94420 + Interview Questions in Chemistry in Class 12 Page 1843 InterviewSolution

92101.	5 milli-moles of a solid A was dissolved in 5 moles of H_(2)O. On adding to the solvent, A starts polymerising into another insolublesolid following zero order kinetics. On adding 6 milli-moles of another solid solute C (after 20 minute) the polymerisation completely stops. The insoluble solid polymer is removed and the resulting solution was cooled to a temperature less then .-0.186^(@)C (freezing point of solution ) to cause solidification of some liquid water . Calculate the value of Xif rate constant for polymerisation reaction is represented as 10^(-X) moles//minute. [K_(f) (H_(2)O)= 1.86 K - Kg "mole"^(-1)]
Answer» ANSWER :4

Discussion

92102.	5 milli-moles of a solid A was dissolved in 5 moles of H_(2)O. On adding to the solvent , A sstarts spolymerising into another insoluble solid following zero order kinetics. On adding 6 milli-moles of another solid solute C (after 20 minute) the polymerisation compeletly stops. The insoluble solid polymer is removed and the resulting solution was cooled tova temperature less then -0.186^(@) C(melting point of solution) to cause solidification of some liquid water. Calculate the value of 'X' if rate constant for polymerisation reaction is represented as 10^(-X)mole//minute. [K_(f)(H_(2)O)=1.86 K-kg "mole"^(-1)]
Answer» ANSWER :4

Discussion

92103.	5 litres of a solution contains 25mg of CaCO_(3). What is its concentration in ppm ?
Answer» ANSWER :5

Discussion

92104.	5 litre solution of 0.4M CuSO_((aq)) is electrolysed using Cu electrode. A currentof 482.5 ampere of passed for 4 minute. The concentrations of CuSO_4 left in solution is
Answer» 0.16 M 0.32 M 0.34 M 0.40 M Solution :LET us suppose `K_1` and `K_2` are the SPECIFIC conductance of solutions 'A' and 'B' REPECTIVELY and CELL constant is 'y'. We know that, Specific conductance = Conductance `xx` Cell constant (A), `k_1` = `1/50xxy`For (B), `k_2` = `1/100xxy` When EQUAL volume of (A) and (B) are mixed, the volume becomes double. then, Specific conductance of mixture = `(k_1+k_2)/2` `:. (k_1 + k_2)/2=1/Rxxy``1/2[y/50+y/100]=1/Rxxy` `1/100+1/200=1/R` R=`200//3`=66.66 ohm

Discussion

92105.	5-Hydroxyhexanal forms a six member hemiacetal, which predominates at equilibrium in aqueous solution. How many stereoisomers are possible for this cyclic hemiacetal?
Answer» Solution :(4) (Two CHIRAL CARBONS HENCE four stereoisomers)

Discussion

92106.	5 gm of CaCO_(3) is made to react with 0.2M, 100ml HCl solution. What will be the volumes of CO_(2)gas evolved at NTP?
Answer» 2240ml 224ml 22.4L 224L Solution :LR `CaCO_(3)+ 2HClrarrCaCl_(2)+H_(2)O+CO_(2)` mole `(5)/(10)=0.05 (0.2xx100)/(1000)=0.02` `n_(CO_(2))` produce `=(n_(HCl))/(2)=(0.02)/(2)=0.01` `V_(CO_(2))NTP=22.4xx0.01` LITRE `=22400xx0.01ml` =224'ml

