94420 + Interview Questions in Chemistry in Class 12 Page 1861 InterviewSolution

93001.	100 mll of 0.01 M metal - chloride completely hydrolysed in excess water , it has been observed that resulting solution required 30 ml of 0.1 M aqueous silver nitrate solution for the complete precipitation of the chloride ion, then expected metal is :
Answer» SN SI PB AL SOLUTION :N/A

Discussion

93002.	100 mL of tap water containing Ca(HCO_(3))_(2) was titrated with N//50HCl with methyl orange as indicator. If 30 mL of HCl were required, calculate the temporary hardness as parts of CaCO_(3) per 10^(6) parts of water.
Answer» 150 ppm 300 ppm 450 ppm 600 ppm Solution :`"30 mL N/50 HCl "-= "30 mL N/50 "Ca(HCO_(3))_(2)` `-=" 30 mL N/50 "CaCO_(3)-="100 mL tap water"` `"MASS of "CaCO_(3)" in 100 mL tap water"` `=(ExxNxxN)/(1000)=(50xx30)/(50xx1000)=0.03g` `rArr"hardness = 300 ppm"`

Discussion

93003.	100 mL of tap water containing Ca(HCO_(3)_(2) was titrated with (N)/(50)HCL with methyl orange as indicaor. If 30 mL of HCL was required, the temporary hardness of water as parts of CaCO_(3) per 10^(6) parts of water was
Answer» 150 ppm 600 ppm 275 ppm 300 ppm Solution :Meq of `CaCO_(3)` = Meq. Of `Ca(HCO_(3))_(2)` = Meq. Of HCl `(w)/(100//2)xx1000=(1)/(50)xx30` `w_(CaCO_(3))=(3xx100)/(5xx2000)=0.03 g` `THEREFORE` 100 g ML of `H_(2)O` CONTAINS `Ca^(2+)=0.03g` `10^(6)mLH_(2)O" contains "Ca^(2+)=(0.03xx10^(6))/(100)=300"ppm"`

Discussion

93004.	100 ml of PH_3 when decomposed produces phosphorus and hydrogen . The change in volume is :
Answer» 50 ML increase 500 ml decrease 900 ml decrease None Answer :A

Discussion

93005.	100 ml of PH_(3) on heating forms P and H_(2). The volume change in the reaction is
Answer» an INCREASE of 50 ML an increase of 100 ml an increase of 150 ml a decrease of 50 ml Solution :`UNDERSET("2 mL")(2PH_(3))(g)rarr2P(s)+underset("3 mL")(3H_(2)(g))` 2 mL `PH_(3)` on decomposition give 3 mL of `H_(2)`, i.e., increase = 1 mL `therefore "100 mL PH"_(3)" will result in increase = 50 mL"`

Discussion

93006.	100 mL of ( M / 1 0) aqueous solution of a monoprotic acid is titrated with a solution of NaOH. When one-third of the acid is neutralised, pH of the solution becomes 3.9. pH of the solution at half-neutralisation point is-
Answer» 5.8 4.2 3.6 6.3 Answer :B

Discussion

93007.	100 mL of O_(2) and H_(2) are kept at same temperature and pressure. What is true about their numberof molecules ?
Answer» `N_(O_(2)) gt N_(H_(2))` `N_(O_(2)) lt N_(H_(2))` `N_(O_(2)) = N_(H_(2))` `N_(O_(2)) + N_(H_(2)) = 1 `mole Solution :ACCORDING to Avogadro.s law, equal volumes of gases contain equal number of molecules under similar CONDITIONS of temperature and pressure.

