94420 + Interview Questions in Chemistry in Class 12 Page 1852 InterviewSolution

92551.	20.2mL of CH_(6)CO OHreacts with 201.mL of C_(2) H_(5) OHto formCH_(3) CO OC_(2) H_(5) (d = 0 . 902 g/mL) by the following reaction CH_(3) CO OH+ C_(2) H_(5) OH to CH_(3) CO OC_(2) H_(5) + H_(2) O(a) Which compoundis the limitingreagent ? If 27.5 mL of pureethyl acetate is produced, whatis the per cent yield ? Densities of CH_(3) CO OHand C_(2) H_(5) OH are 1 . 0 5 g / mL and 0 . 789 g / mL respectively .
Answer» SOLUTION :(a) `CH_(3) C O O H +C_(2) H_(5) OH to CH_(3) CO OH_(2)H_(5) + H_(2) O` Mole of `CH_(3) CO OH = (2 0.2 xx 1.05)/( 60) =0.3535` Mole of `C_(2) H_(5) OH = (20 . 1 xx 0.789)/( 46)= 0.3447` As `CH_(3) CO OH` reacts with `C_(2) H_(5) OH` in a 1 : 1 mole RATIO and mole of `C_(2) H_(5)OH` is less than that of `CH_(3) CO OH, C_(2)H_(5)OH` is the limitingreagent. (b) As ` C_(2)H_(5)OH` is thelimiting reagent, mole of `CH_(3) CO OC_(2)H_(15)` to be produced THEORETICALLY = 0.3447 mole But experimental yield of `CH_(3) CO OC_(2)H_(5) =(27.5 xx 0.902)/(84)` = 0.2953 mole `:.` per centyieldof ethyl acetate = `(0.2953)/( 0.3447) xx 100` ` = 85 . 66 %`

Discussion

92552.	20.27g of benzene containing 0.2965g of benzoic acid (mol.wt. =122)freezes at 0.137^(@) below the freezing point of pure benzene. If benzoic acid exists as dimer in benzene, find its degree of association. K_(f) for benzene is 5.12^(@)C.m^(-1)
Answer» Solution :Molality (calculated) `=(0.2965//122)/(20.27)xx1000` `=0.12` mole//1000g Molality (observed) `(DeltaT_(f))/(K_(f))=(0.317)/(5.12)=0.0619` mole/1000g `:.i=("molality (observed)")/("molality(calculated)")=(0.0619)/(0.12)=0.5158`……(Eqn 10) Since benzoic acid exists as dimer in BENZENE we have `{:("moles BEFOR association",,1"mole",,,,0),(,,2C_(6)H_(5)COOH,,=,,(C_(6)H_(5)COOH)_(2)),("moles after association",,(1-x),,,,(x)/(2)(x="degree of association")):}` `:.i=(1-x+(x)/(2))/(1)=0.5158` ........(Eqn. 10) `x=0.9684` or `96.84%`

Discussion

92553.	200g. Of 5% solution (by mass) of the solute A is mixed with 300 g. of a 10% solution (by mass) of solute B. The mass percent of A and B in the mixtures are respectively
Answer» 3 and 5 5 and 10 2 and 6 6 and 12 Solution :200 g of 5% solution of A CONTAINS 10 g of A & 300 g. of10% solution of B contains 30 g of B. Both these W ts. Are finally present in 500 g of mixture . :. % Mass of `A= (10 XX 100)/500=2 ` "and "% "Mass of" B = (30 xx 100)/500=6`

