94420 + Interview Questions in Chemistry in Class 12

1.	Which of the following is an anhydride of HClO_(4)?
Answer» `Cl_(2)O` `ClO_(2)` `Cl_(2)O_(7)` `Cl_(2)O_(6)` ANSWER :C

Discussion

2.	The anhydride of HClO_4 is-
Answer» `ClO_4^(-)` `Cl_2O_7` `ClO_2` `ClO_3` ANSWER :B

Discussion

3.	The anhydride of carbonic acid H_2CO_3 is :
Answer» `C_2O_2` `CO_2` `CO` `Na_2CO_3` ANSWER :B

Discussion

4.	The angular wave function depends upon quantum numbers.
Answer» N and L l and m n,l and m m and s Answer :B

Discussion

5.	The angular momentum of electron of H atom is proportional to:
Answer» `R^2` 1/r `SQRT r` `1/sqrt r` ANSWER :C

Discussion

6.	The angular momentum of electron in nth orbit is given by:
Answer» NH `H/(2pin)` `nh/(2PI)` `n^2h/(2pi)` ANSWER :C

Discussion

7.	The angular momentum of an electron in a particular orbit of Li^(2+) ion is 5.2728xx10^(-34) kg m^(2)//"sec". Calculate the frequency of the spectral line when electron falls from this level to the level where angular momentum of electron will be 3.1636xx10^(-34)kgm^(2)//"sec".
Answer» ANSWER :`2.1058xx10^(15)HZ`

Discussion

8.	The angular momentum of an electron in a particular orbit of Li^(2+) ion is 5.2728 xx 10^(-34) kg m^(2) // sec. Calculate the frequency of the spectral line when electron falls from this level to the level where angular momentum of electron will be 3.1636xx 10^(-34) kg-m^(2)// sec.
Answer» Solution :Angular momentum of the electron `= 5.2728 xx 10^(-34) kgm^(2) //` sec `( n_(1) h )/( 2pi ) = 5.272 xx 10^(-34)` `n_(1) = 4.99 ~~ 5 ` Angular MOMEMTUM of electron after transition ` =3.1636 xx 10^(-34) kgm^(2) //` sec `(n_(1) h )/( 2pi ) = 3.1636 xx 10^(-34)` `n_(2) = 2.999 ~~ 3 ` `:.` the electron makes transition from `. 5 RARR 3 ` `v = C bar( v ) = c xx R_(h ) xx Z^(2) [ ( 1)/( n_(1)^(2) ) - ( 1)/( n _(2)^(2)) ]` `v= 3 xx 10^(10) xx 109678xx 3^(2) xx [ ( 1)/( 3^(2)) - ( 1)/( 5^(2))] = 2.1058 xx 10^(15) Hz`

Discussion

9.	The angular momentum of an electron in 2p orbitals is
Answer» `(2H)/pi` `H/(sqrt2pi)` `h/(2PI)` None Answer :B

Discussion

10.	The angle strain in cyclopropane is :
Answer» `0^(@) 44'` `24^(@) 44'` `9^(@) 44'` `5^(@) 16'` SOLUTION :`(1)/(2)(019^(@)18-60^(@))=24^(@)44'`

Discussion

11.	The angle strain in cyclobutanne is
Answer» `19^(@)22` `9^(@)44` `24^(@)44` `29^(@)16` Solution :ACCORDING to Baeyer 's STRAIN theory , the amount of the strain is directly proportional to the angle through which a VALENCY bond has deviated from its normal position . i.e., Amount of deviation (d) `=1//2(109^(@)28'-`valency angle) In cyclobutance (d) `=1//2(109^(@)28'-90^(@))=9^(@)44`.