Discussion

93008.	100 mL of HCl gas at 25^@ C and 740 mm pressure was dissolved in 1 L of water. What will be the pH of solution. (Given, vapour pressure of H_2 O at 25^@ C is 23.7 mm.)
Answer» `2.012` `3.241` `2.414` `3.021` Solution :Givenvapourpressureof water ` =23.7` mm `:. P_(HCI _("DRY"))=( 740-23.7 )/( 760) atm= 0.9425 ` atm `V=100/1000= 0.1 L ,T= 25+ 273= 298K` ` therefore ` Mole of ` HCI` i.e.,`""^(n)HCI = (W) /(m ) =(PV )/(RT)= ( 0.9425 xx 0.1 )/( 0.0821 xx 298)` `thereforen= 3.85 xx 10^(-3)` Nowmolarityof HCI= noof molesdissolved in 1 L ofsolution `= (3.85 xx 10^(-3) )/( 1 ) ` ( V in litre ) ` therefore[H^+] = 3.85xx 10^(-3)M` ` thereforepH =- LOG[H^+] =- log ( 3.85 xx 10^(-3))` `pH =2.414`

Discussion

93009.	100 ml of HCl + 35 ml of NaOH, colour methyl orange in the solution will be
Answer» RED Yellow Can't be predicted METHYL orange is not a suitable indicator Solution :According to the questions, 35 ml of NaOH will neutralise 35 ml of HCl and since, in the mixture of these solutions the VOLUMES of HCl is much higher, so, the resultant mixture would be acidic in nature. `HCl_((aq)) + NaOH_((aq)) rarrr NaCl_((aq)) + H_(2)O_((I))` `:.` Methyl orange would impart a red COLOUR to the acidic mixture of solutions.

Discussion

93010.	100 ml of ethyl alcohol [d = 0.92 g/ml] and 900 ml of water [d = 1 g/ml] are mixed to form 1 lit solution. The Molarity and molality of the resulting solution are
Answer» 2M and 2m 2M and 2.415 m 2.2M and 1.1m 2M and 1m Answer :B

Discussion

93011.	100 mL of an O_(2)-O_(3) mixture was passed through turpentine, and 20 mL was reduced. If 100 mL of such a mixture is heated, the increase in volume will be
Answer» 110 mL 10 mL 90 mL 100 mL Answer :B

Discussion

93012.	100 mL of a saturated solution of M""_(2)SO""_(4) is giving 24 disintegrations per hour due to radio active metal "M" (lamda=2xx10""^(-17)hour""^(-1),N""_(0)=6xx10""^(23)a). If the solubility product of the salt is "x"xx10""^(-3y). What is (x+y)?
Answer» Solution :`(-DN)/(dt)=lamdaN,24=2xx10""^(-17)x("moles")6XX10""^(23)` moles of `M""^(+)=2xx10""^(6)`,molarity of `M""^(+)=2xx10""^(-5)`M `M""_(2)SO""_(4)hArr2M""^(+)+SO""_(4)""^(-2)` `2S=10""^(-5)M,K""_(sp)=4S""^(3)=4xx10""^(-15)`

Discussion

93013.	100 ml of an aqueous solution contains 6.023xx10^(21) solute molecules. The solution is diluted to 1 lit. The number of solute moleculespresent in 10ml of the dilute solution is
Answer» `6.0xx10^(20)` `6.0xx10^(19)` `6.0xx10^(18)` `6.0xx10^(17)` Answer :B

Discussion

93014.	100 mL of an aqueous solution of protein contais 6.3 g of protein. Calculate the molar mass of protein if osmatic pressure of the solution at 27^@C is 2.57 xx 10^(-3) bar.
Answer» Solution :Osmotic pressure `pi=2.57xx10^(-3)` BAR , weight of solute = W= 6.3 g Solution constant =S=0.083 L bar `K^(-1) "mol"^(-1)` Volume of solution =V=100mL = 0.1 L : Absolute TEMPERATURE =300 K MOLAR mass =`(wST)/(piV)=(6.3xx0.083xx300)/(2.57xx10^(-3)xx0.1)="610 g mol"^(-1)`

Discussion

93015.	100 ml of an aqueous solution contains 6.023 xx 10^21 solute molecules. The solution is diluted to 1 lit. The number of solute molecules present in 10ml of the dilute solution is
Answer» `6.0xx10^20` `6.0xx10^19` `6.0xx10^18` `6.0xx10^17` ANSWER :B

Discussion

93016.	100 ml of an acid solution is neutralized by 50 ml of NaOH solution containing 0.2 g NaOH. The concentration of acid solution is
Answer» 0.1 N 0.05 N and the solution is basic 0.5 N 0.25 N Solution :`N_(1)=50ml,V_(1)=0.2N,N_(2)=100ML,V_(2)=X` `50xx0.2=100xx x` `x=0.1N`.