Discussion

92554.	20.0dm^(3) of an ideal gas (diatomic C_("v,m")=5R//2) at 673 K and 0.7 MPa expands until pressure of the gas is 0.2 MPa. Calculate q, w, DeltaUandDeltaH for the process if the expansion is : (i) Isothermal and reversible (ii) Adiabatic and reversible (iii) Isothermal and adiabatic (iv) Against 0.2 MPa and adiabatic (v) Against 0.2 MPa and isothermal.
Answer» SOLUTION :(i) `q=-w=17.54kJ,DeltaU=0andDeltaH=0,` (II) `q=0,w=DeltaU=-10.536kJandDeltaH=-14.75kJ` (III) `q=0,w=0,DeltaU=0andDeltaH=0` (IV) `q=0,DeltaU=w=-7.14KJ,DeltaH=-9.996KJ`, (V) `q=-w=10.0KJ,DeltaU=DeltaH=0`

Discussion

92555.	200cm^(3) of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of this solution at 300 K is found to be 2.57xx10^(-3) bar. Calculate the molar mass of the protein.
Answer» Solution :Here, we are given : Mass of solute (protein), `w_(2)=1.26 g` Volume of the solution (V) `=200cm^(3)=0.200L` `pi=2.57xx10^(-3)" bar, T = 300 K, R = 0.083 L bar K"^(-1)"MOL"^(-1)` Substituting these values in the formula, `M_(2)=(w_(2)RT)/(piV),` we get `M_(2)=(1.26gxx0.083 " L bar K"^(-1)"mol"^(-1)xx300K)/(2.57xx10^(-3)" bar"xx0.200L)="61039 g mol"^(-1)`

Discussion

92556.	200.0 mL of calcium chloride solution sotains 3.10xx10^(22)CI^(-) inoc. Calculate the molarity of the solution. Assume that calcium cholride is complately ionised.
Answer» Solution :Setp I. Calculation of mass of `CaCI_(2)` present in solution. lonisation of `CaCI_(2)` and no. of different species in the soluion may be represented as follows : `CaCI(s)overset((aq))TOCA^(2+)(aq)+2CI^(-)(aq)` `(1.505xx10^(22)"MOLECULES")(1.505xx10^(22))(3.01xx10^(22))` `6.022xx10^(22)"molecules of"CaCI_(2) " correspond to mass"=(1.505xx10^(22))/(6.022xx10^(23))xx(111g)=2.774 g` Step II. Calculation of molarity of solution. `"Molarity of soluion (M)"=("Mass of "CaCI_(2)//"MOLAR mass")/("Volume of solution in Litress")` `((2.774g)//(111" g mol"^(-1)))/(200//1000L)=0.125" mol L"^(-1)=0.125 M`

Discussion

92557.	200 mL of water is added to 500 mL of 0.2 M solution. What is the molarity of the diluted solution?
Answer» 0.5010 M 0.2897 M 0.7093 M 0.1428 M Solution :`M_(1)V_(1)=M_(2)V_(2)` `0.2xx100=M_(2)XX(500+200) or M_(2)=0.1428`.

Discussion

92558.	200 tons of Fe_2O_3 containing 40% imurities will give iron by reduction with H_2 equal to :
Answer» 112 TONS 84 tons 560 tons 56 tons Answer :D

Discussion

92559.	200 ml of water is added to 500 ml of 0.2 M solution. What is the molarity of this diluted solution
Answer» 0.5010 M 0.2897 M 0.7093 M 0.1428 M Solution :No. of MILLIMOLES `= 500xx0.2=100` THUS, MOLARITY of DILUTED solution `=(100)/(700)=0.1428 M`

Discussion

92560.	200 mL of an aqueous solution of a protein contains its 1.26g. The osmotic pressure of this solution at 300K is found to be 2.57 xx 10^(-3) bar. The molar mass of protein will be (R = 0.083 L bar mol^(-1)K^(-1))
Answer» 51022 g `mol^(-1)` 122044 g `mol^(-1)` 31011 g `mol^(-1)` 61038 g `mol^(-1)` SOLUTION :According to Van't Hoff equation `M_(B)=(W_(B)xxRxxT)/(pixxV)` `=((1.26g)xx(0.083"L bar"mol^(-1)K^(-1))xx(300K))/((2.57xx10^(-3)"bar")xx(0.2 L))` `=61038 g mol^(-1)`

