94420 + Interview Questions in Chemistry in Class 12

1.	If 50 calorie are added to a system and system does work of 30 calorie on surroundings, the change in internal energy of system is:
Answer» 20 cal 50 cal 40 cal 30 cal Answer :A

Discussion

2.	if 50 atoms of carbon reacts with 200 molecules of oxygen, find out the correct statement(s)
Answer» `O_2` is the limiting reagent `50` MOLECULES of` CO_2` are formed `100` molecules of` CO_2` are formed Both (1) & (2) ANSWER :B

Discussion

3.	If 50 calorie are added to a system and system does wark of 30 calorie on surroundings, the change in internal energy of system is
Answer» 30 cal 50 cal 40 cal 20 cal Answer :D

Discussion

4.	If 5% sugar solution is isotomic with 1% solution of unknown substance, then find out molecular mass of unknown substance.
Answer» ANSWER :68.4 gm/mol

Discussion

5.	If 5 liters of H_(2)O_(2) produce 50 liters of O_(2) at NTP , H_(2) O_(2) is
Answer» `'50V'` `'10V'` `'5V'` `'250 V'` ANSWER :B

Discussion

6.	If 4g of O_(2) diffuse through a very narrow hole, how much H_(2) would have diffused under identical conditions ?
Answer» 1g `1//4g` `16G` 64g Solution :`( 4g//t)/(X//t) = SQRT((2)/( 32))` ( x is the AMOUNT of `H_(2)` diffused and t is the time ) `( 4)/( x) = sqrt((1)/( 16)) = ( 1)/( 4)`

Discussion

7.	If 4.5 gm Aluminium completely reacts with 4gm of oxygen. Then what will be empirical formula of aluminium oxide.
Answer» `Al_(2)O` `Al_(2)O_(3)` `AlO_(2)` `AIO` SOLUTION :`Al+O_(2)rarrAl_(X)O_(y)` `n_(Al)=(4.5)/(27)=(1)/(6) n_(o_(2))=(4)/(52)=(1)/(8)(n_(Al))/(n_(o))=(1)/(6)xx(4)/(1)=(2)/(3)` `n_(o)=2xxn_(o_(2))=(1)/(4) "FORMULA"=Al_(2)O_(3)`

Discussion

8.	If 400 gm radioactive Iodine is taken initially. Then calculate the amount the ""(53)I^(128)(t_((1)/(2))="25 min") left after 75 minutes
Answer» ANSWER :50

Discussion

9.	If 4-methyl-2-pentene is refluxed with dilute sulphuric acid, hydration reaction takes place.On principle, how many different alcohols would be formed ?
Answer» ANSWER :5

Discussion

10.	If 4 g of NaOH dissolved in 36 g of H_(2)O, calcualte the mole fraction of each component in the solution. Also, determine the molarity of solution (specific gravity of solution is "1 g mL"^(-1))
Answer» Solution :`"4 g NAOH"=(4)/(40)"mole"="0.1 mole"` `"36 g "H_(2)O=(36)/(18)"mole = 2 mole"` `THEREFORE " Mole fraction of "H_(2)P=1-0.047=0.953(or(2)/(2+0.1)=0.95)` `"Total mass of solution "="Mass of solvent "+"Mass of solution"=36+4g=40g` `"Volume of solution"=("Mass")/("Sp. gravity")=("40 g")/("1 g mL"^(-1))=40mL=0.040L` `"MOLARITY of solution "=("Moles of solute")/("Volume of solution in L")=("0.1 mole")/("0.040 L")=2.5M`

Discussion

11.	If 36 g. of Al is deposited at the cathode, the number of moles of electrons used is
Answer» 1 2 3 4 Answer :D

Discussion

12.	If 340 g of mixture N_2 and H_2 in the correct raito gave a 20% yield of NH_3. The mass produced would be:
Answer» 16 G 17 g 20 g 68g Answer :D

Discussion

13.	If ‘I’ is the intensity of absorbed light and C is the concentration of AB for the photochemical processthe rate of formation of A + hv to AB^(**) is directly proportional to :
Answer» C I `I^2` `CI` ANSWER :B

Discussion

14.	If 32 g of O_(2)contain 6 022 xx 10^23 molecules at NTP then 32 g of S, under conditions, will contain,
Answer» `6.022 XX 10^(23)` S `3.011 xx 10^(23)` S `12.044 xx 10^(23) S` `1 xx 10^(23) S` ANSWER :A