Discussion

93017.	100 ml of a sample of hard water required 15.1 ml 0.02N H_2 SO_4 for complete reaction. The hardness of water in ppm will be:
Answer» 125 ppm 734 ppm 103 ppm 251 ppm Answer :D

Discussion

93018.	100 ml of a gaseous mixture containing Ne,CO_(2) & H_(2) on complete combustion in just sufficient amount of O_(2) showed contraction of 60 ml at NTP. When the resulting gases were passed through KOH solution, volume reduce by 40%. The volume ratio of V_(co_(2)):V_(Ne):V_(H_(2)) in original mixture is:
Answer» `10:9:2` `6:9:10` `7:3:10` None Solution :`100mlrarrH_(2)+CO+Ne` `H_(2)+1//2O_(2)rarrH_(2)O(l)` x x/2 `DeltaV_("contraction")=x+(x)/(2)=(3X)/(2)=60` `rArrx=40ml` Total initial volume `=V_("mix")+V_(O_(2))` =100+20=120ml Volume of resulting gases=120-60=60ml `40% "of" 60ml=(40)/(100)xx60=24ml` `V_(Ne)=60-40=36` `V_(CO_(2)): V_("Ne"): V_(H_(2))=6:9:10`

Discussion

93019.	100 ml of a mixture of methane and acetylene was exploded with excess of oxygen. After cooling to room temperature the resulting gas mixture was passed through KOH when a reduction of 135 ml in volume was noted. The composition of the hydrocarbon mixture by volume at the same temperature and pressure will be
Answer» 65% `CH_4, 35% C_2H_2` `35% CH_4, 65% C_2H_2` `25% CH_4,75% C_2H_2` `85% CH_4, 15% C_2H_2` ANSWER :A

Discussion

93020.	100 ml of a gaseous mixture containing CO and CH_(4) shows a volume contraction of 65 ml on combustion in excess of O_(2). Then what will be the volume of CO (g).
Answer» `10 mL` `40 mL` `90 mL` `20 mL` SOLUTION :`{:(,2 CO (G),+,O_(2) (g) RARR,,2 CO_(2) (g),,,),("volume",x mL,,(x)/(2) mL,,x mL,,,),(,CH_(4) (g) +,2O_(2) (g),,rarr,CO_(2) (g),+,2H_(2)O (l),),("volume",(100 - x) mL,2 (100 - x) mL,,(100 - x) mL,,2 (100 - x) mL,,):}` Volume contraction `= 2 (100 - x) + (x)/(2) = 65` `x = 90 mL`

Discussion

93021.	100 ml of a buffer of 1M NH_3 and 1M NH_4^(+) are placed in two voltaic cells separately. A current of 1.5A Is passed through both ceils for 20 minutes. If electrolysis of water only takes place (i) 2H_(2)O +O_(2)+4e^(-) to 4OH^(-) (RHS) (ii) 2H_(2)O to 4H^(+) +O_(2(g))+4e^(-) (LHS) then pH of hte
Answer» LHS will increase RHS will increase Both SIDE will increase Both side will decrease Answer :B

Discussion

93022.	100 ml of a buffer of 1M NH_(3) and 1M NH_(4)^(+) are placed in two voltaic cell separately. A current of 1.5 A is passed through both cells for 20 minutes. If electrolysis of water only takes place. (i) 2H_(2)O + O_(2) + 4e^(-) to 4OH^(-) (LHS) (ii) 2H_(2)O to 4H^(+) + O_(2)(g) + 4e^(-) (LHS) then pH of the
Answer» LHS will INCREASE LHS will increase RHS will increase Both SIDE will increase Answer :B