Discussion

92107.	5 g of Na_(2) SO_(4) was dissolved in x g of H_(2)O. The change in froczing point was found to be 3.82^(@)C. If Na_(2)SO_(4) is 81.5% ioniscd, the value of x (K_(f)for water =1.86 ^(@)C kg mol ^(-1)) is approximately: (molar mass of S = 32 g mol ^(-1)) and that of Na=23 g mol ^(-1))
Answer» 15 G 25 g 45 g 65 g Solution :Molarity (cxpcrimental) ` = (DELTA T _(f))/(K _(f)) = (3.82)/( 1.86) =(3.82)/(1.86)= 2.054` mol/1000 g solvent Molarity (thcartical) `= ("mole of solute")/("wt. of solventing(g))xx1000` `= ( 5g//142 g //"mole")/(x) xx1000` `Na_(2) SO_(4) to 2NA^(+) +SO_(4)^(2-)` `{:("Moles before dissociation", 1,0,0),("Moles after dissociation", 1-x, 2x, x):}` Von't FACTOR (i) `= ("Moles after dissociation")/("Moles before dissociation")` `= ((1-x 0 + 2x+x)/( 1)` `Na_(2) SO_(4)` is ionised `81.5%` means `x=0.815` `=((1-0.815) + 2xx0.815 +0.815)/(1) =2.63.` `i = ("Observed molarity ")/("Calculated molarity")` `implies 2.63 =(2.054)/((0.0352)/(x)XX 1000) =45.07g.`

Discussion

92108.	5 g sample contain only Na_(2)CO_(3)and Na_(2)SO_(4)This sample is dissolved and the volume made up to 500 mL. 25 mL of this solution neutralizes 20 mL of 0.1 M H_(2)SO_(4). Calculate the percentage of Na_(2)SO_(4) in the sample.
Answer» Solution :Only `Na_(2)CO_(3)` reacts with `H_(2)SO_(4)` `Na_(2)CO_(3) + H_(2)SO_(4) to Na_(2)SO_(4) + H_(2)CO_(3)` = `20 XX 0.1 =2` no. Of moles of `Na_(2)CO_(3)` in 500 ML solution `=500/25 xx 2= 40` wt. Fo `Na_(2)CO_(3) = 40 xx 106 xx 10^(-3) = 4.24`G = 84.8

Discussion

92109.	5 g of hydrogen reacts with 32 g of oxygen to form moles of water
Answer» 1 2 3 4 Answer :B

Discussion

92110.	5 g each of the following gases at 87^@C and 750 mm pressure are taken. Which of them will have the least volume :
Answer» HF HCl HBr HI Answer :D

Discussion

92111.	5 A current is passed through electrolytic cell filled with AgNO_(3) solution for 2.7 hours. If 1 spoon require 0.01 g silver to get coated, so calculate how many spoon get coated by silver obtained on the cathode ?
Answer» ANSWER :5439 SPOON

Discussion

92112.	4xx10^(-2) moles of a non-volatile solute is added into 10 moles of solvent.The solute plymerises into soluble solid following first order kinetics with rate constant equal to 0.693 min^(-1). If after 2 min, elevation in boiling point is observed to be 0.01^(@) then calculate which form the solute is getting polymerised into? Given : Molar mass of solvent=100 g//mole. K_(b) of solvent =0.5 K-kg mol^(-1) [Instruction : Mark 2 if polymer is dimer, 3 if trimer and so on.]
Answer» ANSWER :3

Discussion

92113.	4xx10^(-3) m^(3)H_(2)O_((i)) is placed in a closed cubic vessel of edge-length of 1 m at 400 K temperature. If the vapour pressure of water is 0.0842 atmosphoere with density 0.89 g cm^(-3) at the same temperature,how much is left behind in liquid state?
Answer» ANSWER :`3.948` LIT

Discussion

92114.	4Zn +10HNO_(3) to 4Zn(NO_(3))_(2) +NH_(4)NO_(3)+3H_(2)O. In this reaction one mole of HNO_(3)is reduced by
Answer» 32g Zn 64G Zn 128g Zn 26g Zn ANSWER :4

Discussion

92115.	4NH_3(g)+5O_2(g)to4NO(g)+6H_2O(g), is in equilibriumin a closed container of volume 1L at a given temperature. If the reaction is started with 1 mol NH_3(g) and 1 mol of O_2(g) and the number of mol of H_2O(g) at equilibrium is 0.6 mol, then at equilibrium-
Answer» `[NH_3]=[NO]` `[NO]LT[O_2]` `[NO]GT[NH_3]` `[O_2]lt[H_2O]` ANSWER :B::D

