94420 + Interview Questions in Chemistry in Class 12 Page 1872 InterviewSolution

93551.	0.1 Msolution of urea, at a given temperature, is isotonic with
Answer» 0.1M NACL solution 0.1 M GLUCOSE solution 0.02 M KCL solution 0.1 `M BaCl_(2)` solution Answer :B

Discussion

93552.	0.1 M NH_4OH is 1% ionized. Find the extent of hydrolysis of 0.1 M NH_4 Cl .
Answer» SOLUTION :`10^(-4)`

Discussion

93553.	0.1 M NaOH was gradually added to 50 ml of aqueous of CH_(3)COOH.Find pH of the solution at the equivalent point of the reaction. [ log 5 = 0.70]
Answer» Solution :CONCENTRATION of the salt formed `(0.1 xx 50)/(100)= (0.1)/(2)= 0.05 [LOG C= log 5 xx 10^(-2)= 0.7 -2.0 = -1.30]` `pH= 7+ (1)/(2) [pKa + log C]= 7 + (1)/(2) [4.7 +(-1.30)] =7 + (1)/(2) [3.40]= 8.70`

Discussion

93554.	0.1 M NaCl and 0.05 M BaCl_(2) solutons are separated by a semi-permeable membrane in a container. For this system choose the correct answer
Answer» There is no movement of any solution ACROSS the membrane Water flows from `BaCl_(2)` solution towards NaCl solution Water flows from NaCl solution towards `BaCl_(2)` solution OSMOTIC pressure of 0.1 M NaCl is lower than the osmotic pressure of `BaCl_(2)` (Assume complete dissociation) Solution :Applying `pi V = iCRT` i=2 for NaCl i=3 for `BaCl_(2)` `pi V_(NaCl)=2xx0.1 RT=0.2 RT` `pi V_(BaCl_(2))=3xx0.05 RT =0.15 RT` As `pi V_(NaCl)gt pi V_(BaCl_(2))`. So, water from `BaCl_(2)` towards `NaCl_(2)` solution.

Discussion

93555.	0.1 M NaCl solution is placed in two different cells having cell constant 0.5 and 0.25 cm^(-1) respectively. Which of the two will have greater value of specific conductance?
Answer» Solution :Specific conductance `kappa=1/rho*1/A` Specific conductance `1/rho` is directly PROPORTIONAL to cell constant `1/A`. `therefore` The cell with higher cell constant has greater value of specific conductance. i.e., The cell with 0.5 `CM^(-1)` cell constant will have greater value of specific conductance.

Discussion

93556.	0.1 M NaCl and 0.1 M CH_(3)COOH are found to have osmotic pressures of P_(1) and P_(2) respectively then what is the correct ststement ?
Answer» `P_(1)gtP_(2)` `P_(1) if P_(2)` `P_(1)`LT`P_(2)` `P_(1)=P_(2)=0` atm. Solution :`NACL` DISSOCIATES completely while `CH_(3)COOH` dissociates to a SMALL extent. Hence, `P_(1) gt P_(2).`

Discussion

93557.	0.1M KI and 0.2 M AgNO_(3) are mixed in 3:1 volume ratio . The deperession in freezing point of the resulting solution will be 0.1 x (Assume Kf of H_(2)O=2Kg//"mol").
Answer» `3.72K` `1.86K` `0.93K` `0.279K` ANSWER :D

Discussion

93558.	0.1 M NaCl and 0.1 M CH_(3)COOH are kept in separate containers. If their osmotic pressures are P_(1) and P_(2) respectively then what is the correct statement ?
Answer» `P_(2)gtP_(2)` `P_(2)=P_(2)` `P_(2)ltP_(2)` `P_(2)=P_(2)=0 ATM, Solution :Since NaCI is a stronger electrolyte than `CH_(3)COOH`, it will DISSOCIATE to freater EXTENT, and release ION (PARTICLE). Therefore, `P_(1)gtP_(2)`.

