94420 + Interview Questions in Chemistry in Class 12 Page 1858 InterviewSolution

92851.	15 gm Ba(MnO_4)_2 sample containing inert impurity is completely reacting with 100 ml of 11.2 V' H_2O_2, then what will be the % purity of Ba(MnO_4)_2 in the sample ? (Atomic mass Ba=137, Mn=55)
Answer» `5%` `10%` `50%` none Solution :Normality of `H_2O_2=11.2/5.6N` milli equivalents `Ba(MnO_4)_2` reacted=milli equivalents of `H_2O_2` reacted =2x100=200 meq=0.2 GRAM eq Moles of `Ba(MnO_4)_2implies0.2/10=0.02 " " :. Wt. of Ba(MnO_4)_2=0.02xx375` % purity of `Ba(MnO_4)_2=(375xx0.02)/15xx100=50%`

Discussion

92852.	15 g of KMnO_(4) in acidic medium equal to
Answer» 0.095 moles 0.477g equivalents 9.54 L of 0.05 N `KMnO_(4)` 10 ML of 0.05 M `KMnO_(4)` Answer :ABC

Discussion

92853.	1.5 g of brass containing Cu and Zn reacts with 3 M HNO_(3) solution, the following reactions take place. Cu + HNO_(3) to Cu^(2+) + NO_(2)(g) + H_(2)O Zn + H^(+) + NO_(3)^(-) to NH_(4)^(+) + Zn^(2+) + H_(2)O The liberated NO_(2) (g) was found to be 1.04 L at 25^(@) C and one atm . (i) Calculate the percentage composition of brass. (ii) How many ml of3 M HNO_(3) will be required for completely reacting 1 g of brass ?
Answer» SOLUTION :N/A

Discussion

92854.	15 g of an unknown molecular substance was dissolved in 450 g of water. The resulting solution freezes at - 0.34^(@)C. what is molar mass of the substance ? (K_(f)" for water" = 1.86k kg " mol"^(-1))
Answer» SOLUTION :`W_(B) = 15 g, W_(A) = 450` g `Delta T_(F) = 0.34^(@) C , K_(f) = 1.86 , M_(B)` = ? `Delta T_(f) = K_(f) xx (W_(B) xx 1000)/(M_(B) xx W_(A))` `0.34 = 1.86(15 xx 1000)/(M_(B) xx 450)` 0.34 = `(61.99)/(M_(B))` `M_(B) = (61.99)/(0.34)` ` = 182.35 g MOL^(-1)`

Discussion

92855.	1.5 g ofBa(NO_(3))_(2) dissolved in 100gofwatershows a depressionin freezing pointequal to 0.28^(@)C . Whatis the percentagedissociationof thesalt ? (k_(f)for water= 1.86 K//m and molar mass of Ba(NO_(3))_(2) = 261).
Answer» SOLUTION :`M_(B) ` (OBSERVED) `= (1.86 XX 1000xx 1.5)/(0.280 xx 100)= 99.64` ` i=(261)/(99.64)= 2.62` `Ba(NO_(3))_(2) hArr Ba^(2+) + 2NO_(3)^(-)` `{:(,"Initial",1,0,0),(,"After dissociaiton ",1-alpha,alpha,2alpha):}` ` i=(1-alpha + alpha+2alpha)/(1) = 2.62` `alpha = 81%`

Discussion

92856.	15 g of an unknown molecular substance was dissolved in 450 g of water. The resulting solution freezes at -0.34^(@)C. What is the molar mass of the substance ? (K_(f) for water ="1.86 K kg mol"^(-1)).
Answer» SOLUTION :`"182.35 G MOL"^(-1)`

Discussion

92857.	15 g of an unknow molecular substance is dissolved in 450 g of water. The reulting solution freezen at -0.34^(@)C. What is the molar mass of the substance (K_(f) for water=1.86 K kg mol^(-1))
Answer» Solution :`W_(B)=15.0g, W_(A)=540g, 0.45kg DeltaT_(f)=0-(-0.34)^(@)C` `=0.34^(@)C=0.34 K, K_(f)=1.86" K kg MOL"^(-1)` `M_(B)=(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))((1.86" K kg mol"^(-1))xx(15.0g))/((0.34K)xx(0.45 kg))=182.35" g mol"^(-1)`