Discussion

92561.	20.0 mL of a mixture of oxygen (O_(2)) and ozone (O_(3)) was heated till ozone was completely decomposed. The mixture of cooling was found to have a volume of 21 mL. Calculate the percentage of ozone by volume in the mixture.
Answer» Solution :Decomposition of OZONE TAKES place as follows : `underset("2 volume")(2O_(3))rarr underset("3 volume")(3O_(2))` SUPPOSE ozone in the mixture = x mL. Then `O_(2)=(20-x)mL` 2 mL of ozone on decomposition give `O_(2)=3mL` `THEREFORE"x mL of ozone on decomposition will give "O_(2)=(3)/(2)x="1.5 x mL"` `therefore" Total volume of mixture after decomposition "=1.5x+(20-x)mL=20+0.5x` `therefore""20+0.5x=21"(Given)or"0.5x=1"or"x=2mL` `therefore"Percentage of ozone in the mixture"=(2)/(20)xx100=10%.`

Discussion

92562.	200 Ml of a gas had a temperature 27^@ C and pressure 750 mm. Find the pressure at which the gas will have a volume 175mL if the temperature is 77^@ C .
Answer» ANSWER :1000 MM

Discussion

92563.	200 mL of a gas at 30 K is compressed by increasing the pressure by 5%. Calculate te volume of compressed gas at 30 K.
Answer» Solution :`V_(1)=200 ML, P_(4) =P" ATM ", T=303 K` `V_(2)=?, P_(2)=P+(5P)/(100) =(105 P)/(100)" atm ", T=303 K` `therefore` From Boyle.s LAW: `P_(1) V_(1)=P_(2)V_(2)` `Pxx200=(105 P)/(100) xx V_(2)` `therefore V_(2)=190.5 mL`

Discussion

92564.	200 ml of 1M H_2 SO_4, 300 ml 3M HCl and 100 ml of 2M HCI are mixed and made up to 1 litre. The proton concentration in the resulting solution is
Answer» 1.25 M 1.5 M 2.5 M 0.75 M Answer :B

Discussion

92565.	200 ml 0.6 N H_(2)SO_(4) and 100 ml of 0.3 N HCl are mixed together. Acidic normality of the resultant solution is
Answer» 0.5 N 0.9 N 0.3 N 0.6 N Solution :`N_(1)V_(1)+N_(2)V_(2)=N_(3)V_(3)` `200xx0.6+100xx0.3=N_(3)V_(3) and V_(3)=V_(2)+V_(1)=300ml` `thereforeN_(3)=0.5N`

Discussion

92566.	20.0 kg of N_(2(g))and 3.0kg of H_(2(g)) are mixed to produce NH_(3(g)). The amount of NH_(3(g)) formed is
Answer» 17 kg 34 kg 20 kg 3 kg Solution :We know that `N_(2)+3H_(2)hArr2NH_(3)` `{:(28g,6g,34g),(14g,3g,17g):}` Here GIVEN `H_(2)` is 3 kg and `N_(2)` is 20 kg but 3 kg of `H_(2)` can only REACT with 14 G of `N_(2)` and thus the obrained `NH_(3)` will be of 17 kg.

Discussion

92567.	20.0 kg of N_2(g) and 3.0 kg of H_2(g) are mixed to produce NH_3(g) . The amount of NH_3 (g) formed is
Answer» 17 g 34 g 20 g 3 kg Solution :`underset(28)(N_2) + underset(2 xx 3)(3H_2) to underset (34)(2NH_3)` 1 mole of `N_2` (28g) combine with 3 moles of `H_2(6g)` ` therefore ` 3KG of `H_2` can react with `28/6 xx 3 = 14 kg ` of `N_2 H_2` is LIMITING reagent . Moles of `NH_3` formed ` = 34/6 xx 3 = 17 kg `.