Discussion

15.	If 30mL of H_(2) and 20mL of O_(2) react to form form water, what is left at the end of the reaction?
Answer» 10mL of `H_(2)` 5 ML of `H_(2)` 10mL of `O_(2)` 5 mL of `O_(2)` ANSWER :A

Discussion

16.	If 30mL of H_(2) and 20mL of O_(2), reacts to form H_(2)O, what is left at the end of the reaction?
Answer» 10mL of `H_(2)` 5ML of `H_(2)` 10mL of `O_(2)` 5mL of `O_(2)` SOLUTION :`H_(2) (G) +(1)/(2) O_(2)(g) to H_(2)O(g)` `{:(,"Initially", 30, 20,0),(,"After the reaction",0,(20-15),30):}`

Discussion

17.	If 3.01xx10^(20) molecules of H_(2)SO_(4) are removed from 98mg of H_(2)SO_(4). Then number of moles of H_(2)SO_(4) left are:
Answer» `0.1xx10^(-3)` mol `0.5xx10^(-3)` mol `1.66xx10^(-3)` mol `9.95xx10^(-2)` mol Solution :Removed number of moles of `H_(2)SO_(4)=(3.01xx10^(20))/(6.02xx10^(23))` `=5xx10^(-4)` `=0.5xx10^(-3)` Initial number of moles of `H_(2)SO_(4)=(0.098)/(98)=0.001=10^(-3)` REMAINING moles of `H_(2)SO_(4)=10^(-3)-0.5xx10^(-3)=0.5xx10^(-3)`

Discussion

18.	If 30g of a solute of molecular weight 154 is dissolved in 250g of benzene, what will be the boiling point of the resulting solution under atmospheric pressure . The molal boiling point elevation constant for benzene is 2.61^(@)C.m^(-1) and the boiling point of pure benzene is 80.1^(@)C
Answer» Solution :Suppose the elevation in B.p. is `DeltaT_(b)` Molality `=(DeltaT_(b))/(K_(b))=(DeltaT_(b))/(2.61)` …………(Eqn . 8) Moles of solute `=(30)/(154)` `:.` molality `=(30)/(154)xx(1000)/(250)` moles / `1000g` of solvent Thus, `(DeltaT_(b))/(2.61)=(30)/(154)xx(1000)/(250)` `DeltaT_(b)=2^(@)` Thus the b.p. of the resulting solution is `80.1+2=82.1^(@)C`

Discussion

19.	If 30% of an organism's DNA is thymine, then :
Answer» 70% is purine 20% is guanine 30% is adenine Both (B) and (C ) are correct Answer :D

Discussion

20.	If 3.01 xx 10^(28) molecules are removed from 98 mg of H_2SO_4, then number of moles of H_2SO_4 left are
Answer» `0.5 xx 10^(-3)` MOL `0.1 xx 10^(-3)` mol `9.95 xx 10^(-2)` mol `1.66 xx 10^(-3)` mol Solution :`"No. of MOLES" H_(2)SO_(4) "in" 98 mg=(98 xx 10^(-3))/(98)` `=1 xx 10^(-3)` No. of moles of `3.01 xx 10^(20)` molecules of `H_(2)SO_(4)` `=(3.01 xx 10^(20))/(6.02 xx 10^(23))=(1)/(2) xx 10^(-3)` `"Moles of" H_(2)SO_(4) "left"=(1 xx 10^(-3)-0.5 xx 10^(-3))` `=0.5 xx 10^(-3)`

Discussion

21.	If 30 ml of H_(2) and 20ml of O_(2) reacts to form water, what is left at the end of the reaction
Answer» 10 ml of `H_(2)` 5 ml of `H_(2)` 10 ml of `O_(2)` 5 ml of `O_(2)` Solution :`H_(2)+(1)/(2)O_(2)toH_(2)O` 1 mole `(1)/(2)` mole 1 mole 1 VOLUME `(1)/(2)` volume 1 ml `H_(2)` REACTS with `(1)/(2)`ml `O_(2)` 30 ml of `H_(2)` reacts with `=(1)/(2)xx30=15ml" "O_(2)` (20-15)=50ml of `O_(2)` will left at the end of the reaction.

