2)CH_(2)CH_(3))`

Discussion

47.	A mixture of 0.5 mole of CO and 0.5 mole of CO_(2) is taken in a vessel and allowed to effuse out through a pin hole into another vessel which has vacum If a total of A mole has effused out in time t show that M_(1)A+M_(2)(1-A)=36 where M_(1) and M_(2) are mean molar masses of the mixture that has effused out and the mixture still remaining in vessel respectively .
Answer»

Discussion

48.	A mixture is to eb anlysed for penicllin. You add 10.0 mg fo penicllin labelled with .^(14)C that has a specific activity of 0.785mu Ci mg^(-1). From this mixture you are able to isotlate only 0.42 mg of pure penicllin. The specific activity of the isolated pen,om od 0.102mu Ci mg^(-1). How much penicllin was in theoriginal mixture?
Answer» ANSWER :`66.96 MG`

Discussion

49.	A mixture of 0.3 mole of H_(2) and 0.3 mole of I_(2) is allowed to react in a 10 lit vessel at 500^(@)C. If K_(C) of H_(2) +I_(2)
Answer» 0.15 mole 0.24 mole 0.03 mole 0.06 mole Solution : `K_(C)=((2X // 10)^(2))/(((0.3Lx)/(10))^(2))=((2x)/(0.3-x))^(2)` `64=((2x)/(0.3-x))^(2) IMPLIES x=0.24` `:.` unreacted `I_(2)=0.3-0.24=0.06`

Discussion

50.	A mixture in which the mole ratio H_(2) and O_(2)is 2: 1 is used to prepare water by the reaction 2 H_(2) (g) + O_(2) (g) to 2 H_(2)O (g) The total pressure of the container is 0*8 " atm " 20^(@) Cbefore the reaction. Determine the final pressure at 120^(@) C after reaction assuming 80% yield of water.
Answer» SOLUTION :` {:(,2H_(2)(g),+,O_(2)(g),to,2H_(2)O(g)),(" Intial moles" ,2a ,,a,,0):}` As pressureare in the ratio of their moles ` :. 2 a + a = 08 "ATM" or 3 a = 08 or a = (08)/3 "atm " ` THEORETICALLY expected yield of `H_(2)O = 2 a` Actual yield ` 2 a xx 90/100 = 16 a` ` :. H_(2) reacted = 16 a " moles ",O_(2) "reacted "= 0 8 a " moles "` Moles after reaction : ` H_(2)= 2 a - 16 a = 0 4 a, O_(2)= a - 08 a = 02 a ` Total no. of moles ` = 04 a + 02 + 16 a = 22a ` HENCE, FINAL pressure` = 22 xx(08)/3= 0* 59 "atm"`

Discussion

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