Discussion

92116.	4NH_3 + 5O_2 to 4NO + 6H_2O. If rate of formation of NO is 6 xx 10^(−4) "atm min"^(-1), calculate the rate of formation of H_2O.
Answer» SOLUTION :`9.0xx10^(-4) "ATM MIN"^(-1)`

Discussion

92117.	4Na+O_2 to 2Na_2O Na_2O+H_2O to 2NaOH In the given reaction, the oxide of sodium is……
Answer» Acidic BASIC Amphoteric Neutral Solution :`Na_2O` from NAOH so that it is basic OXIDE.

Discussion

92118.	4n + 2 series is known as:
Answer» ACTINIUM series thorium series uranium series neptunium series Answer :C

Discussion

92119.	4 L of 0.02 M aqueous solution of NaCl was diluted by adding 1 L of water. The molality of the resultant solution is……..
Answer» 0.004 0.008 0./012 0.016 Solution :`M_(1)V_(1)=M_(2)V_(2)` `(0.02 M)xx(4L)=M_(2)xx(5L)` `M_(2)=((0.0M)xx(4L))/((5L))` =0.16 M

Discussion

92120.	4L of water is added to 2L of 6M HCl. The molarity of the final solution is
Answer» 4M 2M 1M 0.5M Solution :Final VOLUME ` = 2L + 4L = 6L` molarity ` = (6 xx 2)/(6) = 2M`

Discussion

92121.	4HNO_(3)+P_(4)O_(10)toHPO_(3)+ X In the above reaction the product X is:
Answer» `NO_(2)` `N_(2)O_(3)` `N_(2)O_(4)` `N_(2)O_(5)` Answer :D

Discussion

92122.	4L of 0.02 M aqueous solution of NaCl was diluted by adding one litre of water. The molality of the resultant solution is ……………… …
Answer» 0.004 0.008 0.012 0.016 Solution :`M_(1)V_(1)=M_(2)V_(2),0.02xx4=M_(2)XX5 or M_(2)=0.016`

Discussion

92123.	4HBr+O_(2)to2H_(2)O+2Br_(2). The molecularity of the reaction appears as 5, but experimentally, it is 2. Explain.
Answer» Solution :The mechanism of OXIDATION reaction of HBr is given in three STEPS. `HBr+O_(2) to HOOBr, HOOBr+HBr to 2HOBr HOBr+HBr to H_(2)O+Br_(2)` MOLECULARITY of each step is `2`. The overall molecularity of the slowest step `(1st)` is `2`

Discussion

92124.	4g of Substance A, dissolved in 100g H_(2)O depressed the freezing point of water by 0.1^(@)C . While 4g of another substance B, depressed the f.pt by 0.2^(@)C . What is the between molecular weights of A and B.
Answer» `M_(A)=4M_(B)` `M_(A)=M_(B)` `M_(A)=0.5M_(A)` `M_(A)=2M_(B)` ANSWER :D

Discussion

92125.	4g of copper was dissolved in conc. Nitric acid. The copper nitrate so obtained on strong heating gave 5 g of its oxide. The equivalent weight of copper is
Answer» `23` `32` `12` `20` Solution :4g of copper gives 5 g of its OXIDE `:.` 4 g of copper COMBINES with 1 g of oxygen `:.` Mass of copper that combines with 8g of oxygen `8xx4=32` `:.` Eq. mass of copper = 32

Discussion

92126.	4f^(2) 5d^(0) 6s^(0) is the electronic configuration of lanthanoide ion
Answer» only `Eu^(2+)` only `Gd ^(3+) and Eu ^(2+)` only `Eu^(2+) , Gd ^(3+) and Tb ^(4+)` only `Eu^(2+) and Tb ^(2+)` Answer :C

Discussion

92127.	4f^(14) configuratoin is observed in
Answer» YB and Lu Dy and Pm Lu and La Tm and Lu Answer :A