Discussion

93559.	0.1 M HCl and 0.1 M H_(2)SO_(4), each of volume 2 ml are mixed and the volume is made up to 6 ml by adding 2ml of 0.01 N NaCl solution. The pH of the resulting mixture is
Answer» 1.17 `1.0` 0.3 log 2 - log 3 Solution :MILLIMOLES of `H^(+) = 0.1 xx 2 + 0.1 xx 2 xx 2 = 0.6` TOTAL VOLUME in ML = 6 `pH = -log [H^(+)] = -log ((0.6)/(6)) = -log 0.1 = 1.0`.

Discussion

93560.	0.1 M copper sulphate solution in which copper electrode is dipped at 25^@C. Calculate the electrode potential of copper. [Given E_(Cu^(2+)\|Cu)^@ = 0.34]
Answer» Solution :Given that `[CU^(2+)] = 0.1 M` `E_(Cu^(2+)\|Cu)^@ = 0.34` `E_("cell") = ?` Cell reaction is `Cu^(2+) (aq) + 2r^(-) to Cu(s)` `E_("cell") = E^(@) - (0.0591)/n log ([Cu])/([Cu^(2+)]) = 0.34 - (0.0591)/2 log 1/(0.1)` `- 0.34 - 0.0296 = 0.31 V`.

Discussion

93561.	0.1M copper sulphate solution in which copper electrode is dipped at 25^(@)C. Calculate the electrode potential of copper. [Given: E_(Cu^(2+)"\|"Cu)^(@)=0.34]
Answer» SOLUTION :Give that ` [Cu^(2+)] = 0.1M ` ` E_(Cu^(2+\|Cu^(@) = 034 ` ` E_("cell") = ? ` Cell reaction is ` Cu_(aq)^(2+) + 2e^(-) to Cu_((s))` ` E_("cell")= E^(@) - ( 0.0591)/N log ([Cu])/([Cu^(2+)])` ` = 0.34 (0.0591)/2 log 1/(0.1)` = 0.34 - 0.0294 0.31 V

Discussion

93562.	0.1 M acetic acid solution is titrated against 0.1M NaOH solution. What would be the difference in Ph between 1/4 and 3/4 stages of neutralisation of acid :
Answer» 2 LOG 3/4 2log1/4 log1/3 2log3 Answer :D

Discussion

93563.	0.1 litre so.ution is made by mixing 4 gram NaOH, 20 millimoles of H_(2)SO_(4), 40 millimoles of HCl and 20 millimoles of HNO_(3) what is the pH of solution.
Answer» ANSWER :7

Discussion

93564.	0.1 g atom of radioactive isotope ""_ZX^A (half-life 5 days) is taken. How many number of atoms will decay during eleventh day?
Answer» SOLUTION :`N_0 = 0.1 atom` `t = 10" daysand" t_(1//2) = 5` days `LAMBDA = 2.303/t log""N_0/N` `0.693/5 = 2.303/10 log""0.1/N` `:.`Amount left after 10 days = 0.0250 g atom Similarly if t = 11 days `0.693/5 = 2.303/11 log""0.1/N` `:.` Amount left after 11 days = 0.0218 g atom `:.` Amount decayed in 11th DAY =0.0250- 0.0218 `= 3.2xx 10^(-3)` g atom `=3.2 XX 6.023 xx 10^23 xx 10^(-3)" atoms" = 1.93 xx 10^21` atoms

Discussion

93565.	01. ampere current is passed for 10 second through copper and silver voltameters . The metal that is deposited more is
Answer» Cu Ag both (a) and (B) no EFFECT SOLUTION :`w ALPHA E` : provided Q is constant `, E_(Ag)^(@) GT E_(Cu)^(@)`