Discussion

93023.	100 mL of a buffer of 1 M NH_(3) (aq) and 1 M NH_(4)""^( +) (aq) are placed in two voltaic cells separately . A current of 1.5 A is passed through both cells for 20 minutes . If electrolysis of water only takes place 2H_(2)O+ O_(2) + 4 e^(-) rarr 4 OH^(-) (RHS) 2H_(2)O rarr 4H^(+) + O_(2) + 4e^(-) (LHS) then pH og the :
Answer» LHS HALF CELL will increase RHS half-cell will increase both half-cells will increase both half-cells will decrease Answer :B

Discussion

93024.	100 ml of 1M HCI, 200 ml of 2M HCl and 300 ml of 3M HCl are mixed with enough water to get 1M solution. The volume of water to be added is (in ml)
Answer» 600 700 800 125 Answer :C

Discussion

93025.	100 ml of 1M HCl, 200 ml 2 M HCl and 300 ml 3M HCl are mixed. The Molarity of the resulting solution is
Answer» 1M `2.66M` `2.33M` `4.25M` ANSWER :C

Discussion

93026.	100 mL of 1 M H_(2)SO_(4) are mixed with 200 mL of 8 M HCl solution. The noramlity of the resulting solution is
Answer» Solution :`"100 mL of 1 M "H_(2)SO_(4)="100 millimoles"` `"= 200 MILLI eq"` `"200 mL of 8 M HCl = 1600 millimoles"` `"= 1600 milli eq"` `"TOTAL milli eq "=200+1600=1800` `"Total volume "=100+200mL=300mL` `therefore"NORMALITY of resulting solution "=(1800)/(300)=6N.`

Discussion

93027.	100 mL of 1 N NH_4OH (K_b = 5 xx 10^-5) is neutralised to equivalence point by 1 NHCl. Calculate the pH of solution at equivalence point.
Answer» 2 2.5 3 5 Answer :D

Discussion

93028.	100 mL of 1 M HCl is mixed with 50 ml of 2M HCl Hence , [H_3O^+] is :
Answer» 1.00 M 1.50 M 1.33 M 3.00 M Answer :C

Discussion

93029.	100 ml of 1 M H_(2) SO_(4) solution (d_("solution") = 1.5 gm//ml) is mixed with 400 ml of water (d_("water") = 1 gm //ml) then molarity of final solution (d_("final solution") = 1.25 gm//ml) is -
Answer» 0.227 M 2.5 M 0.4 M 2.27 M Solution :Mass of `H_(2)SO_(4)` solution `= 100 xx 1.5 = 150 gm` Mass of water `= 400` Mass (total) `= 550 gm` MOLES of `H_(2)SO_(4) = 0.1` mole VOLUME finally `= (550)/(1.25) = 440 mL` `M = (0.1)/(440) xx 1000 = 0.227 M`

Discussion

93030.	100 mL of 0.5 N NaOH solution is added to 10 mL of 3 N H_(2)SO_(4) solution and 20 mL of 1 N HCl solution . The mixture is
Answer» ACIDIC ALKALINE NEUTRAL none of these Answer :C

Discussion

93031.	100 ml of 0.2 M H_(2)SO_(4) is added to 100 ml of 0.2 M NaOH. The resulting solution will be
Answer» Acidic Basic Neutral Slightly basic Solution :M.eq. of 0.2 M `H_(2)SO_(4) = (2 xx 0.2 M)/(1000) xx 100 = 0.04 m//l` M.eq. of 2M NaOH `= (0.2)/(1000) xx 100 = 0.02 m//l` LEFT `[H^(+)] = .04 - .02 = .02` Total volume `= 200 = (.02)/(200) =.0001 = 10^(-4)M` pH =4.