Discussion

92568.	200 gm of an Oleum sample labeled as 109 % is mixed with 518 gm water then the molality of final mixture is -
Answer» 5.5 m 4 m 4.45 m 1 m Solution :`(109) % = (100 + 9) % = 109 gm H_(2)SO_(4)` for each 100 gm Oleum From 200 gm Oleum `H_(2)SO_(4)` produced `= 109 + 109 = 218 gm` WATER Remaining `= 518 = 500 gm` Molality `(m) =(218)/(98) XX (1000)/(500) = 4.45 m`

Discussion

92569.	20.0 g of chlorine are evolved in 6 hours from sodium chloride by the current of
Answer» 5 amp 10 amp 2.5 amp 50 amp Solution :`W=Zit = (E)/(96,500)It` `THEREFORE I = (Wxx96,500)/(Fxxt)=(20xx96,500)/(35.5xx6xx3,600) =2.5`

Discussion

92570.	20.0 grams of CaCO_(3)(s) were placed in a closed vessel, heated & maintained at 727° C under equilibrium CaCO_(3)(s)hArrCaO(s) + CO_(2)(g) and it is found that 75% of CaCO_(3) was decomposed. What is the value of K_(p) ? The volume of the container was 15 litres.
Answer» ANSWER :0.0821 ATM

Discussion

92571.	200 cm^(3) of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300K is found to be 2.57xx10^(-3) bar. Calculate the molar mass of the protein.
Answer» Solution :The VARIOUS quantities known to us are as follows : `pi = 2.57xx10^(-3)` bar, `V=200 CM^(3)=0.200` litre T = 300 K `R = 0.083 " L bar mol"^(-1)K^(-1)` Putting above value in equation : `M_(2)=(w_(2)RT)/(pi V)` `M_(2)=(1.26 g XX 0.083 " L bar K"^(-1)mol^(-1)xx 300K)/(2.57xx10^(-3)"bar"xx0.200 L)` `= 61,022 g mol^(-1)`

Discussion

92572.	200 cm^(3) of an aqueous solution of a protein contains 1.26 g of protein. The osmotic pressure of such a solutionat 300 K is found to be 2.57 xx 10^(-3) bar. Calculate the molar mass of the protein. (R = 0.083 L bar "mol"^(-1)K^(-1))
Answer» SOLUTION :`M_(2)=61.038.91g"G "MOL^(-1)`

Discussion

92573.	200 cm^3 of aqueous solution ofa protein contain 1.26g of protein. The osmotic pressure of solution at 300K is found to be 8.3xx10^(-2)bar. Calculate the molar mass of protein, (R=0.083 l bar K^(-1)mol^(-1)
Answer» SOLUTION :`pi`V = nRT, MOLARMASS,`M_B = (W_BRT)/(pi V) = (1.26xx0.083xx300)/(0.2xx8.3xx10^(-2)` = 1890g`mol^(-1)`

Discussion

92574.	Around 20% surface sites have adsorbed N_(2). On heating N_(2) gas evolved form sites and were collected at 0.001 atm and 298 K in a container of volume 2.46cm^(3) the density of surface sites is 6.023xx10^(14)cm^(-2) and surface area is 1000cm^(2) find out the number of surface sites occupied per molecule of N_(2).
Answer» Solution : No. of SURFACE sites PER MOLECULE of `N_2= (" no. of surface sites used to adsorb "N_2 "on a surface AREA of" 1000 cm^2)/(" total no. of adsorbed molecules ")` 2

Discussion

92575.	2^(0) structure of protein have .....................
Answer» HYDROGEN BOND Covalent bond Ionic bond Di-sulphide bond SOLUTION :Hydrogen bond