Discussion

92128.	4f^(14) 5d^(0) 6s^(0) represents which of the followig ?
Answer» `YB ^(3+)` `LU^(3+)` `Nb^(3+)` `Lr ^(3+)` ANSWER :B

Discussion

92129.	4f^(0) , 5d^(0) 6s^(0) represents which one of the following
Answer» `Ce ^(3+)` `AC ^(3+)` `LA ^(3+)` `Ce ^(+2)` ANSWER :C

Discussion

92130.	4f and 5f-series has total number of elements
Answer» 13 12 14 28 Answer :D

Discussion

92131.	4Cu+ 10HNO_(3) to Cu(NO_(3))_(2) + X + H_(2)O Mention the substance 'X'
Answer» `NO_(2)` `N_(2)O` NO `N_(2)O_(3)` SOLUTION :`4Cu_(g) + underset("dilute")(10HNO_(3))(g) to 4Cu(NO_(3))_(2)(aq) + 5H_(2)O (L) + N_(2X) O_(g)`

Discussion

92132.	4CH_(3)CH_(2)underset(O)underset(\|\|)CH_(3) underset(H_(2)O)overset(NaBH_(4))to X what is X?
Answer» `4CH_(3)CH_(2)UNDERSET(OH)underset(\|)CH_(2)CH` `4CH_(3)CH_(2)CH_(2)CH_(2)OH` `4CH_(3)-underset(OH)underset(\|)CH-CH_(3)` `4CH_(3)CH_(2)CH_(2)OH` Solution :`4CH_(3)CH_(2)-underset(O)underset(\|\|)C- CH_(3) underset(H_(2)O)overset(NaBH_(4))to underset(" (Butan-2-ol)")(4CH_(3)-CH_(2)-underset(OH)underset(\|)CH-CH_(3))`

Discussion

92133.	4CH_(3)CH_(2)CH_(2)CHO underset(H_(2)O)overset(NaBH_(4))to X what is X?
Answer» `4CH_(3)CH_(2)underset(OH)underset(\|)CH_(2)CH` `4CH_(3)CH_(2)CH_(2)CH_(2)OH` `4CH_(3)CH_(2)CH_(2)underset(O)underset(\|\|)CH_(3)` `4CH_(3)- CH_(2)-CH_(2)-CH_(3)` SOLUTION :`4CH_(3)CH_(2)CH_(2)CH_(2)OH`

Discussion

92134.	Be^7_4 Captures K electrons into its nucleus. The mass no and atomic no. of new nucleus formed are
Answer» 7, 5 7, 4 6, 3 7, 3 Answer :D

Discussion

92135.	4Ag_((s)) + 8CN_((aq))^(-) + 2H_2O_((l)) = O_(2(g)) to X + OH_((aq))^(-) then X = ………
Answer» `[AG(CN)_(2)]^(-)` `[Ag(CN)_(4)]^(-)` `AgCN` `[4Ag(CN)_2]^(-)` ANSWER :D

Discussion

92136.	4.90 g of KClO_(3) when heated produced 1.92 g of oxygen and the residue (KCl) left behind weighs 2.96 g. Show that these results illustrate the law of conservation of mass.
Answer» Solution :Mass of `KClO_(3)` taken = 4.90 g TOTAL mass of the product `(KCl+O_(2))=2.96+1.92=4.88g` Difference betweenthe mass of the reactant and the total mass of the PRODUCTS = `4.90-4.88=0.02g` This small difference may be DUE to EXPERIMENTAL error. THUS, law of conservation of mass holds good within experimental errors.