Discussion

93566.	0.092 g of a compound with the molecular C_(3)H_(8)O_(3) on reaction with an excess CH_(3)MgI gives 67.00mL of methone at STP. The number of active hydrogen atoms present in a molecule of the compound is
Answer» four three two one Solution :`ROH+CH_(3)MgIrarrCH_(4)+Mg(OR)I` According to the above reaction EVERY one mole of a compound contaning one active `H` atom PER molecule releases one mole of `CH_(4)` GAS at `STP`. Thus, number of moles of `CH_(4)` gas realesed per mole of compound should be an indicator of the number of active HYDROGEN atoms per molecule of compound. `n_(CH_(4))=(Vol. of CH_(4) at STP)/(22,400 mL mol^(-1))` `=0.003 mol` `n_("compound")=("mass")/("molar mass)"=(0.092g)/(92 g mol^(-1))` `=0.001 mol` This implies that `1` mol of compound releases `3` mols of `CH_(4)` gas i.e. there are `3` active hydrogen atoms present in a molecule of the compound.

Discussion

93567.	0.0833 mol of carbohydrate of empirical formula CH_(2)O contain 1 gof hydrogen. The molecular formula of the carbohydrate is
Answer» `C_(5)H_(10)O_(5)` `C_(3)H_(4)O_(3)` `C_(12)H_(22)O_(11)` `C_(6)H_(12)O_(6)` Solution :`:' 0.0833` mole carbohydrate has hydrogen `= 1g` `:.` 1 mole carbohydrate has hydrogen `= 1/(0.0833) =12g` EMPIRICAL Formula `(CH_(2)O)` has hydrogen `= 2g` Hence ` n = (12)/(2)= 6` Hence molecular formula of carbohydrate `= 2g` Hence `n = 12/2 =6` Hence molecular formula of carbohydrate `= (CH_(2)O)_(6)` `= C_(6)H_(12)O_(6)`.

Discussion

93568.	0.0852g of an organic halide (A) when dissolved in 2.0g of camphor, the melting point in the mixture was found to be 167^(@)C. Compound (A) when heated with sodium gives a gas (B). 280mL of gas (B) at STP weight 0.375g. What would be 'A' in the whole process? K_(f) for comphor=40, m.pt. of camphor=179^(@)C.
Answer» `C_2H_5Br` `CH_3I` `(CH_3)_2CHI` `C_3H_7Br` Solution :`DeltaT=179-167=12, w=0.0852g, W=2g, K_f`=40, molecular WEIGHT of (A) `=(1000xxK_fxxw)/(DeltaTxxW)=(1000xx40xx0.0852)/(12xx2)=142` (A) undergoes Wurtz REACTION to form (B) i.e. `(A) overset(Na)to(B)+NaX` (B)is an ALKANE say `C_nH_(2n+2)` `:'` 280 mL of (B) weighs 0.375 g at NTP `:.` 22400 mL of (B) weighs `=(0.375xx22400)/280` = 30 g at NTP `:.` M.wt. of (B) =30,12n+2n+2n+2=30,n=2 Thus (B) si ethane and THEREFORE (A) is `CH_3X`. The m.wt.of `CH_3X=142` At. wt. of X = 127 `:.` X is iodine Therefore alkyl ahlide is `CH_3I`. This reaction is `underset((A))(2CH_3I)underset("either")overset(Na)tounderset((B))(C_2H_6)`

Discussion

93569.	0.06% (w/v) aqueous solution of urea is isotonic with:
Answer» 0.1 M glucose SOLUTION 0.01M glucose solution 0.6% glucose solution 0.06% glucose solution Solution :0.06% (W/V) = 0.06 G in 100 mLi.e. 0.6 g in 1000 ML `M =0.6/60 = 0.01M`

Discussion

93570.	0.06 mole og KNO_(3) solid is added to 100 cm^(3) of water at 298 K. The enthalpy of KNO_(3) aq solution is 35.8 kJ mol^(-1). After the solute is dissolved the temperature of the solutions will be
Answer» 293 K 298 K 301 K 304 K SOLUTION :On APPLYING `mC_(p)Delta T`.