Discussion

93032.	100 mL of 0.1N I_2 oxidizes Na_2S_2O_3 in 50 ml solution to Na_2S_4O_6.The normality of this hypo solution against KMnO_4(Which oxidezes it to Na_2SO_4) would be :
Answer» 0.1 0.2 `1.0` 1.6 Solution :NA

Discussion

93033.	100 ml of 0.15 N H_(2)O_(2) is completelyoxidized by
Answer» 150ML of 0.1 N `KMnO_(4)` solution `2.5 xx 10^(-3)` moles of `K_(2) Cr_(2)O_(7)` in ACIDIC MEDIUM `15 xx 10^(_3)` moles of `KMnO_(4)` in basic medium 15 moles of `O_(3)` in acidic medium Answer :ABC

Discussion

93034.	100 ml of 0.1N FeSO_4 solution will be completely oxidised by 'x' gms of K_2Cr_2O_7 in acidic medium (Mol.wt = 294). The value of 'x' is
Answer» `4.9` `2.94` `0.49` `1.47` ANSWER :C

Discussion

93035.	100 ml of 0.1 N NaOH is mixed with 50 ml of 0.1 N H_2SO_4 . The pH of the resulting solution is
Answer» `LT 7` `gt 7` `=7` Cannont be predicted Answer :C

Discussion

93036.	100 ml of 0.1 N NaOH are mixed with 100 ml of 0.2 N HCl solution and the whole solution is made upto 1 litre. The pH of the resulting solution is :
Answer» 1 2 3 4 Answer :B

Discussion

93037.	100mL of 0.1 M solution of solute A are mixed with 200 mL of 0.1 M solution of solute B. If A and B are non-reacting substances, the molarity of the final solution will be
Answer» 0.3M 0.4M 0.1M 0.15M Solution :`M_(1) V_(1) + M_(2) V_(2) = M_(3) V_(3)` (MIXTURE) :. `100 xx 0.1 + 200 xx 0.1 = 300 xx M_(3)` or `M_(3)=0.1`

Discussion

93038.	100 ml of 0.1 M solutio of AB (d = 1.5 gm/m) is mixed with 100ml of 0.2 M solution of CB_(2) (d = 2.5 gm/m). Calculate the molarity of B^(-) in final solution if the density of final solution is 4gm/m. Assuming AB and CB_(2) are non reacting & dissociates completely into A^(+), B^(-), C^(2+)
Answer» ANSWER :0.5

Discussion

93039.	100 ml of 0.1 M solutionA is mixed with 20ml of 0.2 M solution B. The final molarity of the solution is :
Answer» 0.12 M 0.15 M 0.18 M 0.21 M Solution :Moles of A in 100 ML `=(0.1xx100)/(1000)=0.01` Moles of B in 20 ml `=(0.2xx20)/(1000)=0.004` Total moles `= 0.01+0.004=0.014` Molarity `= (0.014)/(120)xx1000=0.117=0.12` M

Discussion

93040.	100 mL of 0.1 M acetic acid is completely neutralized using a standard solution of NaOH. The volume of ethane obtained at STP after the complete electrolysis of the resulting solution is
Answer» 112 mL 56 mL 224 mL 560 mL Answer :A

Discussion

93041.	100 mL of 0.1 M acetic acid is completely neutralized using a standard solution of NaOH . The volume of ethane obtained at STP after the complete electrolysis of the resulting solution is
Answer» 112 mL 56 mL 224 mL 560 mL Solution :`CH_(3) CO OH + NaOH to CH_(3) CO O Na 2 CH_(3) CO O Na + 2 H_(2) O to CH_(3) - CH_(3) + 2 CO_(2) + 2 NaOH + H_(2)` NUMBER of moles of acetic acid = Number of moles of sodium acetate = `0.1 XX 100= 10` m moles = `10 xx 10^(-3)` moles 2 moles of `CH_(3) CO ONa` on electrolysis produces 22400 mL of ETHANE . `therefore 10 xx 10^-3` moles of `CH_(3) CO ONa ` will produce = `(22400 xx 10 xx 10^(-3))/(2) = 112` mL of ethane

Discussion

93042.	100 mL of 0.01 M solution of NaOH is diluted to 1 litre. The pH of resultant solution will be
Answer» 3 12 11 8 Answer :C

Discussion

93043.	100 ml of 0.04 N HCl aqueous solution is mixed with 100 ml of 0.02 N NaOH solution. The pH of the resulting solution is
Answer» `1.0` 1.7 `2.0` 2.3 Solution :`N_(1)V_(1) = 0.04 XX 100 = 4` `N_(2)V_(2) = 0.02 xx 100 =2` `N_(1)V_(1) - N_(2)V_(2) = N_(3)V_(3)` `4-2 = N_(3) xx 200, N_(3) = 10^(-2) M` `pH = LOG 10 (1)/(10^(-2)) = 2`.