Discussion

92576.	20% of surface sites are occupied by N_(2) molecules. The density of surface sites is 6.023xx10^(14)cm^(-2) and total surface area is 1000 cm^(2). The catalyst is heated to 300 K while N_(2) is completely desorbed into a pressure of 0.001 atm and volume 2.46cm^(3). Find hte active sites occupied by each N_(2) molecule.
Answer» SOLUTION :Step I. Calculation of total no. of surface sites Density of surface sites `=6.023xx10^(14)cm^(-2)` Total surface area `=1000 cm^(2)=10^(3)cm^(2)` `therefore` Total no. of surface sites `=(6.023xx10^(14))xx(10^(3))=6.023xx10^(17)` Step 2. Calculation of surface sites OCCUPIED by `N_(2)` moolecules. Surface sites occupied by `N_(2)` molecules `=20/100xx6.023xx10^(17)=1.2046xx10^(17)` Step 3. Calculation of total no of `N_(2)` molecules `P=0.001atm, V=2.46cm^(3)=2.46xx10^(-3)L, T=300K` `PV=nRT orn=(10^(-3)ATM xx2.46xx10^(-3)L)/(0.0821L atm K^(-1)MOL^(-1)xx300K)=10^(-7)`mole `therefore` No. of molecules `=(10^(-7))xx(6.023xx10^(23))=6.023xx10^(16)` Step 4. Calculation of no. of sites occupied by each `N_(2)` molecule `=("No. of sites occupied")/("No. of"N_(2)"molecules")=(1.2046xx10^(17))/(6.023xx10^(16))=2`

Discussion

92577.	20% of N_2O_4molecules are dissociated in a sample of gas at 27^@C and 760 torr. Calculate the density of the equilibrium mixture.
Answer» SOLUTION :3.12 g/L

Discussion

92578.	20%of a first order reaction was found to be completed at 10 a.m at 11.30 a.m on the samedat, 20% of the reaction was foundto be remaining . The half life period in minutes of the reaction is
Answer» 90 45 60 30 Solution :a = initial concentration Set - I Given, (a - X) = conc.At time t. Amount LEFT` = (a - x) at t_(10_(A.M)) = a` `(for t_(11.30 A.M))= 100 - 20 = 80`(initial) = ( a for set - 2) Set - 2 ` (a - x) at t_(11.30_(A.M))` = 100 - 80 = 20 (Final) ( t = 11.30 (A.M) - 10 (A.M)= 1.50` H xx 60 = 90 min`) `:. K = (2.303)/(t) "log"(a)/(a -x)` `K = (2.303)/(90) "log"(80)/(20)` `K =(2.303)/(90) log 4 ` ` = (2 .303 xx 0 . 6020)/(90)` K = 0 015 min `:.t_(1//2) = (0.693)/(K)` `t_(1//2) = (0.6930)/(0.015) = 46 .2 ~~ 45 min`

Discussion

92579.	20% N_(2)O_(4) molecules are dissociated in a sample of a gas at 27^(@)C and 760 torr pressure. The density of the equilibrium mixture is :
Answer» `3.1 G L^(-1)` `6.2 g L^(-1)` `12.4 g L^(-1)` `18.0 g L^(-1)` ANSWER :C

Discussion

92580.	2^(0)- nitroparaffinson acid hydrolysis will give
Answer» aldheydes CARBOXYLIC acids KETONES AMIDE ANSWER :C

Discussion

92581.	2.0 molar solution is obtained, when 0.5 mole solute is dissolved in
Answer» 250 ml solvent 250 g solvent 250 ml solution 1000 ml solvent Solution :We KNOW that MOLARITY = `("Number of MOLES of solute")/("Volume of solution in litre")` `therefore 2.0 =(0.5)/("Volume of solution in litre")` `therefore` Volume of solution in litre `=(0.5)/(2.0)=0.250` litre = 250 ml.