Discussion

93571.	0.05M NaOH solution offered a resistance of 31.69Omegain a conductivity cell at 298K. If the cell constant of the conductivity cell is 0.367cm^(-1)the molar conductivity of the NaOH solution is
Answer» `232 Omega^(-1) cm^2 "mol"^(-1)` `234 Omega^(-1) cm^2 "mol"^(-1)` `231 Omega^(-1) cm^2 "mol"^(-1)` `230 Omega^(-1) cm^2 "mol"^(-1)` ANSWER :A

Discussion

93572.	0.05 moles of AICI_3 is required to coagulate 500 ml of As_2S_3 sol. If its coagulation value is 10^x find x
Answer» SOLUTION :`500 to 0.05 ` MOLES `1000 to ?` `0.1` moles = 100 MILLI moles = `10^2`

Discussion

93573.	0.05 m urea solution will have freezing point (K_(f) = 1.86 K kg "mol"^(-1) )
Answer» 273.093 K 273.186 K 272.907 K 272.814 K Answer :B::C

Discussion

93574.	0.05 M NaOH solution offered a resistance of 31.6 ohm in a conductivity cell. If the cell constant of cell is 0.367cm^(-1) , calculate the molar conductivity of the solution,
Answer» Solution :Molarity =0.05 `R=31.6Omega` cell constant, `(l)/(a)=0.367cm^(-1)` `^^_(m)=(Kxx1000)/("Molarity")` `=((1)/(R)xx(l)/(a)xx1000)/("MOLARTIY")` `=(1)/(31.6)xx(0.367xx1000)/(.05)` `=(0.367xx1000)/(31.6xx0.05)` `=232.2ohm^(-1)cm^(2)MOL^(-1)`

Discussion

93575.	0.05 M NaOH solution offered a resistance of 31.6 ohm in a conductivity cell. If the cell constant of cell is 0.367cm^(-1), calculate the molar conductivity of the solution.
Answer» Solution :MOLARITY =0.05 `R=31.6Omega` Cell consatant, `(L)/(a)=0.367cm^(-1)` MOLAR conductivity, `Delta_(m)=(Kxx1000)/("Molarity")` `=((1)/(R)xx(l)/(a)xx100)/("Molarity")` `=(1)/(31.6)xx(0.367xx1000)/(0.5)` `=(0.367xx1000)/(31.6xx0.05)` `=232.2ohm^(-1)cm^(2)MOL^(-1)`

Discussion

93576.	0.04 N solution of a weak acid has a specific conductance 4.23 xx10^(-4)" mho.cm"^(-1). The degree of dissociation of acid at this dilution is 0.0612. Calculate the equivalent conductance of weak acid at infinite solution.
Answer» Solution :Specific CONDUCTANCE `kappa=4.23xx10^(-4)"mho.cm"^(-1)` `lambda_(C )=(kappa1000)/(C )=(4.23xx10^(-4)xx1000)/(0.04)` `=10.575"mho.cm"^(2).EQ^(-1)`. `alpha=0.0612=(lambda_(C ))/(lambda_(oo))=(10.575)/(lambda_(oo))` `therefore lambda_(oo)="172.79 mho. cm"^(2)."gm. equiv."^(-1)`.

Discussion

93577.	Specific conductance of 1 M KNO_(3) solution is observed to be 5.55 times 10^(-3)mho cm^(-1). What is the equivalent conductance of KNO_(3) when one litre of the solution is used?
Answer» Solution :Specific conductance `kappa=4.23xx10^(-4)"mho.CM"^(-1)` `lambda_(C )=(kappa1000)/(C )=(4.23xx10^(-4)xx1000)/(0.04)` `=10.575"mho.cm"^(2).eq^(-1)`. `alpha=0.0612=(lambda_(C ))/(lambda_(oo))=(10.575)/(lambda_(oo))` `THEREFORE lambda_(oo)="172.79 mho. cm"^(2)."GM. equiv."^(-1)`.