Discussion

12.	The angle strain in cyclobutane is
Answer» `19^@22'` `9^@44'` `24^@44'` `29^@16'` Solution :ACCORDING to Baeyer's strain theory, the amount of the strain is directly PROPORTIONAL to the angle through which a valency bond has deviated from its normal position . i.e., Amount of deviation (d)=1/2 (`109^@28'` -valency angle ) in cyclobutane (d)=1/2 `(109^@28'-90^@)=9^@44'`

Discussion

13.	The angle corresponding to maximum diffra ction of x-rays on solid crystal is determined by electrometre reading in
Answer» TRIANGLE RECTANGLE Tetrahedron parallelogram ANSWER :D

Discussion

14.	The analysis of an organic compound gave the following data: (a) 0.4020g gave 0.6098g and 0.2080g of CO_(2) and H_(2)O respectively. (b) 1.01g by Kjeldahl method produced amonia whise was neutralised by 23.2mL of N/2HCl (c ) 0.1033g of the compound gave 0.2772g of BaSO_(4) (d) 0.1015g when vaporised in Victor Meyer's apparatus displaced 27.96mL of air at 15^(@)C and 766 mm pressure Calculate the molecular formula. (Aqueous tension at 15^(@)C=16mm)
Answer» Solution :Moles of `C=1 xx` moles of `CO_(2)=1 xx (0.6098)/(44)= 0.0138` Moles of `H= 2xx` moles of `H_(2)O = 2 xx (0.2080)/(18)= 0.0231` Calculation of moles of N: m.e. of `HCl= 23.2 xx (1)/(2)=11.6` `therefore` m.e. of `NH_(3)=11.6` m mol of `NH_(3)=11.6` Moles of `NH_(3)= (11.6)/(1000)= 0.0116` `therefore` moles of `N =1 xx` moles of `NH_(3)=0.0116` Moles of N in 0.4020g of compound `=(0.0116)/(1.01)xx 0.4020` =0.0046 Moles of `S=1 xx` moles of `BaSO_(4)= (0.2772)/(233.3)= 0.00119 (BaSO_(4)=233.3)` Moles of S in 0.4020g of compound `=(0.00119)/(0.1033) xx 0.4020` = 0.0046 `therefore` mole of `C: H: N: S= 0.0138: 0.0138: 0.0046: 0.0046` `=3:5:1:1` Hence the empirical formula is `C_(3)H_(5)NS` Calculation of molecular weight: PRESSURE due to dry AIR only `=766-16=750mm` Volume of vapour of 0.1015g of compound at NTP `=(750 xx 27.96 xx 273)/(288 xx 760)=26.155mL` `therefore` moles of vapour `=(26.155)/(22400)=0.00116` Molecular weight `=("weight in grams")/("no. of moles") = (0.1015)/(0.00116)=86.97 ~~87` Since empirical formula weight is also equal to 87, the molecular formula is `C_(3)H_(5)NS`

Discussion

15.	The angle between the bonding orbitals of a molecules AX_(3) with zero dipole moment is
Answer» `120^(º)` `109^(º)` `104^(º)` `180^(º)` ANSWER :A

Discussion

16.	The analysis of a uranium reveals that ratio of mole of .^(206)Pband .^(238) U in sample is 0.2 . If effective decay constant of process .^(238)U rarr .^(206) Pb is lambda then age of rock is
Answer» `(1)/(lambda) In (5)/(4)` `(1)/(lambda) In ((5)/(1))` `(1)/(lambda) ln(4)/(1)` `(1)/(lambda) In ((6)/(5))` Solution :`t=(1)/(lambda)In[(n_(PB+n_(U))/(n_(U)))] ""...(i)` GIVEN `(n_(Pb))/(n_(U))=0.2=(1)/(5) rArr (n_(Pb+n_(U))/(n_(U)))=(1)/(5)+1=((6)/(5))"" ...(II)` SUBSTITUTING (ii)in(i) `t=(1)/(lambda) In ((6)/(5))`

Discussion

17.	The anaesthetic which is administrated by injection is-
Answer» diethy ether divinyl ether liquid NITROUS oxide MORPHINE Solution :morphine

Discussion

18.	The amountphosphoric acid, (H_(3)PO_(4)) required to prepare 550 mL of 0.40 N solution (i) assuming complete neutralization (ii) assumingreduction to HPO_(3)^(2-)are respectively :
Answer» `7.19g , 10.89 g` `10.8 g, 7.19 g` `7.19 g, 7.19 g` `10.8 g , 10.8 g` SOLUTION :(i) `H_(3)PO_(4) + 3NaOH to Na_(3)PO_(4) + 3H_(2)O , . E = M/3` Meq . of `H_(3)PO_(4) = 550 xx 0.4` `W/(98//3) xx 1000= 550 xx 0.4 rArrW = 7.19 ` g (ii) ` P^(5+) + 2e to P^(3+) "" (. :. E = M/2) ` Meqof `H_(3)PO_(4) = 550xx 0.4` ` W/(98//2) xx 1000= 550xx 0.4 rArrW = 10.8 g`