Discussion

92582.	If 20 mL of 0.1 M NaOH is added to 30 mL of 0.2 M CH_(3)COOH(pK_(a)=4.74), the pH of the resulting solution is :
Answer» 3.44 4.01 4.44 4.71 Answer :C

Discussion

92583.	20 mL of methane is completely burnt using 50 mL of oxygen . The volume of the gas left after cooling to room temperature is
Answer» 80 mL 40 mL 60 mL 30 mL SOLUTION :`CH_(4) + O_(2) to CO_(2) + 2 H_(2)O` 22400 mL of methane requires 22400 mL of oxygen for burning . `therefore` 20 mL of methane will require = 20 mL of oxygen. This means that 20 mL of methane will burn completely using 20 mL of oxygen . `therefore` Volume of the gas LEFT will be of oxygen only = (50 - 20) = 30 mL .

Discussion

92584.	20 mL of methane is completely burnt using 50 mL of oxygen. The volume of the gas left after cooling to room temperature is
Answer» 80 mL 40 mL 60 mL 30 mL Solution :`underset("1 mL")(CH_(4))+underset("2 mL")(2O_(2))rarr underset("1 mL")(CO_(2))+2H_(2)O(l)` 1 mL of `CH_(4)` reacts with 2 mL of `O_(2)` and produces 1 mL of `CO_(2)` `therefore " 20 mL of "CH_(4)" will react with 40 mL "O_(2) and " PRODUCE 20 mL of "CO_(2)` `therefore O_(2)" left = 10 mL and "CO_(2)` produced = 20 mL `therefore" Total voluem of gas after COOLING"= 10+20` =30 mL

Discussion

92585.	20 mL of methane are burnt with 50 mL of oxygen, the volume of the gas left after cooling to room temperature will be
Answer» 80 mL 60 mL 30 mL 20 mL Solution :`underset(20 mL)underset(1 vol) (CH_(4)) + underset(40 mL)underset(2 vol)(2O_(2)) to underset(20 mL)underset(1 vol) (CO_(2)) + underset(vol.)underset("NEGLIGIBLE")(2 H_2O)` VOLUME of `CO_2` formed = 20 mL Volume of `O_2` USED = 40 mL . `therefore` Volume of `O_2` left unused = 50 - 40 = 10 mL Volume of GAS left after cooling = volume of `CO_2` formed + volume of `O_2` left unused = 20 + 10 = 30 mL

Discussion

92586.	20 mL of a weak monobasic acid (HA) requires 20 mL 0.2 M NaOH for complete titration. If pH of solution upon addition of 10 mL of this alkali to 25 mL of the above solution of HA is 5.8, then pK_(a) of the weak acid is :
Answer» `6.1` `5.8` `5.98` `5.58` ANSWER :C

Discussion

92587.	20 ml of CO is exploded with 30 ml of O_2 at constant temperature and pressure. Final volume of the gases in ml will be
Answer» 35 40 50 60 Answer :B

Discussion

92588.	20 ml of acetic acid reacts with 20 ml of ethyl alcohol to form ethyl acetate. The density of acid and alcohol are 1 g/ml and 0.7 g/ml respectively. The limiting reagent in this reaction is :
Answer» ACETIC acid Ethyl ALCOHOL Acetic acid and ethyl alcohol Ester. Solution :`CH_3COOH + C_2H_5OH to CH_3COOC_2H_5 + H_2O` moles of `CH_3COOH= (20 xx 1)/(60) = 0.333 `moles molesof `C_2H_5OH = (20 xx 0.7)/(46) = 0.304` moles ` therefore ` Ethyl alcohol `(C_2H_5OH)` is the LIMITING reagent .

Discussion

92589.	20 ml of a solution of a weak monobasic acid neutralizes 22.18 ml of solution of NaOH and 20 ml of N/10 HCl neutralizes 21.5 ml of the same NaoH solution. The normality for the acid is nearly
Answer» 10 N 1 N 0.10 N 100 N SOLUTION :`N_(1)V_(1)=N_(2)V_(2)` `N_(1)xx10=N_(2)xx22.18` `N_(1)=(N_(2)xx22.18)/(20)` NAOH solution=HCl solution `N_(2)xx21.5=(1)/(10)xx20` `N_(2)=(20)/(10xx21.5)` By EQ. (i) and (ii) `N_(1)=(20xx22.18)/(20xx10xx21.5)=(22.18)/(215)=0.1N`.