Discussion

92858.	1.5 g of an impure sample of sodium sulphate dissolved in water was treated with excess of barium chloride solution when 1.74 g of BaSO_(4) were obtained as dry precipitate. Calculate the percentage purity of the sample.
Answer» Solution :Given : 1.5 g of impure `Na_(2)SO_(4) overset("TREATED with")UNDERSET(BaCl_(2)"gave")rarr 1.7 g` of `BaSO_(4)` The chemical equation representing the REACTION is : `underset(=142 g)underset(2xx23+32+4xx16)(Na_(2)SO_(4))+BaCl_(2)rarr underset(=233 g)underset(137+32+4xx16)(BaSO_(4))` Step 1. To calculate the mass of `Na_(2)SO_(4)` which PRODUCES 1.74 g of `BaSO_(4)`. From the chemical equation, 233 g of `BaSO_(4)` are produced from `Na_(2)SO_(4)=142 g` `therefore` 1.74 g of `BaSO_(4)` are produced from `Na_(2)SO_(4)=(142)/(233)xx1.74=1.06 g` This is the mass of pure `Na_(2)SO_(4)` present in 1.5 g of impure sample. Step 2. To calculate the percentage purity of impure sample. 1.5 g of impure sample contains pure `Na_(2)SO_(4)=1.06 g` `therefore 100 g` of the impure sample will CONTAIN pure `Na_(2)SO_(4)=(1.06)/(1.5)xx100=70.67 g` Thus, percentage purity of impure sample = 70.67.

Discussion

92859.	1.5 g of an impure sample of sodium sulphate dissolved in water was treated with excess of barium chloride solution when 1.74 g of BaSO_(4) were obtained as dry precipitate . Calculate the percentage purity of sample.
Answer» SOLUTION :N/A

Discussion

92860.	1.5 g of a non-volatile, non-electrolyte is dissolved in 50 g benzene (K_b = 2.5 kg mol^(-1)). The elevation of the boiling point of the solution is 0.75 K. The molecular weight of the solute in g mol^(-1) is :
Answer» 200 50 75 100 Answer :D

Discussion

92861.	(15)/(16)th of a radioactive sample decays in 40 days. Half-life of the sample is
Answer» 100 days 10 days 1 day `log_(E) 2` days Solution :Quantity of radioactive ELEMENT decayed = `(15)/(16)` Quantity LEFT `= 1 - (15)/(16) = (1)/(16)` `(1)/(16) = 1 XX ((1)/(2))^(n) or ((1)/(2))^(4) = ((1)/(2))^(n)` one half-life `= (40)/(4) = 10` days.

Discussion

92862.	""^(14)C of A appears in (B) at…………………..
Answer» ANSWER :6

Discussion

92863.	1.44 g of titanium (Ti) reacte with excess of O_(2) and produced x gm of a nonstoichiometric compound Ti_(1.44)O_(1). The value of x is:
Answer» 1.44 1.77 2 none of these. Solution :In the reaction: `Ti+O_(2)rarrTi_(1.44)O_(1)` NUMBER of MOLES of titanium =Number of moles of `Ti_(1.44)O_(1)` `(1.44)/(48)=(x)/(48xx1.44+16)` x=1.77g

Discussion

92864.	142 g of chlorine represents
Answer» 4 mol of chlorine ATOMS 2 G mol of chlorine 2 mol of Cl atoms both A and B Solution :142G of `Cl_2` = 2 mol of `Cl_2` or= 4 mol of Cl atoms

Discussion

92865.	14.2gm "Na"_(2)"SO"_(4) is dissolved in 400 mLwater. Find out (i) formality (ii) Normalityof solution.
Answer» ANSWER :0.25,0.75