Discussion

93578.	0.04 g of pure NaOH is dissolved in 10 litof distilled water. The pH of the solution is:
Answer» 9 10 11 12 Answer :B

Discussion

93579.	0.037 g of an alcohol was added to methyl magnesium iodide and the gas evolved at STP, occupies the volume 11.2 cm^(3). On dehydration, the alcohol gives an alkene. The alkene upon ozonolysis produces acetone as one of its products. The alcohol gives the carboxylic acid upon oxidation with same number of carbon atom. The alcohol is ...
Answer» n-butyl ALCOHOL Isopropyl alcohol Isobutyl alcohol Sec. Butyl alcohol Solution :The molecular mass of `R- OH= (22400 xx 0.037)/(11.2)= 74 g` `:. C_(n)H_(2n+2)O= 74 IMPLIES 12n +(2n+2)+16= 74` `implies n=4` `to` The possible alcohols are n-butyl alcohol, isobutyl alcohol, sec. Butyl alcohol and tertiary butyl alcohol. As the alcohol gives the carboxylic acid with same number of carbon atoms UPON oxidation, the option (d) is ruled out. Also, option (b) is ruled out as the alcohol is of 4-carbon atoms. The one of the products of ozonolysis of Alkene is acetone. Thus, alkene has = `C(CH_(3))_(2)` GROUP. Thus, alkene must be isobutylene, i.e., `(CH_(3))_(2) = CH_(2)`. So, the given alcohol is isobutyl alcohol.

Discussion

93580.	0.037 g of an alcohol, ROH was added to CH_(3)MgI and the gas evolved measured 11.2 cm^(3) at N.T.P. What is the molar mass of alcohol ? On dehydration, ROH gave an alkene which on ozonolysis gave acetone as one of the products. ROH on oxidation easily gave an acid containing the same number of carbon atoms. give the structure of ROH and of acid with proper reasoning.
Answer» Solution :(i) The molar mass of alcohol can be calculated as follows : `11.2 cm^(3)` of `CH_(4)` are evolved from alcohol `=0.037 g` `22400 cm^(3)` of `CH_(4)` are evolved from alcohol `=((0.037 g))/((11.2 cm^(3)))xx(22400 cm^(3))=74.0 g`. (ii) General formula of alcohol `=C_(n)H_(2n+1)-OH` `:.""12n+2n+1+17=74` or `14n+18=74` `:.""14n=74-18=56` or `n=56//14=4`. The molecular formula of alcohol `= C_(4)H_(9)OH`. (iii) As the alcohol can be easily oxidised to ACID with the same number of carbon atoms, it is PRIMARY alcohol. Since on DEHYDRATION alcohol gives an alkene which upon ozonolysis gives acetone as ONE of the products, the alcohol must have `(CH_(3))_(2)CH-` group. Thus, the alcohol is isobutyl alcohol. The reactions involved a+re as follows : `{:(""CH_(3)""CH_(3)""CH_(3)""H),("\|""\|""\|""\|"),(CH_(3)-CH-CH_(2)OH underset(-H_(2)O)overset(Conc. H_(2)SO_(4)//"heat")(rarr)CH_(3)-C=CH_(2)underset(Zn//H_(2)O)overset(O_(3))(rarr)CH_(3)-C=O+H-C=O),("Isobutyl alcohol""Isobutylene""Acetone""Formaldehyde"):}` `{:(""CH_(3)""CH_(3)""CH_(3)),("\|""\|""\|"),(CH_(3)-CH-CH_(2)OH underset(-H_(2)O)overset((O))(rarr)CH_(3)-CH-CHO overset((O))(rarr)CH_(3)-CH-COOH),("Isobutyl alcohol""Isobutyraldehyde""Isobutyric acid"):}`