Discussion

92590.	20 ml of a mixture of methane and a gaseous compound of acetylene series were mixed with 100 mol of oxygen and exploded to complete combustion. The volume of the products after colling to original room temperature and pressure, was 80 ml and on treatment with potash Sol. a further contraction of 40 ml was observed. Calculate (a) the molecular formula of the hydrocarbon, (b) the percentage composition of the mixture.
Answer» SOLUTION :Oxygen is PRESENT in excess. This is because the combustion has been done completely (which means methane and the other gas are consumed completely) and on `CO_(2)` absorption 40 ml of a gas is still left over. This let over gas is oxygen. `therefore` Oxygen used up =100-40=60 ml `CO_(2)` produced =40 ml If the molecular formula of the gas of acetylene series is `C_(N)H_(2n-2)` and its volume is a ml, then the volume of `CH_(4)` is (20-a) ml. On combustion `CH_(4)(g) +2O_(2)(g) rarr CO_(2)(g)+2H_(2)O(l)` `C_(n)H_(2n-2)(g)+((3n-1))/(2)O_(2)(g) rarr nCO_(2)+(n-1)H_(2)O(l)` Oxygen used up `=2(20-a) +((3n-1))/(2)a=60` (from `CH_(4)`) (from `C_(n)H_(2n-2)`) `CO_(2)` produced =(20-a)+na =40 (from `CH_(4)`) (from `C_(n) H_(2n-2)`) Solving a=10 ml and n=3 `therefore` Formula of the gas is `C_(3)H_(4)` % of `C_(3)H_(4)=(10)/(20)xx100=50%` % of `CH_(4)=50%`

Discussion

92591.	20 ml of '20 vol' H_(2)O_(2) solution is diluted to 80 ml. The final volume strength of solution is
Answer» 80 VOL' 25 vol' 5 vol' 8 vol' Answer :C

Discussion

92592.	20 ml of 1N HCl, 10 ml of N/2 H_2SO_4 and 30ml of N/3 HNO_3 are mixed together and volume made to 1000 ml. Find out the normality of H^+ ions in the resulting solution
Answer» `7/100N` `7/200N` `7/18N` `7/9N` ANSWER :B

Discussion

92593.	20 ml of 1 N solution of KMnO_(4) just reacts with 20 ml of a solution of oxalic acid. The weight of oxalic acid crystals in 1N of the solution is
Answer» 31.5 G 126 g 63 g 6.3 g Answer :C

Discussion

92594.	20 mL of 10 N HCl are mixed with 10 mL of 36 N HCland the mixture is made 1L. Normality of the mixture will be
Answer» 0.56 N 0.50 N 0.40 N 0.35 N Solution :`N_1V_1 + N_2V_2 = N_3V_3` `10 xx 20 + 36 xx 10 = N_3 xx 1000` ` 200 + 360 = 1000 N_3` `N_3 = 560/1000 = 0.56 N`

Discussion

92595.	20 ml of 0.5 N HCl and 35 ml of 0.1 NaOH are mixed. The resulting solution will
Answer» Be neutral Be basic Turn phenolphthalein solution pink Turn methyl orange red Solution :(i) 20 ml of 0.5 N HCl `0.5 N rArr 1000 m` 0.5 mole HCl is present in 20 ml `= (20 xx 0.5)/(1000) = 1.0 xx 10^(-2)` (ii) 35 ml of 0.1 NaOH `0.1 N rArr 1000 ml` of 0.1 mole NaOH is 35 ml `= (35 xx 0.1)/(1000) = 0.35 xx 10^(-2)` `rArr (1.0 - 0.35) 10^(-2) = 0.65 xx 10^(-2)` mole HCl `HCl = H^(+) + Cl^(-) rArr [HCl] = [H^(+)] + [Cl^(-)]` 55 ml contains `0.65 xx 10^(-2)` mole of `H^(+)` ions `1000 ml - (0.65 xx 10^(-2) xx 10^(3))/(55) = (6.5)/(55)` `pH = -log [H^(+)] = -log (6.5//55)` `= log 55 - log 6.5 = 9.2` Due to acidic nature of solutions the colour of methylorange becomes red.