Discussion

92866.	1.42g of a mixture of caCO_(3) and MgCO_(3)was dissolved in 200 mL of0.2 N HCl solutionwhich was then diluted to 250 mL . 10 mL of this solution was neutralised by 12 mL of (N/30) Na_(2)CO_(3) .Find out the percentage of eachin the mixture. ( Ca = 40 , Mg = 24 , C = 12 , O = 16)
Answer» Solution :Let the wt. of `CaCO_(3)` be x G . ` :. ` wt . Of `MgCO_(3) = (1.42- x ) g ` ` :. ` eq of `CaCO_(3) = x/50` and eq. of `MgCO_(3) = (1.42 - x)/42 ""…(Eqn.4)` `( " eq. wt of " CaCO_(3) = 100/2 = 50 " eq. wt of " MgCO_(3) = 84/2 = 42)` Total m.e of `CaCO_(3) and MgCO_(3) = x/50 xx 1000 + (1.42 - x)/42 xx 1000 ...(Eqn.3)` m.e of HCl = `0.2 xx 200 = 40 ` From the givenquestion it is clearthat m.eof HCl is greater than those of `caCO_(3) and MgCO_(3)` `:. ` m.e of excess HCl = m.e of HCl - m.e of `CaCO_(3) and MgCO_(3)` ` = 40 - { 1000 (x/50 + (1.42 -x)/42 )}` ` :. ` the m.e of the resulting solution does not change on dilution . ` :. ` normality of excess HClin the diluted resulting solution ` = (m.e)/250 = (4 - {1000(x/50+(1.42-x)/42)})/250` ` :. ` m.e of 10 ML of the resulting solution ` (4 - {1000(x/50+(1.42-x)/42)})/250xx10` m.e of `na_(2)CO_(3)` solution ` = 1/30 xx12 ` ...(Eqn.1) ` :. (40 - {1000(x/50+(1.42-x)/42)})/250 xx 10 = 1/ 30 xx 12 "" ( Eqn. 2)` ` :. x = 1 ` ` :. % of CaCO_(3) = 1/(1.42) xx 100 = 70.4 % ` % of `MgCO_(3) = 100 - 70.4 = 29 .6 % `

Discussion

92867.	1.4 g of acetone dissolved in 100 g of benzene gave a solution which freezes at 277.12 K. Pure benzene freezes at 278.4 K.2.8 of solid (A) dissolved in 100 g of benzene gave a solution which froze at 277.76 K. Calculate the molecular mass of (A).
Answer» ANSWER :232

Discussion

92868.	1.4 g of an organic compound was digestedaccording to Kjeldahl's method and the ammonia evolved was absorbed in 60 mL of M/10 H_(2)SO_(4) solution. The excess sulphuric acid required 20 mL of M/10 NaOH solution for neutralization. Thepercentage of nitrogen in the compound is :
Answer» 24 5 10 3 Answer :C

Discussion

92869.	1.4 g of a sample of chalk (CaCO_(3)) containingclay as impurity were treated with excess of dilute hydrochloric acid. Volume of CO_(2) evolved when measured at 15^(@)C and 768 mm pressure was 282cm^(3). Calculate the percentage purity of the sample.
Answer» SOLUTION :`underset(100g)(CaCO_(3))+2HCl rarrCaCl_(2)+H_(2)O+underset(22400cm^(3)"at S.T.P.")(CO_(2))` First CONVERT the GIVEN volume to volume at S.T.P. Calculate the MASS of `CaCO_(3)` from which this volume of `CO_(2)` at S.T.P. is obtained. Then calculate `%` age purity.

Discussion

92870.	1,4-dimethylbenzene on heating with anhydrous AlCl_3 and HCl produces-
Answer» 1,2-dimethylbenzene 1,3-dimethylbenzene 1,2,3-trimethylbenzene ethylbenzene Solution :When, 1,4-dimethylbenzene is HEATED with ANHYDROUS `AlCl_(3)` in the PRESENCE of HCl , it undergoes ISOMERISATION to form thermodynamically more stable 1,3-dimethylbenzene.