Discussion

22.	If 30 ml of H_(2) and 20 ml of O_(2) react to form water, what is left at the end of the reaction?
Answer» `"10 ML of "H_(2)` `"5 ml of "H_(2)` `"10 ml of "O_(2)` `"5 ml of "O_(2)` Solution :`UNDERSET("2 ml")(2H_(2))+underset("1 ml")(O_(2))rarr2H_(2)O` `"2 ml of "H_(2)" REACT with 1 ml of "O_(2)." Hence, 30 ml of "H_(2)" will react with 15 ml of "O_(2)`

Discussion

23.	If 30 mL of H_(2) and 20 mL of O_(2) reacts to form water, what is left at the end of the reaction?
Answer» 10 mL of `H_(2)` 5 mL of `H_(2)` 10 mL of `O_(2)` 5 mL of `O_(2)` Solution :`UNDERSET(2vol)(2H_(2))+underset(1vol)(O_(2))to2H_(2)O` 2 mole of `H_(2)` reacts with 1 vol of `O_(2)` to give water. Therefore, 30 mL of `H_(2)` will reacts with 15 mL of `O_(2)` to give water. Here, `H_(2)` is the LIMITING REACTANT. Volume of `O_(2)` left unreacted = 20-15=5 mL

Discussion

24.	If 30% of a first order reaction is completed in 12 mins, what percentage will be completed in 65.33 mins?
Answer» Solution :Give data : Time taken for COMPLETION of 30% of the reaction = 12 mins. a = 100 , x = 30 , a - x = 70 and t = 12 min Formula : `K =(2.303)/(t)"log"(a)/(a-x)` Solution : For a first order reaction, `=(2.303)/(12)"log"(100)/(70)` `=(2.303)/(12)xx0.1.549=0.02972 "min"^(-1)` `k=2.97xx10^(-2)"min"^(-1)` If t = 65.33 minutes, x = ? `k=(2.303)/(65.33)"log"(100)/(100-x)` `0.02972 = (2.303)/(65.33)"log"(100)/(100-x)` `"log"(100)/(100-x)=(0.02972xx65.33)/(2.303)=0.8430` `(100)/(100-x)` = Antilog of 0.8430 `100 = 6.966 (100 -x)` `100 = 696.6-6.966 x` `-6.966 x = 100 - 696.6=596.6` `x=(596.6)/(6.966)=85.62%` The reaction completed in 65.33 minutes

Discussion

25.	If 30 ml of H_2and 20 ml of O_2react to formwater, what is left at the end of the reaction ?
Answer» 10 ml of `H_2` 5 ml of `H_2` 10 ml of `O_2` 5 ml of `O_2` Solution :`underset("2vol")(2H_2) + underset("1 vol")(O_2) to 3H_2O` 2 vol. of `H_2` react with 1 vol of `O_2` to give WATER ` therefore ` 30 ml of `H_2` will react with 15 ml of `O_2` Hence of `H_2` is a limiting REACTANT vol of `O_2` left unreacted = 20 - 15 = 5 ml

Discussion

26.	If 3 gm of glucose (mol. wt. 180) is dissolved in 60 gm of water at 15^(@)C. Then the osmotic pressure of this solution will be
Answer» 0.34 ATM 0.65 atm 6.57 atm 5.57 atm SOLUTION :`pi=CRT =(3xx1000)/(180xx60)xx0.0821xx288=6.56` atm.

Discussion

27.	If 3/4 quantity of a radioactive element disintegrates in two hours, its half-life would be
Answer» 1 hours 45 minutes 30 minutes 15 minutes Solution :`N = (25)/(100) N_(0)` (at t = 2 HR) Thus `t = (2.303)/(0.693) XX t_(1//2) "log"(N_(0))/(N)`

Discussion

28.	If 3//4 quantity of a radioactive element disintegrates in two hours, its half life would be
Answer» 1 hours 45m 30m 15m Answer :A

Discussion

29.	If 2L of Cl_(2) gas and 2L of ClF_(3) gas react to form 6L of a pure gaseous compound at the same conditions of temperature and pressure, what is the molecular formula of the compound formed?
Answer» `ClF` `Cl_(2)F_(2)` `Cl_(3)F_(3)` `Cl_(2)F_(5)` ANSWER :A