Discussion

92137.	4.9 g of H_(2) SO_(4) is present in500 cm^(3) of the solution has molarity.
Answer» 0.2 0.1 0.01 0.02 Solution :`M= "Mass"/("Molar mass" xx V (DM^(3)))`=`4.9/(98 xx 5)=1/10=0.1`

Discussion

92138.	48 g of SO_(3) , 12.8 g of SO_(2) and 9.6 g of O_(2) are present in one litre . The respective active masses will be
Answer» 1.0 , 05 and 0.3 0.6 , 0.2 and 0.3 0.6 , 0.4 and 0.2 1.0 , 0.5 and 1.5 Answer :2

Discussion

92139.	""_(4)^(7)Be captures a K-electron into its nucleus. What will be the mass number and atomic number of the nuclide formed?
Answer» SOLUTION :When a nucleus captures a K-electron, a proton is converted to a NEUTRON. So the mass number does not change but the ATOMIC number REDUCES by 1 unit. Thus the mass number and atomic number of the resulting nuclide will be 7 and 3 respectively.

Discussion

92140.	4.6 gm of liquid ethanol (C_(2)H_(5)OH) is taken in 12 litre container and at 27^(@)C, 40% of ethanol is vaporised till equilibrium. Now if volume of container is halved and system is allowed to attain equilibrium then find [Give: R=0.08atm/litre mole-k]
Answer» Solution :`C_(2)H_(5)OH(L)hArrC_(2)H_(5)OH(g)` Molet=0 0.1 0 `t=t_("eqm") 0.1-0.04 0.04` =0.06 For gas PV=nRT `rArrP_(C_(2)H_(5)OH("gas"))=(nRT)/(V)=(0.04xx0.08xx300)/(12)` `V.P_("gas")=0.08` atm (at `eq^(m)`) =0.08xx100=8

Discussion

92141.	4.6g of a polydric alcoholwas treated with an excess of methyl magnesium bromide to produce 3.36 liter of CH_(4) at STP. Calculate number of -OH(molecular weightof alcohol = 92)
Answer» SOLUTION :N//A

Discussion

92142.	4.58 g of NaCl was dissolved in 129 g of H_(2)O at 280 K. The osmotic pressure of the solution was found to be 2.606 magapascal. Density = 1.02 g/cc. Calculate osmotic coefficient.
Answer» ANSWER :I=1.906

Discussion

92143.	450 waves of a energy radiation passes in one minute from a point then what will be thewavelength of radiation :-
Answer» `6.66xx10^(-5)` m `4XX10^(7)` m `2.25xx10^9` m `2.5xx10^(-8)` m SOLUTION :`lambda=C/v` `=(3xx10^8)/(450/60)=(3xx10^8xx6)/45` `=4xx10^7` m

Discussion

92144.	45.4 L of dinitrogen reacted with 22.7 L of oxygen and 45.4 L of nitrous oxide was formed. The reaction is : 2N_(2)(g)+O_(2)(g) rarr 2N_(2)O(g) Which law is being obeyed in this experiment ? Write the statement of the law?
Answer» Solution :As gases are reacting together to form GASEOUS products and the ratio of the volumes of `N_(2):O_(2):N_(2)O=45.4:22.7:45.4=2:1:2` which is a simple whole NUMBER ratio. Hence, the EXPERIMENT proves Gas LUSSAC's LAW of gaseous volumes.

Discussion

92145.	4.5 moles each of hydrogen and iodine heated in a seated in a sealed ten litre vessel. At equlibrium, 3 moles of HI were found. The equilibrium constnt for H_(2(g))hArrPCl_(3(g))+Cl_(2(g)) is
Answer» 1 10 5 `0.33` Solution :`H_(2)+I_(2)hArr2HI` `{:("INITIAL conc.",4.5,4.5,0),("at Equlibrium",4.5-x,4.5-x,2x):}` From euestion `2x=3` `x=3/2=1.5` So conc. At eqm. `4.5-1.5of H_(2)` `=4.5-1.5of I_(2)and 3 of HI` `K=([H]^(2))/([I_(2)][H_(2)])=(3xx3)/(3xx3)=1.`