Discussion

92871.	1,4-Dimethylbenzene on heating with anhydrous AlCl_(3) and HCl produces
Answer» 1,2-dimethylbenzene 1,3-dimethylbenzene 1,2,3-trimethylbenzene ethylbenzene Answer :B

Discussion

92872.	1,4-Dichlorobutane to hexane-1,6- diamine
Answer» Solution :`Cl-(CH_(2))_(4)-Cl overset(KCN)rarr CN-(CH_(2))_(4)-CN overset("Reduction")rarr underset("(Hexamethylene DIAMINE)")(NH_(2)-(CH_(2))_(6)-NH_(2))`

Discussion

92873.	._(13)Al^(28) when radiated by suitable projectile gives ._(15)P^(31) and neutron. The projectile used is
Answer» Proton Neutron Alpha particle Deuteron Solution :`._(13)Al^(28) + ._(2)He^(4) rarr ._(15)P^(31) + ._(0)N^(1)`

Discussion

92874.	139.18 g of glucose is added to 178.2 g of water the vapour pressure of water for this aqueous solution at 100^@ C is
Answer» 700.7 torr 759 torr 7.6 torr 76 torr Answer :A

Discussion

92875.	138 gm of N_2O_4(g) is placed in 8.2L container at 300 K.The equilibrium vapour density of mixture was found to be 30.67.Then (R=0.082 L atm "mol"^(-1) K^(-1))
Answer» `ALPHA`=DEGREE of DISSOCIATION of `N_2O_4=0.25` `K_p` of `N_2O_4 hArr 2NO_2` (g) will be 9 atm Total pressure at equilibrium =6.75 atm The density of equilibrium mixture will be 16.83 gm/litre. Solution :`{:(,N_2O_4""hArr,2NO_2),(t=0,a,0),(t,a(1-alpha),2aalpha):}` vapour density`=46/(1+alpha)=30.67` so `1+alpha=1.5=0.5=50%` Total pressure`=(1.5xx1.5xx0.082xx300)/8.2=6.75` atmSo, `K_p=(4alpha^2)/(1-alpha^2)P=9` atm and for density of mixture `=138/8.2` gm/L=16.83 gm/L

Discussion

92876.	138 g of N_(2)O_(4(g)) is placed in 8.2 L container at 300 K. The equilibrium vapour density of mixture was found to be 30.67. The (R = 0.082 L atm mol^(-1)K^(-1))
Answer» the total pressure at equilibrium = 4.5 atm the degree of DISSOCIATION of `N_(2)O_(5)=0.25` the total number of moles at equilibrium is 1.5 `K_(p)` of `N_(2)O_(4)iff2NO_(2(g))` will be 6 atm Solution :`N_(2)O_(4)iff2NO_(2)` `{:("INITIAL moles",1,0),("Eq. moles",(1-alpha),2ALPHA):}` Total no. of moles at equilibrium = `1-alpha+2alpha=1+alpha` `alpha=(1)/(n-1)((D-d)/(d))` where, D = Theoretical vapour density `=("MOL. mass")/(2)=(92)/(2)=46` d = Observed vapour density n = No. of moles of products formed from the dissociation of 1 MOLE of `N_(2)O_(4)` `therefore alpha=(1)/(2-1)((46-30.67)/(30.67))=0.4998~=0.5` Thus, total no. of moles at equilibrium = `1+0.5=1.5` Total pressure `=(1.5xx0.082xx300)/(8.2)=4.5` atm So, `K_(p)=(4alpha^(2))/(1-alpha^(2))xx4.5=6` atm