Discussion

30.	If 2.7 gm aluminum metal is deposited on electrodes when two different electrolytic cell having molten Cu(NO_(3))_(2) and Al(NO_(3))_(3) respectively are arranged in series, then how much copper metal is produced ? (Cu=63.5,Al=27.0" gm "mol^(-1)).
Answer» 9.525 gm 31.75 gm 63.5 gm 190.5 gm Solution :`AL^(3+)+3e^(-) to Al` So, 3 F electricity GIVES 27 gm of Al, but Al is produced only 2.7 gm So, FARADAY `=(2.7)/(27)xx3=0.3` Faraday Reaction : `Cu^(2+)+2E^(-) to Cu` So 2 F electricity gives 63.5 gm of Cu So, 0.3 F electricity gives `=(0.3xx63.5)/(2)` `=9.525gm` Cu

Discussion

31.	If 250 ml of 0.25 M NaCl solution is diluted with water to voluem of 500 ml, the new concentration of the solution is :
Answer» 0.167 M 0.125 M 0.50 M 0.0625 M Solution :`M_(1)V_(1)=M_(2)V_(2)` `0.25 M xx 250 ML M_(2)xx500ml` `:.M_(2)=(0.25xx250)/(500)=0.125M`

Discussion

32.	If 25 gram of Na_(2)SO_(4) is dissolved in 10^(3) kilogram solution, then concentration will be …….
Answer» 25 ppm 0.25 ppm 250 ppm 2.5 ppm SOLUTION :ppm `= (0.025)/(10^(3))xx10^(6)=25` ppm.

Discussion

33.	If 224 ml of a triatomic gas has a mass of 1 g at N.T.P., then the mass of one atom is :
Answer» `9.3 xx 10^(-23) G ` `2.08 xx 10^(-23) g` `5.53 xx 10^(-23) g ` `9.62 xx 10^(-23) g ` Solution :224 ml of gas weigh = 1g 22400 ml of gas weigh ` = (1)/(224) xx 22400= 100 g ` ` THEREFORE ` molecular mass of gas= 100 mass of one molecule ` = (100)/(6.02 xx 10^23)` mass of atom ` = (100)/(6.02 xx 10^23) xx 1/3 ` (triatomic gas) = `5.53 xx 10^(-23) g `

Discussion

34.	If 22.4 ml of a triatomic gas has a mass of 0.048 gat 273K and 1 atm pressure, then the mass of one atom is:
Answer» `7.9 xx 10^(-23) g ` `2.6 xx 10^(-23) g ` `2.4 xx 10^(-22) g ` `2.4 xx 10^(-23) g ` Solution :22400 ml of gas weight ` = (0.048)/(22.4) xx 22400 = 48 g ` ` therefore ` 1 MOL of gas = 48 g Mass of ONE atom `= (48)/(3 xx 6.02 xx 10^23) = 2.6 xx 10^(-23) g `

Discussion

35.	If 2.0g of a radioactive isotope has a half-life of 20 hr, the half-life of 0.5 g of the same substance is
Answer» 20 hr 80 hr 5 hr 10 hr Solution :Half-life PERIOD is a CHARACTERISTIC of RADIOACTIVE ISOTOPE which is independent of initial concentration.

Discussion

36.	If 201 persons are sitting in a row AB and if we release NO_(2) (laughing gas) from the side A and tear gas (molecular weight - 176) from side B, then which person will have a tendency to laugh and weep simultaneously?
Answer» `151^(ST)` from A `67^(TH)` from A `134^(th)` from A `134^(th)` from B Answer :C

Discussion

37.	If 200ml of 0.2M HgCl_(2) solution is added to 800ml of 0.5M Kl (100% dissociated) solution. Assuming that the following complex formation taken place to 100% extent.
Answer» Solution :`{:(HgCl_(2),+,4KI,rarr,K[HgI_(4)],+,2KCl.),(40,,400,,0,,0),(0,,400-160,,40,,80),(,,(240)/(1000),,(40)/(1000),,(80)/(1000)):}` `PI=(i_(1)C_(1)+i_(2)C_(2)+i_(3)C_(3))RT`. `=(0.24xx2+3xx0.04+0.08xx2)0.082xx300.=18.69` atm. Other METHOD : `{:(HgCl_(2),+,4KI,hArr,K[HgI_(4)],+,2KCl.),(40,,400,,0,,0),(0,,400-160,,40,,80),(,,(240)/(1000),,(40)/(1000),,(80)/(1000)):}` `pi=(i_(1)C_(1)+C_(2)+i_(3)C_(3))RT`. `=(0.24xx2+xx0.04+0.08xx2)0.082xx300.=18.69` atm.