Discussion

93044.	100 ml O_(2) and H_(2) kept at same temperature and pressure. What is true about their number of molecules
Answer» `N_(O_(2)) gt N_(H_(2))` `N_(O_(2)) lt N_(H_(2))` `N_(O_(2))=N_(H_(2))` `N_(O_(2))+N_(H_(2))=1` mole Solution :This is Avogardo's HYPOTHESIS. According to this, equal volume of all gases contain equal no. of molecules under similar CONDITION of TEMPERATURE anf PRESSURE.

Discussion

93045.	100 ml Fe(OH)_(3) is coagulated by 10 ml 1N Na_(2)SO_(4). Then what will becoagulation value of Na_(2)SO_(4)?
Answer» 5 10 50 100 Solution :100 ML `FE(OH)_3 ` is coagulated by =`10xx1/2`=5 milli MOL 1 mol `Fe(OH)_3` is coagulated by = `5/100` 1000 ml `Fe(OH)_3` is coagulated by`=(1000xx5)/100`=50

Discussion

93046.	100 ml each of 1M AgNO_3 and 1M NaCl are mixed. The nitrate ion concentration in the resulting solution is
Answer» 1 m 0.5 M 0.75 M 0.25 M ANSWER :B

Discussion

93047.	100 ml each of 0.5 NaOH , N/5 HCl and N/10 H_2SO_4 are mixed together . The resulting solution will be :
Answer» Acidic Neutral Alkaline None of these Answer :C

Discussion

93048.	100 ml 0.5 M H_(2)SO_(4) (strong Acid) is neutralised by 200 ml 0.2 M NH_(4)OH . In a constant pressure calorimeter which results in temperature rise of 1.4 ^(@) C. If heat capacity of calorimeter constant is 1.5 kJ// ^(@) C . Which statement is /are correct . Given HCl + NaOH to NaCl + H_(2)O + 57 kJ CH_(3) COOH + NH_(4) OH to CH_(3) COO NH_(4) + H_(2) O + 48.1 kJ Ethalpy of dissociation of CH_(3)COOH is 4.6 kJ/mol
Answer» Solution :`57 - (x +y) = 48.1` `IMPLIES x + y = 8.9` `implies 4.5 + y = 8.9 implies y = 4.4` `implies ` ENTHALPY of dissociation of `CH_(3) COOH` `=4.4 ` kJ/mol

Discussion

93049.	100 ml 0.5 M H_(2)SO_(4) (strong Acid) is neutralised by 200 ml 0.2 M NH_(4)OH . In a constant pressure calorimeter which results in temperature rise of 1.4 ^(@) C. If heat capacity of calorimeter constant is 1.5 kJ// ^(@) C . Which statement is /are correct . Given HCl + NaOH to NaCl + H_(2)O + 57 kJ CH_(3) COOH + NH_(4) OH to CH_(3) COO NH_(4) + H_(2) O + 48.1 kJ DeltaH for 2H_(2)O (l) to 2 H^(+) (aq) + 2 OH^(-) (aq) is 114 kJ
Answer» Solution :As we know `H^(+) + OH^(-) to H_(2) O "" Delta H = -57` `IMPLIES 2 H^(+) + 2OH^(-) to 2H_(2)O Delta H = -57 XX 2` `implies 2 H_(2) O to 2H^(+) + 2 OH^(-) Delta H = 114 kJ`

Discussion

93050.	100 ml 0.25 M H_(2)SO_(4) (strong acid) is neutralised with 200 ml 0.2M NH_(4)OH in a constant pressure Calorimeter which results in temperature rise of 1.4^(@)C. If heat capacity of Calorimeter constent is 1.5 kJ//^(@)C. Which statement is/are correct {:("Given :",HCl+NaOH rarr NaCl+H_(2)O+57 kJ),(,CH_(3)COOH+NH_(4)OH rarr CH_(3)COONH_(4)+H_(2)O+48.1 kJ):}
Answer» Enthalpy of neutralisation of `HCl` v/s `NH_(4)OH` is `-52.5` kJ/mol Enthalpy of dissociation (ionization) of `NH_(4)OH` is `4.5` kJ/mol Enthalpy of dissociation of `CH_(3)COOH` is `4.6` kJ/mol `DeltaH` for `2H_(2)O(l) rarr 2H^(+) (AQ.)+2OH^(-) (aq.)` is `114` kJ Answer :A::B::D

Discussion

Explore topic-wise InterviewSolutions in .