Discussion

92146.	45 grams of ethylene glycol (C_(2)H_(6)O_(2)) is mixed with 600 grams of water. What is the depression in freezing point (K_(f) for water = 1.86 kg "mole"^(-1)) (C = 12, O = 16, H = 1 g/mol).
Answer» 2.25 K 22.5 K 3.25 K 32.5 K Solution :M.F `= C_(2)H_(6)O_(2)` M.wt `= 2(12)+6(1)+2(16)` `= 62 g mol^(-1)=M_(2)` Weight of ETHYLENE glycol = 45 g `= W_(2)` Weight of solvent (water = 600 g `= W_(1)` `K_(f)` (water) = 1.86 kg `mol^(-1)` `Delta T = K_(f)m` `= (K_(f)XX W_(2)xx1000)/(M_(2)xx W_(1))` `= (1.86xx 45" gram" xx 1000)/(62 g mol^(-1)xx 600 " gram")` = 2.25 K

Discussion

92147.	4.5 g of PCl5 on vapourisation occupied a volume of 1700 mL at 1 atmosphere pressure and 227 °C temperature. Its degree of dissociation is
Answer» 0.0921 0.0091 0.921 None of these Solution :Let the degree of dissociation be `alpha`, then `PCl_(5)toPCl_(3)+Cl_(2)` `(1-alpha)""alpha""alpha` (M) Molecular weight of `PCl_(5)` = 31 + 5 `xx` 35.5 = 208.5 Total number of MOLECULES before dissociation= 1 Total number of molecules after dissociation `=1-alpha+alpha+alpha=1+alpha` THUS each gram mole changes to (1 + a) gram-mole. Thus `PV=W/M(1+alpha)RT` ( V = volume after dissociation) or `1xx1700/1000=(4.5)/(208.5)(1+alpha)RT` [1700ml`=1700/1000L`] or `alpha=0.921` `therefore%` age dissociation =92.1%

Discussion

92148.	4.5 gm of aluminium (atomic mass 27 amu) get deposited on cathode by passing fixed amount of electricity through Al^(3+) solution. So what is the volume of hydrogen gas at STP by passing same amount of electricity through H^(+) solution.
Answer» 44.8 L 22.4 L 11.2 L 5.6 L Solution :NUMBER of grams in equivalence of `H^(+)`=Number of grams in equivalence of `Al^(3+)` EQUIVALENT weight of `Al^(3+)=(27)/(3)=9` `Al^(3+)=(4.5)/(9)=0.5` Number of grams in equivalence of `H^(+)`=number of moles of `H^(+)` So, equivalent weight of `H^(+)=0.5xx1.0gm` `=0.5gm` of `H_(2)` As we know that, 2 gm `H_(2)` at STP=22.4 L So, at STP, 0.5 gm `H_(2)=(22.4)/(2)xx0.5=5.6L`

Discussion

92149.	45 g of ethylene glycol (C_(2)H_(6)O_(2)) is mixed with 650 g of water. The freezing point depression (in K) will be (K_(f)"for water "=1.86 "K kg mol"^(-1)).
Answer» Solution :Molality of solution (m) `=("No of moles of "C_(2)H_(6)O_(2))/("Mass of water in kg")` `=(((45G))/((62" G mol"^(-1))))/(6.65kg)=((0.73 mol))/((0.65kg))=1.12"mol kg"^(-1)` `DeltaT_(f)=K_(f)xxm=(1.86" K kg mol"^(-1))xx(1.12 "mol kg"^(-1))` `=2.08 K~~2K.`

Discussion

92150.	4.5 g of PCl_(5)on vapourisation occupied a volume of 1700mL at atmosphere pressure and 227°C temperature. Calculate it: degree of dissociation.
Answer» Solution :Let the degree of dissociation be `alpha`, then `underset(1-alpha)(PCl_(5)) to underset(alpha)(PCl_(3)) + underset(alpha)(Cl_(2))` (M) MOLECULAR weight of `PCl_(5) = 31+5 xx 35.5 = 208.5` Total number of molecules before dissociation = 1 Total number of molecules ALTER dissociation `=1- alpha + alpha + alpha =1 + alpha` Thus each gram mole CHANGES to `(1+ alpha)` gram-mole. Thus, `PV =W/M (1+alpha) RT` (V = Volume after dissociation) or `1 xx 1700/1000 = 4.5/(208.5) (1+ alpha) RT [1700 ml = 1700/1000 L]` or `alpha = 0.921`

Discussion

Explore topic-wise InterviewSolutions in .