Discussion

92596.	20 ml of 10 N HCl are mixed with 10 ml of 36 N H_(2)SO_(4) and the mixture is made one litre. Normality of the mixture will be
Answer» 0.56 N 0.50 N 0.40 N 0.35 N Solution :`NV=N_(1)V_(1)+N_(2)V_(2)` `N=(10xx20+36xx10)/(1000)=(200+360)/(1000)rArr N = (560)/(1000)=0.56 N`

Discussion

92597.	20 ml of 0.5 N HCl and 35 ml of 0.1 N NaOHare mixed. The resulting solution will
Answer» be neutral be basic turn PHENOLPHTHALEIN solution pink turn methyl orange red Solution : (i) 20 ml of 0.5 N HCI `0.5 N implies` 1000 ml 0.5 mole HC L is present in`20ml=(20xx0.5)/(1000)=1.0xx10^(-2)` (II) 35 ml of 0.J N NaOH 0.1 N implies 1000 ml of 0.1 mole of NaOH `implies35ml=(35xx0.1)/(1000)=0.35xx10^(-2)` Total `= 20 + 35 = 55 ml` ` implies (1.0 - 0.35)10^(-2) = 0.65 xx 10^(-2) mole HCI` `HCI = H^(+) + CI^(-)` `implies[HCI] = [H\^(+)] + [CH^(-)]` 55 ml contains `0.65 xx 10^(-2)` mole of `H^(+)` ions `1000ML-(0.65xx10^(-2)xx10^(3))/(55)=(6.5)/(55)` `pH=-log[H^(+)]=-log(6.5//55)` `=log55-log6.5=0.92` Due to acidic nature of solution the colour of phenolphthalein becomes pink.

Discussion

92598.	20 mL of 0.1 M HCl is divided into two equal parts and kept in two separate beakers. To one beaker 10 mL of 0.06 M NaOH is added and to the other 10 mL of 0.02 M NaOH is added. Two hydrogen electrodes are placed in the two solution which are linked through a salt bridge. what will be the emf of the cell formed?
Answer» Solution :Each beaker initially contains 10ML of 0.1 M HCl 10 mL of 0.1 M HCl contains HCl=`10xx0.1`Millimole=1 millimole 10 mL of 0.06 M NAOH contains NaOH=`10xx0.06=0.6` millimole HCl left unneutralized in beaker1=1-0.6=0.4 millimole TOTAL volume=20 mL `THEREFORE`In beaker 1, [HCl] or `[H^(+)]=(0.4)/(20)=0.02M` 10 mL of 0.02 M NaOH contains NaOH=`10xx0.02=0.2` millimole `thereforeHCl` left unneutralized in beaker`2=1-0.20` millimole=0.80 millimole total volume=20mL `therefore`In beaker 2, [HCl] or `[H^(+)]=(0.80)/(2)=0.04M` Thus, we have a concentration cell for which `E_(cell)=(0.0591)/(1)"log"(0.04)/(0.02)=0.0178V=17.8mV`

Discussion

92599.	20 ml of 0.1 M acetic acid is mixed with 50 ml of potassium acetate. K_(a) of acetic acid = 1.8 xx 10^(-5) at 27^(@)C. Calculate concentration of potassium acetate if pH of the mixture is 4.8
Answer» 0.1 M 0.04 M 0.4 M 0.02 M Answer :B

Discussion

92600.	20 litre of hydrogen gas weigh about
Answer» 44.8g 20g 12.2 G 1.8 g ANSWER :D

Discussion

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