Discussion

19.	The amount of urea to be dissolved in 500 ml of water (K = 18.6 K "mole"^(-1) in 100 g solvent) to produce a depression of 0.186^(@)C in freezing point is
Answer» 9 G 6 g 3 g 0.3 g Solution :`Delta T_(f)=(100xx K xx W)/(m xx W) therefore 0.186 = (100xx18.6xx w)/(60xx500)` w = 3 g

Discussion

20.	The amount of urea to be dissolved in 500mL of water (K=18.6 K mol^-1 100 g solvent) to produce a depression of 0.186^@C in freezing point is:
Answer» 0.3 g 3 g 6 g 9 g Answer :B

Discussion

21.	The amount of substance deposited by the passage of 1 amp of current for 1 second is equal to
Answer» EQUIVALENT mass Molecular mass Electrochemical equivalent Specific equivalent Solution :W=Zit(Z is the ECE) when i=1 AMP, t=1 SEC, W=Z

Discussion

22.	The amount of solute (molar mass 60 g mol^(-1)) that must be added to 180 g of water so that the vapour pressure of water is lowered by 10% is
Answer» 30 g 60 g 120 g 12 g Solution :RELATIVE LOWERING of vapour pressure is given by the formula : `(p^(@)-p_(s))/(p^(@))=(omega_(A))/(m_(A))xx(m_(B))/(omega_(B))` Given, relative lowering of vapour pressure `=(p^(@)-p_(s))/(p^(@))=(10)/(100)` `m_(A)=60, m_(B)=18, omega_(B)=180, omega_(A)=x` `THEREFORE (10)/(100)=(x//60)/(180//18)rArr(1)/(10)=(x//60)/(10)rArr x = 60` Thus, 60 g of the solute must be added to 180 g of water so that the vapour pressure of water is lowered by 10%.

Discussion

23.	The amount of substance which gives 3.7xx10^(7) dps is :
Answer» ONE BECQUEREL One Curie One millicurie One RUTHERFORD. ANSWER :C

Discussion

24.	The amount of solute required to prepare 10 L of a decimolar solution is :
Answer» 0.01 mol 0.2 mol 0.1 mol 1 mol Solution :Molarity `=("Moles of solute")/("Volume")xx1000` Moles of solute `=("Molarity" XX "Volume")/(1000)=(0.1xx10000)/(1000)=1`

Discussion

25.	The amount of solute ( molar mass 60 g "mol"^(-1) ) that must be added to 180 g of water so that the vapour pressure of water is lowered by 10% is
Answer» <P>30 g 60 g 120 g 12 g Solution :`(p_(A)^(@)-p_(A))/(p_(A)^(@))=(w_(B)M_(A))/(w_(A)M_(B))` `:.(10)/(100)=(w_(B)xx18)/(180xx60)` `:.w_(B)=(10)/(100)xx(180xx60)/(18)=60G`

Discussion

26.	The amount of solute (molar mass "60 g mol"^(-1)) that must be added to 180 g of water so that the vapour pressure of water is lowered by 10% is
Answer» <P>30 g 60 g 120 g 12 g Solution :If `p^(@)="100 mm, "p_(s)=100-10="90 mm"` `(p^(@)-p_(s))/(p^(@))=(w_(2)//M_(2))/(w_(1)//M_(1))=(w_(2))/(M_(2))xx(M_(1))/(w_(1))` `(100-90)/(100)=(w_(2))/(60)xx(18)/(180)` `"or"w_(2)=(10)/(100)xx60xx10=60g.`