5 litres of a solution contains 25mg of CaCO_(3). What is its concentration in ppm ?

5 litre solution of 0.4M CuSO_((aq)) is electrolysed using Cu electrode. A currentof 482.5 ampere of passed for 4 minute. The concentrations of CuSO_4 left in solution is

5-Hydroxyhexanal forms a six member hemiacetal, which predominates at equilibrium in aqueous solution. How many stereoisomers are possible for this cyclic hemiacetal?

5 gm of CaCO_(3) is made to react with 0.2M, 100ml HCl solution. What will be the volumes of CO_(2)gas evolved at NTP?

5 g of Na_(2) SO_(4) was dissolved in x g of H_(2)O. The change in froczing point was found to be 3.82^(@)C. If Na_(2)SO_(4) is 81.5% ioniscd, the value of x (K_(f)for water =1.86 ^(@)C kg mol ^(-1)) is approximately: (molar mass of S = 32 g mol ^(-1)) and that of Na=23 g mol ^(-1))

5 g sample contain only Na_(2)CO_(3)and Na_(2)SO_(4)This sample is dissolved and the volume made up to 500 mL. 25 mL of this solution neutralizes 20 mL of 0.1 M H_(2)SO_(4). Calculate the percentage of Na_(2)SO_(4) in the sample.

5 g of hydrogen reacts with 32 g of oxygen to form moles of water

5 g each of the following gases at 87^@C and 750 mm pressure are taken. Which of them will have the least volume :

5 A current is passed through electrolytic cell filled with AgNO_(3) solution for 2.7 hours. If 1 spoon require 0.01 g silver to get coated, so calculate how many spoon get coated by silver obtained on the cathode ?

4xx10^(-3) m^(3)H_(2)O_((i)) is placed in a closed cubic vessel of edge-length of 1 m at 400 K temperature. If the vapour pressure of water is 0.0842 atmosphoere with density 0.89 g cm^(-3) at the same temperature,how much is left behind in liquid state?

4Zn +10HNO_(3) to 4Zn(NO_(3))_(2) +NH_(4)NO_(3)+3H_(2)O. In this reaction one mole of HNO_(3)is reduced by

4NH_3(g)+5O_2(g)to4NO(g)+6H_2O(g), is in equilibriumin a closed container of volume 1L at a given temperature. If the reaction is started with 1 mol NH_3(g) and 1 mol of O_2(g) and the number of mol of H_2O(g) at equilibrium is 0.6 mol, then at equilibrium-

4NH_3 + 5O_2 to 4NO + 6H_2O. If rate of formation of NO is 6 xx 10^(−4) "atm min"^(-1), calculate the rate of formation of H_2O.

4Na+O_2 to 2Na_2O Na_2O+H_2O to 2NaOH In the given reaction, the oxide of sodium is……

4n + 2 series is known as:

4 L of 0.02 M aqueous solution of NaCl was diluted by adding 1 L of water. The molality of the resultant solution is……..

4L of water is added to 2L of 6M HCl. The molarity of the final solution is

4HNO_(3)+P_(4)O_(10)toHPO_(3)+ X In the above reaction the product X is:

4L of 0.02 M aqueous solution of NaCl was diluted by adding one litre of water. The molality of the resultant solution is ……………… …

4HBr+O_(2)to2H_(2)O+2Br_(2). The molecularity of the reaction appears as 5, but experimentally, it is 2. Explain.

4g of Substance A, dissolved in 100g H_(2)O depressed the freezing point of water by 0.1^(@)C . While 4g of another substance B, depressed the f.pt by 0.2^(@)C . What is the between molecular weights of A and B.