Discussion

93581.	0.037 g of an alcohol, ROH, was added to CH_3 MgIand the gas evolved measured 11.2 cm^3 at STP. What is the molecular weight of ROH? On dehydration, ROH gives an alkene which on ozonolysis gives acetone as one of the products. ROH on oxidation easily gives an acid containing the same number of carbon atoms. Give structures of ROH and the acid.
Answer» SOLUTION :`C_4H_9OH, C_3H_2COOH`

Discussion

93582.	0.025 g of starch sol is required to prevent the coagulation of 10 mL of gold sol when 1 mL of 10% NaCl solution is aded. What is the gold number of starch sol ?
Answer» SOLUTION :Amount of starch sol needed to PREVENT the coagulation of gold sol = 0.025 g = 25 mg. By DEFINITION, this REPRESENTS the gold NUMBER. Thus, gold number of starch = 25

Discussion

93583.	0.023g of sodium metal is reacted with 100 cm^(3) of water. The pH of the resulting solution is :
Answer» 10 11 9 12 Solution :`M=(0.023xx1000)/(23xx100)=0.01M` `NA^(+)+OH^(-)hArrNaOH` `[OH^(-)]=0.01M=10^(-2)M," "pOH=-log(10^(-2))=2` `pH+pOH=14` `:." "pH=14-2rArrpH=12`.

Discussion

93584.	0.02 mole of [Co(NH_(3))_(5)Br]Cl_(2) and 0.02 mole of [Co(NH_(3))_(5) Cl] SO_(4) are present in 200 cc of a solution X . The number of moles of the precipitates Y and Z that are formed when the solution X is treated with excess silver nitrate and excess barium chloride are respectively
Answer» 0.02 , 0.02 0.01 , 0.02 0.02 , 0.04 0.04 , 0.02 Solution :`[Counderset(0.02 "mole") underset(1"mole") ((NH_(3))_(5))Br]Cl_(2) + 2 underset(2 "MOLES")(AgNO_(3)) to [Co(NH_(3))_(5)underset(1 "mole")(Br)](NO_(3))_(2) + 2 Agunderset(2 "moles")(Cl)_((PPT)) (Y)` `0.02 xx 2 = 0.04 ` mole `[Co(NH_(3))_(5)underset(1 "mole")(Cl)]SO_(4) + underset(0.02 "moles")(BaCl_(2)) to [Co(NH_(3))_(5)underset(1 "mole")(Cl)] Cl_(2) + BaSO_(4)_("(ppt.)") underset(0.02 "moles")((Z))`

Discussion

93585.	0.02 M monobasic acid dissociates 2% hence, pH of the solution is
Answer» 0.3979 1.3979 1.699 3.3979 Solution :`[H^(+)] = alpha.C = (2)/(100) xx 0.2, [H^(+)] = 4 xx 10^(-4)M` `pH = -log [H^(+)] = 4 -log 4 , pH = 33979`.

Discussion

93586.	0.02 g equivalent of Ag was deposited in an electrolysis experiment. Find the quantity of charge passed. If the same charge is passed through a gold solution, 1.314 g of gold is deposited. Find the oxidation state of gold (Given atomic mass of Au =197 amu)
Answer» SOLUTION :(a) CALCULATION of quantity of charge passed : The reaction at cathode is : `Ag^(+)(aq)+UNDERSET((96500" C"))underset(1Faraday)(E^(-)) to underset(1g equivalent) (Ag(s))` 1g equivalent of Ag was deposited on passing charge `=96500" C"`. 0.02g equivalent of Ag was deposited on passing charge `=((96500C))/((1g" equiv".))xx(0.02g " equiv".)=1930" C"` (b) Calculation of oxidation state of gold 1.314g of gold is deposited on passing charge`=1930" C"` 197g of gold is deposited on passing charge `=((1930 C))/((1.314 g))xx(197 g)=289353" C"` `=(289353)/(96500)=3F`. `:.` The oxidation state of gold is +3 `AU^(3+)(aq)+underset((3F))(3e^(-)) to underset((197g))(Au(s))`

Discussion

93587.	0.02 g of gelatin is required to protect 10 mL of gold sol from 10% NaCl, then find the gold number for gelatin. Report your answer by multiplying by 100.
Answer» Solution :Gold NUMBER is MINIMUM amount of protective colloid which can protect 10 mL standard gold SOL from COAGULATION when 1 mL of `10% NaCl` is added.