Discussion

38.	If 200 ml of 0.1 MH_2SO_4 salution is mixed with 200 ml of 0.15 M NaOH solution then heat evolve during the neutralization process is
Answer» 411 cal 548 cal 959 cal 822 cal Answer :A

Discussion

39.	If 20.0 g of CaCO_(3) is treated with 20.0 g of HCl, how many grams of CO_(2) will be produced ?
Answer» Solution :`UNDERSET(40+12+3xx16=100g)(CaCO_(3))+underset(2(1+35.5)=73g)(2HCl)rarrCaCl_(2)+H_(2)O+underset(12+2xx16=44g.)(CO_(2))` Here, `CaCO_(3)` will be the limiting reactant.

Discussion

40.	If 20% nitrogen is present in a compound its molecular weight will be ,
Answer» 0.144 28 100 140 Answer :D

Discussion

41.	If "20.0 g mL"^(-1) of "1.0 M CaCl"_(2) and "60.0 cm"^(3) of "0.20 M CaCl"_(2) are mixed, what will be the molarity of the final solution?
Answer» SOLUTION :`M_(1)V_(1)+M_(2)V_(2)=M_(3)V_(3)" ,"1xx20 +0.2 xx60=M_(3)(20+60)or M_(3)=(32)/(80)=0.40M`

Discussion

42.	If 20.0 g of CaCO_3is treated with 200 g of HCI, how many grams of CO_2can be obtained according to the following reaction : CaCO_3(s) + 2HCl(aq)toCaCl_2(aq) + H_2O (l) + CO_2(g)
Answer» 8.80 G 27.4 g 4.20 g 13.7 g Solution :`CaCO_3 `is the limiting REAGENT 100 g of `CaCO_3` produces `CO_2 = 44g ` 20g of `CaCO_3` will GIVE `CO_2 = 44/100 xx 20` = 8.8 g

Discussion

43.	If 20% nitrogen is present in a compound, its minimum molecular weight can be:
Answer» 144 70 100 140 Answer :B

Discussion

44.	If 20 ml of0.5M Na_(2).SO_(4) is mixed with 50ml of 0.2M H_(2)SO_(4) & 30 ml of 0.4M Al_(2)(SO_(4))_(3) solution. Calculate [Na^(+)],[H^(+)].[SO_(4^(2-))],[Al^(3+)]. [Assuming 100% dissociation]
Answer» Solution :Molarity`=("moles")/("VOLUME")=10M`. Moles of `Na_(2)SO_(4)` `rArr20m`. Moles of `Na^(+)` (i) `therefore[Na^(+)]=(20)/(100)=0.2M` (ii) `[H^(+)]=?` `10m`. Moles `H_(2)SO_(4)` 20M. Moles `H^(+)` `[H^(+)]=(20)/(100)=0.2M` (iii) `[SO_(4)^(2-)]=(10+10+36)/(100)=(56)/(100)=0.56M` (iv) `[Al^(3+)]=(24)/(100)=0.24M` `rarr` Derive a relationship between molarity of a solution in which `w GM` of solute of molar MASS `M g//mol` is dissolved in ` W g` solvent `&` density of resulting solution`=d' g//ml`. say `1 L` solution taken, mass of `1 lit` solution `=(1000 d) g` moles of solute `=("molarity") x m` `therefore=(("molarity")xx1000)/(1000d-"molarity"xxM.Wt)` [Where no, of moles of solute `= "molarity")`

Discussion

45.	If 20 g. of sodium hydroxide is dissolved in 1 dm^(3) of water, the molarity of the solution will be
Answer» 0.5 0.25 0.1 0.2 Solution :`"Molarity"= "MASS"/("MOLAR mass" xx DM^(3))=20/(40 xx 6)=0.5`

Discussion

46.	If 20 ml of 0.25N strong acid and 30 ml of 0.2 N of strong base are mixed, then the resulting solution is
Answer» 0.25 N basic 0.2 N acidic 0.25 N acidic 0.2 N basic Solution :`V_(1)=20ML,N_(1)=0.25N,V_(2)=30ml,N_(2)=0.2N` `thereforeN=(N_(1)V_(1)+N_(2)V_(2))/(V_(1)+V_(2))=(20(0.25)+30(0.2))/(20+30)` `=(5+6)/(50)=(11)/(50)=0.2N` basic.