Discussion

92877.	13.6eV is needed for ionisation of a hydrogen atom. An electron in a hydrogen atom in its ground state absorbs 1.50 times as much energy as the minimum energy required for it to escape ffrom the atom. What is the wavelength of the emitted electron? (m_(e)= 9.109 xx 10^(-31)kg, e= 1.602 xx 10^(-19) coulomb, h= 6.63 xx 10^(-34) J.s)
Answer» SOLUTION :1.5 times of 13.6eV, i.e., 20.4eV, is absorbed by the hydrogen atom out of which 6.8eV (20.4-13.6) is CONVERTED to kinetic energy. KE = 6.8eV= 6.8 (`1.602 xx 10^(-19)` coulomb)(1 volt)= `1.09 xx 10^(-18)J`. Now, KE `=(1)/(2) MV^(2)` or `v= sqrt((2KE)/(m))= sqrt((2(1.09 xx 10^(-18)J))/((9.109 xx 10^(-31)kg)))= 1.55 xx 10^(6) m//s` `therefore lamda= (h)/(mv)= ((6.63 xx 10^(-34) J.s))/((9.109 xx 10^(-31)kg) (1.55 xx 10^(6) m//s))` `=4.70 xx 10^(-10)` metres.

Discussion

92878.	13.5 g of Al getdepositedwhen electricity is passed through solution of AlCl_3 . The number of faradays used are
Answer» `2.00` `1.00` `1.50` `0.50` ANSWER :C

Discussion

92879.	13.5 g of Al get deposited when electricity is pass through the solution of AICI_3. The number of Faradays used are :
Answer» `0.50` `1.00` `1.50` `2.00` ANSWER :C

Discussion

92880.	13.4 gm of a sample of unstable hydrated salt : Na_(2)SO_(4)xxH_(2)O was strongly heated . Weight loss on heating is found to be equal to 6.3 gm .Calculate the value of x.
Answer» 6 5 7 8 Answer :C

Discussion

92881.	13.5 g of Al get deposited when electricity is passed through the solution of AICI3. The number of Faradays used are :
Answer» `0.50` `1.00` `1.50` `2.00` ANSWER :C

Discussion

92882.	._(13)^(27)Al is a stble isotope. ._(13)^(29)Al is expected disintegrate by:
Answer» `ALPHA "EMISSION"` `BETA` "emission" POSITRON emission proton emission Answer :B

Discussion

92883.	""_(13)^(27)Al is a stable isotope. ""_(13)^(29)Al is expected to disintegrate by
Answer» `ALPHA`-EMISSION `BETA`-emission positron emission proton emission Answer :B

Discussion

92884.	._(13)^(27)Al is a stable isotope. ._(13)^(29)Al is expected to disintegrate by
Answer» `alpha-` emission `BETA-emission Positron emission Proton emission Solution :The `(n)/(p)` RATIO of `._(13)Al^(29)` places it above the belt of STABILITY and thus it emits `beta-` particles.

Discussion

92885.	1.30 Lit N_2 gas at 2atm and 300K in a container is exposed to 4g of solid surface. After complete adsorption the pressure of N_2 is reduced by 30% calculate the value of x/m
Answer» 0.22 0.56 0.32 0.43 Solution :PRESSURE is reduced by 30% `=(2 XX 30)/(100) 0.6 ATM` `PV = W/(MRT) , 0.6 xx 1.3 = W/28 xx 0.0821 xx 300 , wt = 0.8732 , x/m = (0.8732)/(4) = 0.22`

Discussion

92886.	1,3- Pentadiene and 1,-4 - pentadiene are compared with respect to their intrinsic stability and reaction with HI . The correct statement is
Answer» 1,3 pentadiene is more STABLE and more REACTIVE than 1,4- pentadiene 1,3 pentadiene is LESS stable and less reactive than 1,4- pentadiene 1,3 pentadiene is more stable and less reactive than 1,4- pentadiene 1,3 pentadiene is less stable and less reactive than 1,4- pentadiene ANSWER :A

Discussion

92887.	13 g of a hybrocarbon contains 1.0g of hydrogen.Its formula is :
Answer» `C_2H_2` `C_2H_3` `C_3H_4` `C_4H_7` ANSWER :A

Discussion

92888.	1,3-Dichloropropane reacts with Zn and Nal and gives
Answer» Propane propene cyclopropane n-propyl iodide. Solution :