Discussion

93588.	0.0125 mol of sucrose is dissolved in 100 gm of water and it undergo partial inversion according to following equation C_12H_22O_11+H_2OtoC_6H_12O_6+C_6H_12O_6 If elevation in boiling point of solution is 0.104^@C calculate 1/10 mol percentage of sugar inverted (K_b,_(H_2O)=0.52).
Answer» SOLUTION :`underset("Mol")(C_12H_22O_11)+underset(0.0125)(H_2O)tounderset(0)(C_6H_12O_6)+underset(0)(C_6H_12O_6)` `0.0125-x " " x ""x` `DeltaT_b=m_1K_b+m_2K_b+m_3K_b` `m_1+m_2+m_3=0.104/0.52=0.2` `(0.0125-x+x+x)/100 xx1000=0.2` x=0.0075 mol % `=0.0075/0.0125xx100=60%` `1/10th` of mol% =`60/10=6`

Discussion

93589.	0.01 molar solutions of glucose, phenol and potassium chloride were prepared in water. The boiling points of
Answer» `GLUCOSE solution = Phenol solution = Potassium CHLORIDE solution ` `Potassium chloride solution gt Glucose solution gt Phenol solution ` `Phenol solution gt Potassium chloride solution gt Glucose solution ` `Potassium chloride solution gt Phenol solution gt Glucose solution ` Solution :`underset("Boiling point DECREASING order"rarr.)(KCl gt C_(6)H_(5)OH gt C_(6)H_(12)O_(6))` Potassium chloride is ionic compound and phenol formed phenoxide ion hence it shows greater boiling point then glucose.

Discussion

93590.	0.01 M solution of KCl and BaCl_(2) are prepared in water. The freezing points of KCl is found to be -2^(@)C. What is the freezing point of BaCl_(2) to be completely ionized
Answer» `-3^(@)C` `+3^(@)C` `-2^(@)C` `-4^(@)C` Solution :i for KCl = 2, i for `BaCl_(2)=3 because Delta T_(F) PROP i` `(Delta T_(f)(KCl))/(Delta T_(f)(BaCl_(2)))=(2)/(3)rArr Delta T_(f)(BaCl_(2))=(3)/(2)xx2=3^(@)C` `therefore` Freezing point of `BaCl_(2)=-3^(@)C`.

Discussion

93591.	0.01 M solution each of urea, common salt and sodium sulphate are taken, the ratio of depression in freezing point of these solutions is
Answer» `1:1:1` `1:2:1` `1:2:1` `1:2:3` Solution :It is the RATIO of numberof PARTICLEIN solution Urea = 1, common salt =2 , sodium sulphat = 3.

Discussion

93592.	0.01 M solution each of urea, common salt and Na_(2)SO_(4) are taken, the ratio of depression of freezing point is :
Answer» `1:1:1` `1:2:1` `1:2:3` `2:2:3` ANSWER :C

Discussion

93593.	0.01 M NaCl solution is placed in two different cells having cell constant 0.5 and 0.25 cm^(-1) respectively. Which of the two will have greater value of specific conductance.
Answer» SOLUTION :The SPECIFIC CONDUCTANCE values are same. Because the reaction (CATION) of CELL constant does not change.