100 mll of 0.01 M metal - chloride completely hydrolysed in excess water , it has been observed that resulting solution required 30 ml of 0.1 M aqueous silver nitrate solution for the complete precipitation of the chloride ion, then expected metal is :

100 mL of tap water containing Ca(HCO_(3))_(2) was titrated with N//50HCl with methyl orange as indicator. If 30 mL of HCl were required, calculate the temporary hardness as parts of CaCO_(3) per 10^(6) parts of water.

100 mL of tap water containing Ca(HCO_(3)_(2) was titrated with (N)/(50)HCL with methyl orange as indicaor. If 30 mL of HCL was required, the temporary hardness of water as parts of CaCO_(3) per 10^(6) parts of water was

100 ml of PH_3 when decomposed produces phosphorus and hydrogen . The change in volume is :

100 ml of PH_(3) on heating forms P and H_(2). The volume change in the reaction is

100 mL of ( M / 1 0) aqueous solution of a monoprotic acid is titrated with a solution of NaOH. When one-third of the acid is neutralised, pH of the solution becomes 3.9. pH of the solution at half-neutralisation point is-

100 mL of O_(2) and H_(2) are kept at same temperature and pressure. What is true about their numberof molecules ?

100 mL of HCl gas at 25^@ C and 740 mm pressure was dissolved in 1 L of water. What will be the pH of solution. (Given, vapour pressure of H_2 O at 25^@ C is 23.7 mm.)

100 ml of HCl + 35 ml of NaOH, colour methyl orange in the solution will be

100 ml of ethyl alcohol [d = 0.92 g/ml] and 900 ml of water [d = 1 g/ml] are mixed to form 1 lit solution. The Molarity and molality of the resulting solution are

100 mL of an O_(2)-O_(3) mixture was passed through turpentine, and 20 mL was reduced. If 100 mL of such a mixture is heated, the increase in volume will be

100 mL of a saturated solution of M""_(2)SO""_(4) is giving 24 disintegrations per hour due to radio active metal "M" (lamda=2xx10""^(-17)hour""^(-1),N""_(0)=6xx10""^(23)a). If the solubility product of the salt is "x"xx10""^(-3y). What is (x+y)?

100 ml of an aqueous solution contains 6.023xx10^(21) solute molecules. The solution is diluted to 1 lit. The number of solute moleculespresent in 10ml of the dilute solution is

100 mL of an aqueous solution of protein contais 6.3 g of protein. Calculate the molar mass of protein if osmatic pressure of the solution at 27^@C is 2.57 xx 10^(-3) bar.

100 ml of an aqueous solution contains 6.023 xx 10^21 solute molecules. The solution is diluted to 1 lit. The number of solute molecules present in 10ml of the dilute solution is

100 ml of an acid solution is neutralized by 50 ml of NaOH solution containing 0.2 g NaOH. The concentration of acid solution is

100 ml of a sample of hard water required 15.1 ml 0.02N H_2 SO_4 for complete reaction. The hardness of water in ppm will be:

100 ml of a gaseous mixture containing CO and CH_(4) shows a volume contraction of 65 ml on combustion in excess of O_(2). Then what will be the volume of CO (g).