Discussion

47.	If 20 ml of methane (CH_4) is burnt using 50 ml of oxygen. The volume of the gases left after cooling to room temperature will be:
Answer» 60 ml 70 ml 30 ml 50 ml Answer :C

Discussion

48.	If 20 cc of methane in burnt using 50 cc of oxygen, the volume of the gases left after cooling to room temperature is
Answer» 60 cc 70 cc 30 cc 50 cc Answer :C

Discussion

49.	If 2 moles of MnO_(4)^(-) completely oxidizes 2.5 moles of M^(+), where the MnO_(4)^(-) reduces to Mn^(2+), then the product M^(+) is
Answer» `MO_(3)^(-)` `MO_(2)` `MO_(4)^(-)` `M_(2)O_(2)` SOLUTION :N/A

Discussion

50.	If 2 faraday current is pass through CuSO_(4) solution then find out how much copper can be deposited on cathode ? [Cu=63.5 gm "mole"^(-1)]
Answer» 0 gm 2 gm 63.5 gm 127 gm Solution :`CuSO_(4)toCu_((aq))^(2+)+SO_(4(aq))^(2-)`. . . Ionization `Cu_((aq))^(2+)+2E^(-) to Cu_((S))` . . . (Cathodic reduction) 2 mole `e^(-) to ` 1 MOL COPPER So 2F current produce 63.5 gm copper

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

If 50 calorie are added to a system and system does work of 30 calorie on surroundings, the change in internal energy of system is:

if 50 atoms of carbon reacts with 200 molecules of oxygen, find out the correct statement(s)

If 50 calorie are added to a system and system does wark of 30 calorie on surroundings, the change in internal energy of system is

If 5% sugar solution is isotomic with 1% solution of unknown substance, then find out molecular mass of unknown substance.

If 5 liters of H_(2)O_(2) produce 50 liters of O_(2) at NTP , H_(2) O_(2) is

If 4g of O_(2) diffuse through a very narrow hole, how much H_(2) would have diffused under identical conditions ?

If 4.5 gm Aluminium completely reacts with 4gm of oxygen. Then what will be empirical formula of aluminium oxide.

If 400 gm radioactive Iodine is taken initially. Then calculate the amount the ""(53)I^(128)(t_((1)/(2))="25 min") left after 75 minutes

If 4-methyl-2-pentene is refluxed with dilute sulphuric acid, hydration reaction takes place.On principle, how many different alcohols would be formed ?

If 4 g of NaOH dissolved in 36 g of H_(2)O, calcualte the mole fraction of each component in the solution. Also, determine the molarity of solution (specific gravity of solution is "1 g mL"^(-1))

If 36 g. of Al is deposited at the cathode, the number of moles of electrons used is

If 340 g of mixture N_2 and H_2 in the correct raito gave a 20% yield of NH_3. The mass produced would be:

If ‘I’ is the intensity of absorbed light and C is the concentration of AB for the photochemical processthe rate of formation of A + hv to AB^(**) is directly proportional to :

If 32 g of O_(2)contain 6 022 xx 10^23 molecules at NTP then 32 g of S, under conditions, will contain,

If 30mL of H_(2) and 20mL of O_(2) react to form form water, what is left at the end of the reaction?

If 30mL of H_(2) and 20mL of O_(2), reacts to form H_(2)O, what is left at the end of the reaction?

If 3.01xx10^(20) molecules of H_(2)SO_(4) are removed from 98mg of H_(2)SO_(4). Then number of moles of H_(2)SO_(4) left are:

If 30g of a solute of molecular weight 154 is dissolved in 250g of benzene, what will be the boiling point of the resulting solution under atmospheric pressure . The molal boiling point elevation constant for benzene is 2.61^(@)C.m^(-1) and the boiling point of pure benzene is 80.1^(@)C

If 30% of an organism's DNA is thymine, then :

If 3.01 xx 10^(28) molecules are removed from 98 mg of H_2SO_4, then number of moles of H_2SO_4 left are

If 30 ml of H_(2) and 20ml of O_(2) reacts to form water, what is left at the end of the reaction

If 30 ml of H_(2) and 20 ml of O_(2) react to form water, what is left at the end of the reaction?