Discussion

92889.	1,3-dibromopropane reacts with metallic zinc to form :
Answer» Propene Cydopropane Propane None Answer :B

Discussion

92890.	1,3-Butadiene when treated with Br_(2) gives
Answer» 1,4-dibromo-2-butene<BR>1,3-bibromo-2-butene 3,4-dibromo-1-butene 2,3-dibromo-2-butene Solution :The intermediate `2^(TH)` CARBOCATION shows resonance `CH_(2)=CH-CH=CH_(2) overset(Br^(+))to [CH_(2)Br-overset(+)CH-CH=CH_(2)]` `harr CH_(2)Br-CH=CH-overset(+)(CH_(2))]` `overset(Br^(+))to underset(1,4-"dibromo-2-butene")(CH_(2)Br-CH=CH-CH_(2)Br)`

Discussion

92891.	1,3-butadiene has:
Answer» Only sp-hybridised C-atoms Only `sp^2` -hybridised C-atoms sp, `sp^2 and sp^3`-hybridised C-atoms sp and `sp^2` -hybridised C-atoms Answer :B

Discussion

92892.	1,3-butadiene reacts with ethylene to form
Answer» Benzene Cyclohexane Cyclohexene 2, 3 DIMETHYL butane Answer :C

Discussion

92893.	1,3-butadiene has :
Answer» only sp-hybridized C-ATOMS only `sp^(3)` hybridized C-atoms only `sp^(2)` hybridized carbon atoms both `sp^(2)` and `sp^(3)` hybridized C-atoms. ANSWER :C

Discussion

92894.	12g of urea is dissolved in 1 litre of water a 68.4g of sucrose is dissolved in 1 litre of water The lowering of vapour pressure of first case is
Answer» equal to second greater than second less than second DOUBLE that of second SOLUTION :12g UREA = 12/60 mol = 0.2 mol, 86.4g sucrose =68.4/342=0.2 mol. As mole fraction of SOLUTE is same. Hence, lowering of V.P. is same .

Discussion

92895.	1280Ã·2.0=? The correct answer to this problem in proper number of significant digits is
Answer» `64` `6.40xx10^(2)` `640.0` `640` Solution :`UNDERSET("Three//four S.F.")(1280)Ã·underset("Two S.F.")(2.0)=underset("Three//two S.F.")(640)` The answer should two significant DIGITS.

Discussion

92896.	12.8 gm mixture of CO and CO_(2) exerts a pressure of 6atm at 300K in 1.642 litre container. If all the oxygen of this mixture is used to form H_(2)O (in gas) formed will be? (Given: R=0.0821 amt-L/mole-K)
Answer» Solution :`n=(PV)/(RT)=(6xx0.821xx2)/(0.0821xx300)=0.4` mole `n_(CO)+n_(CO_(2))=0.4` ` x xx 28+(0.4-x)xx44=12.8` (total mass of MIXTURE) Solve x=0.3 0.3 mole CO and 0.1 mole `CO_(2)` mole of O-atom `=0.3+0.2=0.5` mole of `H_(2)O` formed =0.5 mass of `H_(2)O` formed =9gm.

Discussion

92897.	127 ml of a certain gas diffuse in the same time as 100 ml of chlorine under the same conditions. Calculate the molecular mass of the gas.
Answer» Solution :If t is the time taken for diffusion of each gas, then by Graham.s law of diffusion `(r_(x))/(rCl_(2))=(127//t)/(100//t)=SQRT((M_(Cl_(2)))/(M_(x)))=sqrt((71)/(M_(x)))` Or `(71)/(M_(x))=((127)/(100))^(2)` or `M_(x) =44 u`

Discussion

92898.	1.250 g of metal carbonate (MCO_(3)) was treated with 500 mL of 0.1 M HCl solution. The unreacted HCl required 50.0 mL of 0.500 M NaOH solutionfor neutralization, Identify the metal M
Answer» Mg Ca Sr Ba Answer :B