100 ml of a buffer of 1M NH_3 and 1M NH_4^(+) are placed in two voltaic cells separately. A current of 1.5A Is passed through both ceils for 20 minutes. If electrolysis of water only takes place (i) 2H_(2)O +O_(2)+4e^(-) to 4OH^(-) (RHS) (ii) 2H_(2)O to 4H^(+) +O_(2(g))+4e^(-) (LHS) then pH of hte

100 ml of 1M HCI, 200 ml of 2M HCl and 300 ml of 3M HCl are mixed with enough water to get 1M solution. The volume of water to be added is (in ml)

100 ml of 1M HCl, 200 ml 2 M HCl and 300 ml 3M HCl are mixed. The Molarity of the resulting solution is

100 mL of 1 M H_(2)SO_(4) are mixed with 200 mL of 8 M HCl solution. The noramlity of the resulting solution is

100 mL of 1 N NH_4OH (K_b = 5 xx 10^-5) is neutralised to equivalence point by 1 NHCl. Calculate the pH of solution at equivalence point.

100 mL of 1 M HCl is mixed with 50 ml of 2M HCl Hence , [H_3O^+] is :

100 ml of 1 M H_(2) SO_(4) solution (d_("solution") = 1.5 gm//ml) is mixed with 400 ml of water (d_("water") = 1 gm //ml) then molarity of final solution (d_("final solution") = 1.25 gm//ml) is -

100 mL of 0.5 N NaOH solution is added to 10 mL of 3 N H_(2)SO_(4) solution and 20 mL of 1 N HCl solution . The mixture is

100 ml of 0.2 M H_(2)SO_(4) is added to 100 ml of 0.2 M NaOH. The resulting solution will be

100 mL of 0.1N I_2 oxidizes Na_2S_2O_3 in 50 ml solution to Na_2S_4O_6.The normality of this hypo solution against KMnO_4(Which oxidezes it to Na_2SO_4) would be :

100 ml of 0.15 N H_(2)O_(2) is completelyoxidized by

100 ml of 0.1N FeSO_4 solution will be completely oxidised by 'x' gms of K_2Cr_2O_7 in acidic medium (Mol.wt = 294). The value of 'x' is

100 ml of 0.1 N NaOH is mixed with 50 ml of 0.1 N H_2SO_4 . The pH of the resulting solution is

100 ml of 0.1 N NaOH are mixed with 100 ml of 0.2 N HCl solution and the whole solution is made upto 1 litre. The pH of the resulting solution is :

100mL of 0.1 M solution of solute A are mixed with 200 mL of 0.1 M solution of solute B. If A and B are non-reacting substances, the molarity of the final solution will be

100 ml of 0.1 M solutio of AB (d = 1.5 gm/m) is mixed with 100ml of 0.2 M solution of CB_(2) (d = 2.5 gm/m). Calculate the molarity of B^(-) in final solution if the density of final solution is 4gm/m. Assuming AB and CB_(2) are non reacting & dissociates completely into A^(+), B^(-), C^(2+)

100 ml of 0.1 M solutionA is mixed with 20ml of 0.2 M solution B. The final molarity of the solution is :

100 mL of 0.1 M acetic acid is completely neutralized using a standard solution of NaOH. The volume of ethane obtained at STP after the complete electrolysis of the resulting solution is

100 mL of 0.1 M acetic acid is completely neutralized using a standard solution of NaOH . The volume of ethane obtained at STP after the complete electrolysis of the resulting solution is

100 mL of 0.01 M solution of NaOH is diluted to 1 litre. The pH of resultant solution will be

100 ml of 0.04 N HCl aqueous solution is mixed with 100 ml of 0.02 N NaOH solution. The pH of the resulting solution is

100 ml O_(2) and H_(2) kept at same temperature and pressure. What is true about their number of molecules

100 ml Fe(OH)_​(3) is coagulated by 10 ml 1N Na_​(2)​SO​_(4)​. Then what will becoagulation value of Na_(​2)​SO​_(4)​?

100 ml each of 1M AgNO_3 and 1M NaCl are mixed. The nitrate ion concentration in the resulting solution is

100 ml each of 0.5 NaOH , N/5 HCl and N/10 H_2SO_4 are mixed together . The resulting solution will be :

100 ml Fe(OH)_(3) is coagulated by 10 ml 1N Na_(2)SO_(4). Then what will becoagulation value of Na_(2)SO_(4)?