If 30 mL of H_(2) and 20 mL of O_(2) reacts to form water, what is left at the end of the reaction?

If 30% of a first order reaction is completed in 12 mins, what percentage will be completed in 65.33 mins?

If 30 ml of H_2and 20 ml of O_2react to formwater, what is left at the end of the reaction ?

If 3 gm of glucose (mol. wt. 180) is dissolved in 60 gm of water at 15^(@)C. Then the osmotic pressure of this solution will be

If 3/4 quantity of a radioactive element disintegrates in two hours, its half-life would be

If 3//4 quantity of a radioactive element disintegrates in two hours, its half life would be

If 2L of Cl_(2) gas and 2L of ClF_(3) gas react to form 6L of a pure gaseous compound at the same conditions of temperature and pressure, what is the molecular formula of the compound formed?

If 2.7 gm aluminum metal is deposited on electrodes when two different electrolytic cell having molten Cu(NO_(3))_(2) and Al(NO_(3))_(3) respectively are arranged in series, then how much copper metal is produced ? (Cu=63.5,Al=27.0" gm "mol^(-1)).

If 250 ml of 0.25 M NaCl solution is diluted with water to voluem of 500 ml, the new concentration of the solution is :

If 25 gram of Na_(2)SO_(4) is dissolved in 10^(3) kilogram solution, then concentration will be …….

If 224 ml of a triatomic gas has a mass of 1 g at N.T.P., then the mass of one atom is :

If 22.4 ml of a triatomic gas has a mass of 0.048 gat 273K and 1 atm pressure, then the mass of one atom is:

If 2.0g of a radioactive isotope has a half-life of 20 hr, the half-life of 0.5 g of the same substance is

If 201 persons are sitting in a row AB and if we release NO_(2) (laughing gas) from the side A and tear gas (molecular weight - 176) from side B, then which person will have a tendency to laugh and weep simultaneously?

If 200ml of 0.2M HgCl_(2) solution is added to 800ml of 0.5M Kl (100% dissociated) solution. Assuming that the following complex formation taken place to 100% extent.

If 200 ml of 0.1 MH_2SO_4 salution is mixed with 200 ml of 0.15 M NaOH solution then heat evolve during the neutralization process is

If 20.0 g of CaCO_(3) is treated with 20.0 g of HCl, how many grams of CO_(2) will be produced ?

If 20% nitrogen is present in a compound its molecular weight will be ,

If "20.0 g mL"^(-1) of "1.0 M CaCl"_(2) and "60.0 cm"^(3) of "0.20 M CaCl"_(2) are mixed, what will be the molarity of the final solution?

If 20.0 g of CaCO_3is treated with 200 g of HCI, how many grams of CO_2can be obtained according to the following reaction : CaCO_3(s) + 2HCl(aq)toCaCl_2(aq) + H_2O (l) + CO_2(g)

If 20% nitrogen is present in a compound, its minimum molecular weight can be:

If 20 ml of0.5M Na_(2).SO_(4) is mixed with 50ml of 0.2M H_(2)SO_(4) & 30 ml of 0.4M Al_(2)(SO_(4))_(3) solution. Calculate [Na^(+)],[H^(+)].[SO_(4^(2-))],[Al^(3+)]. [Assuming 100% dissociation]

If 20 g. of sodium hydroxide is dissolved in 1 dm^(3) of water, the molarity of the solution will be

If 20 ml of 0.25N strong acid and 30 ml of 0.2 N of strong base are mixed, then the resulting solution is

If 20 ml of methane (CH_4) is burnt using 50 ml of oxygen. The volume of the gases left after cooling to room temperature will be:

If 20 cc of methane in burnt using 50 cc of oxygen, the volume of the gases left after cooling to room temperature is

If 2 moles of MnO_(4)^(-) completely oxidizes 2.5 moles of M^(+), where the MnO_(4)^(-) reduces to Mn^(2+), then the product M^(+) is

If 2 faraday current is pass through CuSO_(4) solution then find out how much copper can be deposited on cathode ? [Cu=63.5 gm "mole"^(-1)]