Discussion

92899.	(12.5)^2upto correct number of significant figures is:
Answer» 156 156.2 156.25 `1.562 xx 10^2` Solution :`(12.5)^2 = 156.25 " or" = 156 ` (upto 3 significant DIGITS)

Discussion

92900.	1.25 of sample of C Cl_2F_2 was cooled at a constant pressure of 1 atm. From 320 K to 293 K. During cooling, the volume decreased from 274 to 248 mL.Delta H and DeltaU" for " CCl_2 F_2 " is " (C_p = 80. 7 J mol^(-1) K^(-1)) :
Answer» `-22.5 J,- 19.88J` `19.88 J, - 25.51 J` `108.26 J, -4659 J` `- 26.7 J, 89.5 J`. ANSWER :A

Discussion

Explore topic-wise InterviewSolutions in .

15 gm Ba(MnO_4)_2 sample containing inert impurity is completely reacting with 100 ml of 11.2 V' H_2O_2, then what will be the % purity of Ba(MnO_4)_2 in the sample ? (Atomic mass Ba=137, Mn=55)

15 g of KMnO_(4) in acidic medium equal to

15 g of an unknown molecular substance was dissolved in 450 g of water. The resulting solution freezes at - 0.34^(@)C. what is molar mass of the substance ? (K_(f)" for water" = 1.86k kg " mol"^(-1))

1.5 g ofBa(NO_(3))_(2) dissolved in 100gofwatershows a depressionin freezing pointequal to 0.28^(@)C . Whatis the percentagedissociationof thesalt ? (k_(f)for water= 1.86 K//m and molar mass of Ba(NO_(3))_(2) = 261).

15 g of an unknown molecular substance was dissolved in 450 g of water. The resulting solution freezes at -0.34^(@)C. What is the molar mass of the substance ? (K_(f) for water ="1.86 K kg mol"^(-1)).

15 g of an unknow molecular substance is dissolved in 450 g of water. The reulting solution freezen at -0.34^(@)C. What is the molar mass of the substance (K_(f) for water=1.86 K kg mol^(-1))

1.5 g of an impure sample of sodium sulphate dissolved in water was treated with excess of barium chloride solution when 1.74 g of BaSO_(4) were obtained as dry precipitate. Calculate the percentage purity of the sample.

1.5 g of an impure sample of sodium sulphate dissolved in water was treated with excess of barium chloride solution when 1.74 g of BaSO_(4) were obtained as dry precipitate . Calculate the percentage purity of sample.

1.5 g of a non-volatile, non-electrolyte is dissolved in 50 g benzene (K_b = 2.5 kg mol^(-1)). The elevation of the boiling point of the solution is 0.75 K. The molecular weight of the solute in g mol^(-1) is :

(15)/(16)th of a radioactive sample decays in 40 days. Half-life of the sample is

""^(14)C of A appears in (B) at…………………..

1.44 g of titanium (Ti) reacte with excess of O_(2) and produced x gm of a nonstoichiometric compound Ti_(1.44)O_(1). The value of x is:

142 g of chlorine represents

14.2gm "Na"_(2)"SO"_(4) is dissolved in 400 mLwater. Find out (i) formality (ii) Normalityof solution.

1.42g of a mixture of caCO_(3) and MgCO_(3)was dissolved in 200 mL of0.2 N HCl solutionwhich was then diluted to 250 mL . 10 mL of this solution was neutralised by 12 mL of (N/30) Na_(2)CO_(3) .Find out the percentage of eachin the mixture. ( Ca = 40 , Mg = 24 , C = 12 , O = 16)

1.4 g of acetone dissolved in 100 g of benzene gave a solution which freezes at 277.12 K. Pure benzene freezes at 278.4 K.2.8 of solid (A) dissolved in 100 g of benzene gave a solution which froze at 277.76 K. Calculate the molecular mass of (A).