Discussion

93594.	0.01 M aqueous glucose solution is more concentrated than0.01 M aqueous glucose solution.
Answer»

Discussion

93595.	0.01 m aqueous solution of K_(3)[Fe(CN)_(6)] freezes at -0.062^(@)C. What is the apparent percentage of dissociation ? (K_(f)" for water = 1.86 K kg mol"^(-1))
Answer» Solution :`DeltaT_(f)"(observed)"=0.062^(@)C` `DeltaT_(f)"(CALCULATED)"=K_(f)xxm=1.86xx0.01=0.0186^(@)C` `therefore""i=(DeltaT_(f)"(observed)")/(DeltaT_(f)"(calculated)")=(0.062)/(0.0186)=3.33` `{:(,""K_(3)[Fe(CN)_(6)]hArr3K^(+)+[Fe(CN)_(6)]^(3-)),("Initial","1 MOL "),("After disso.",""1-ALPHA""3alpha ""alpha "Total "= 1+3alpha):}` `therefore""i=(1+3alpha)/(1) or alpha=(i-1)/(3)=(3.33-1)/(3)=0.777=77.7%`.

Discussion

93596.	0.004 M Na_(2)SO_(4) is isotonic with 0.01 M glucose. Degree of dissociation of Na_(2)SO_(4) is
Answer» 0.75 0.5 0.25 0.85 Solution :`(1+2alpha)0.004=0.01`

Discussion

93597.	0.004M Na_(2)SO_(4) is isotonic with 0.01M glucose. Degree of dissociation of Na_(2)SO_(4) is :
Answer» `25%` `50%` `75%` `85%` Answer :C

Discussion

93598.	0.001 mole of strong electrolyte Zn(OH)_2 is present in 200 mL of an aqueous solution. The pH of this solution is
Answer» 2 4 12 10 Solution :`Zn(OH)_2hArrZn^(2+) + 2OH^(-)` `[OH^-] = 0.001xx2xx 1000/200` `[OH^-]=1 XX 10^(-2)mol L^(-1)` `POH=- log[OH^(-) ]= log(1 xx 10^(-2) )= + 2 LOG10` `pOH=2, pH+POH= 14` `pH +2=14 impliespH = 12`

Discussion

93599.	0.001 mole each of Fe^(2+) and Cd^(2+) is present in one litre of 0.02M HCl, saturated with H_2S . Find whether each of these ions shall precipitate as sulphides. Calculate [Cd^(2+)] in the solution at equilibrium. K_a(H_2S) = 1 xx 10^(-21), K_(sp) (Cds) = 8 xx 10^(-27) , K_(sp) (FeS) = 3.7 xx 10^(-19)
Answer» Solution :CDS precipitates , `[Cd^(2+) ] = 3.86 X 10^(-8) M`

Discussion

93600.	0.00012% MgSO_(4) and 0.000111% CaCl_(2) is present in water. What is the measured hardness of water and millimoles of washing sodarequired to purigy water 1000 L water ?
Answer» Solution :Basis of calculation = 100 g hard water `MgSO_(4) = 0.00012g = (0.00012)/(120) ` mole `CaCl_(2)=0.00012 g = (0.000111)/(111) ` mole `:. ` EQUIVALENT moles of `CaCO_(3)=((0.00012)/(120)+(0.000111)/(111))` mole `:.` mass of `CaCO_(3)=((0.00012)/(120)+(0.000111)/(111))xx100=2xx10^(-4)g` Hardness (in terms of ppm of `CaCO_(3)`)=`(2xx10^(-4))/(100)xx10^(6)=2`ppm `CaCl_(2)+Na_(2)CO_(3)rarrCaCO_(3)+2NaCl` `NaSO_(4)+Na_(2)CO_(3)rarrMgCO_(3)+Na_(2)SO_(4)` `:.` Required `Na_(2)CO_(3)` for 100g of water `=((0.00012)/(120)+(0.000111)/(111))` mole `=2xx10^(-6)` mole `:.` Required `Na_(2)CO_(3)` for 1000 litre water `=(2xx10^(-6))/(100)xx(2)/(100)` mole (`:.` d=1g/mL) `=(20)/(1000)` mole = 20 m mole

Discussion

Explore topic-wise InterviewSolutions in .

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