4g of copper was dissolved in conc. Nitric acid. The copper nitrate so obtained on strong heating gave 5 g of its oxide. The equivalent weight of copper is

4f^(2) 5d^(0) 6s^(0) is the electronic configuration of lanthanoide ion

4f^(14) configuratoin is observed in

4f^(14) 5d^(0) 6s^(0) represents which of the followig ?

4f^(0) , 5d^(0) 6s^(0) represents which one of the following

4f and 5f-series has total number of elements

4Cu+ 10HNO_(3) to Cu(NO_(3))_(2) + X + H_(2)O Mention the substance 'X'

4CH_(3)CH_(2)underset(O)underset(||)CH_(3) underset(H_(2)O)overset(NaBH_(4))to X what is X?

4CH_(3)CH_(2)CH_(2)CHO underset(H_(2)O)overset(NaBH_(4))to X what is X?

Be^7_4 Captures K electrons into its nucleus. The mass no and atomic no. of new nucleus formed are

4Ag_((s)) + 8CN_((aq))^(-) + 2H_2O_((l)) = O_(2(g)) to X + OH_((aq))^(-) then X = ………

4.90 g of KClO_(3) when heated produced 1.92 g of oxygen and the residue (KCl) left behind weighs 2.96 g. Show that these results illustrate the law of conservation of mass.

4.9 g of H_(2) SO_(4) is present in500 cm^(3) of the solution has molarity.

48 g of SO_(3) , 12.8 g of SO_(2) and 9.6 g of O_(2) are present in one litre . The respective active masses will be

""_(4)^(7)Be captures a K-electron into its nucleus. What will be the mass number and atomic number of the nuclide formed?

4.6 gm of liquid ethanol (C_(2)H_(5)OH) is taken in 12 litre container and at 27^(@)C, 40% of ethanol is vaporised till equilibrium. Now if volume of container is halved and system is allowed to attain equilibrium then find [Give: R=0.08atm/litre mole-k]

4.6g of a polydric alcoholwas treated with an excess of methyl magnesium bromide to produce 3.36 liter of CH_(4) at STP. Calculate number of -OH(molecular weightof alcohol = 92)

4.58 g of NaCl was dissolved in 129 g of H_(2)O at 280 K. The osmotic pressure of the solution was found to be 2.606 magapascal. Density = 1.02 g/cc. Calculate osmotic coefficient.

450 waves of a energy radiation passes in one minute from a point then what will be thewavelength of radiation :-

45.4 L of dinitrogen reacted with 22.7 L of oxygen and 45.4 L of nitrous oxide was formed. The reaction is : 2N_(2)(g)+O_(2)(g) rarr 2N_(2)O(g) Which law is being obeyed in this experiment ? Write the statement of the law?

4.5 moles each of hydrogen and iodine heated in a seated in a sealed ten litre vessel. At equlibrium, 3 moles of HI were found. The equilibrium constnt for H_(2(g))hArrPCl_(3(g))+Cl_(2(g)) is

45 grams of ethylene glycol (C_(2)H_(6)O_(2)) is mixed with 600 grams of water. What is the depression in freezing point (K_(f) for water = 1.86 kg "mole"^(-1)) (C = 12, O = 16, H = 1 g/mol).

4.5 g of PCl5 on vapourisation occupied a volume of 1700 mL at 1 atmosphere pressure and 227 °C temperature. Its degree of dissociation is

4.5 gm of aluminium (atomic mass 27 amu) get deposited on cathode by passing fixed amount of electricity through Al^(3+) solution. So what is the volume of hydrogen gas at STP by passing same amount of electricity through H^(+) solution.

45 g of ethylene glycol (C_(2)H_(6)O_(2)) is mixed with 650 g of water. The freezing point depression (in K) will be (K_(f)"for water "=1.86 "K kg mol"^(-1)).

4.5 g of PCl_(5)on vapourisation occupied a volume of 1700mL at atmosphere pressure and 227°C temperature. Calculate it: degree of dissociation.