1.4 g of an organic compound was digestedaccording to Kjeldahl's method and the ammonia evolved was absorbed in 60 mL of M/10 H_(2)SO_(4) solution. The excess sulphuric acid required 20 mL of M/10 NaOH solution for neutralization. Thepercentage of nitrogen in the compound is :

1.4 g of a sample of chalk (CaCO_(3)) containingclay as impurity were treated with excess of dilute hydrochloric acid. Volume of CO_(2) evolved when measured at 15^(@)C and 768 mm pressure was 282cm^(3). Calculate the percentage purity of the sample.

1,4-dimethylbenzene on heating with anhydrous AlCl_3 and HCl produces-

1,4-Dimethylbenzene on heating with anhydrous AlCl_(3) and HCl produces

1,4-Dichlorobutane to hexane-1,6- diamine

._(13)Al^(28) when radiated by suitable projectile gives ._(15)P^(31) and neutron. The projectile used is

139.18 g of glucose is added to 178.2 g of water the vapour pressure of water for this aqueous solution at 100^@ C is

138 gm of N_2O_4(g) is placed in 8.2L container at 300 K.The equilibrium vapour density of mixture was found to be 30.67.Then (R=0.082 L atm "mol"^(-1) K^(-1))

138 g of N_(2)O_(4(g)) is placed in 8.2 L container at 300 K. The equilibrium vapour density of mixture was found to be 30.67. The (R = 0.082 L atm mol^(-1)K^(-1))

13.5 g of Al getdepositedwhen electricity is passed through solution of AlCl_3 . The number of faradays used are

13.5 g of Al get deposited when electricity is pass through the solution of AICI_3. The number of Faradays used are :

13.4 gm of a sample of unstable hydrated salt : Na_(2)SO_(4)xxH_(2)O was strongly heated . Weight loss on heating is found to be equal to 6.3 gm .Calculate the value of x.

13.5 g of Al get deposited when electricity is passed through the solution of AICI3. The number of Faradays used are :

._(13)^(27)Al is a stble isotope. ._(13)^(29)Al is expected disintegrate by:

""_(13)^(27)Al is a stable isotope. ""_(13)^(29)Al is expected to disintegrate by

._(13)^(27)Al is a stable isotope. ._(13)^(29)Al is expected to disintegrate by

1.30 Lit N_2 gas at 2atm and 300K in a container is exposed to 4g of solid surface. After complete adsorption the pressure of N_2 is reduced by 30% calculate the value of x/m

1,3- Pentadiene and 1,-4 - pentadiene are compared with respect to their intrinsic stability and reaction with HI . The correct statement is

13 g of a hybrocarbon contains 1.0g of hydrogen.Its formula is :

1,3-Dichloropropane reacts with Zn and Nal and gives

1,3-dibromopropane reacts with metallic zinc to form :

1,3-Butadiene when treated with Br_(2) gives

1,3-butadiene has:

1,3-butadiene reacts with ethylene to form

1,3-butadiene has :

12g of urea is dissolved in 1 litre of water a 68.4g of sucrose is dissolved in 1 litre of water The lowering of vapour pressure of first case is

1280Ã·2.0=? The correct answer to this problem in proper number of significant digits is

12.8 gm mixture of CO and CO_(2) exerts a pressure of 6atm at 300K in 1.642 litre container. If all the oxygen of this mixture is used to form H_(2)O (in gas) formed will be? (Given: R=0.0821 amt-L/mole-K)

127 ml of a certain gas diffuse in the same time as 100 ml of chlorine under the same conditions. Calculate the molecular mass of the gas.

1.250 g of metal carbonate (MCO_(3)) was treated with 500 mL of 0.1 M HCl solution. The unreacted HCl required 50.0 mL of 0.500 M NaOH solutionfor neutralization, Identify the metal M

(12.5)^2upto correct number of significant figures is:

1.25 of sample of C Cl_2F_2 was cooled at a constant pressure of 1 atm. From 320 K to 293 K. During cooling, the volume decreased from 274 to 248 mL.Delta H and DeltaU" for " CCl_2 F_2 " is " (C_p = 80. 7 J mol^(-1) K^(-1)) :