Discussion

27.	The amount of sodium hydrogen carbonate, NaHCO_(3), in an antacid tablet is to be determinedby dissolving the tablet in water and titrating the resulting solution with hydrochloric acid. Which indicator is the most appropriate for this titration ? {:("Acid", ""K_(a)),(H_(2)CO_(3),2.5xx10^(-4)),(HCO^(-3),2.5xx10^(-8)):}
Answer» METHYL ORANGE, `pK_(In)=3.7` bromothymol blue, `pK_(In)=7.0` phenolphthalein,`pK_(In)=9.3` ALIZARIN yellow `pK_(In)=12.5` Answer :a

Discussion

28.	The amount of sodium deposited by 5 ampere current for 10 minute from fused NaCl is
Answer» `0.517` G `71.5` g `5.17` g `0.715` g Answer :D

Discussion

29.	The amount of silver deposited on passing 2 faraday of charge though an aqueous solution of AgNO_3 is :
Answer» 54 g 108 g 216 g 324 g Answer :C

Discussion

30.	The amount of sodium deposited by 5 ampere current for 10 minute from fused NaCI is :
Answer» `0.715` G `71.5 g` `5.17g` `0.517 g` ANSWER :B

Discussion

31.	The amountof silver deposited on passing 2 F of electricty throughtaqueous solutionof AgNO_(3) is
Answer» 54 G 108 g 216 g 324 g Solution :Accrodingto faraday 's LAW m=`z xx Q=(ExxQ)/(96500)` `=(108)/(96500)xx2xx986500=216 g`

Discussion

32.	The amount of silver (at. mass=108) deposited from a solution of silver nitrate, when a current of 9650 coulombs was passed is:
Answer» 10.8gm 0.108gm 1.08gm `1.08 XX 10^3` gm Answer :A

Discussion

33.	The amount of lime, Ca(OH)_(2) required to remove the hardness in 60 L of pond water containing 1.62 mg of calcium bicarbonate per 100 ml of water, will be :
Answer» `4.44g` `0.222g` `2.22 G` `0.444 g` ANSWER :D

Discussion

34.	The amount of KMnO_4 required to prepare 100 ml of 0.1 N solution in alkaline medium
Answer» 0.31 G `0.52g` `3.16` `1.58` ANSWER :D

Discussion

35.	The amount of ice that will separate out from a solution containing 25 g of ethylene glycol in 100 g of water that is cooled to -10^(@)C, will be [Given : K_(f) "for" H_(2)O = 1.86K mol^(-1) kg]
Answer» 50.0 G 25.0 g 12.5 g 30.0 g ANSWER :B

Discussion

36.	The amount of ice that will separate out on cooling the solution containing 50g of ethylene glycol (CH_2OH)_2 in 200 g of water at -9.3^@C is
Answer» 276 g 188 g 138 g 38.71 g Answer :D

Discussion

37.	The amount of ice that will separate out on cooling a solution containing 50 g of ethylene glycol in 200 g water to -9.3^@C is: (K_f = 1.86 K molality^-1)
Answer» 38.71 g 37.71 mg 42 g 42 mg Answer :A

Discussion

38.	The amount of hydrazine (N_(2)H_(4)) oxidized to N_(2)by 19.4 gK_(2)CrO_(4) which itself reduces to Cr(OH)_(4)^(-) is :
Answer» `2.4 g` `2.8 g` `3.0` g `2.0 g` Solution :`N_(2)^(4-) to OVERSET(o) N_(2) + 4e^(-)` ` Cr^(6+) + 3eto Cr^(3+)` Eq. of `K_(2)CrO_(4) = " Eq. of " N_(2)H_(4)` ` :.(19.4)/(194//3) = W/(32//4) rArrW = 2.4g`

Discussion

39.	The amount of heat required to raise the temperature of 75 kg of ice at 0^oC to water at 10^oC is (latent heat of fusion of ice is 80 cal/g, specific heat of water is 1 cal/g^oC)
Answer» 750 KCAL `6.75xx10^3kcal` `60xx10^3kcal` 6.75 kcal Answer :B

Discussion

40.	The amount of heat measured for a reaction in a bomb calorimeter is
Answer» `DELTAG` `DeltaH` `DELTAE` `PDeltaV` Solution :The BOMB calorimeter is used to measure `DeltaE`.

Discussion

41.	The amount of heat evolved when 500cm^(3) of 0.1M HCl is mixed with 200cm^(3) of 0.2 M NaOH is
Answer» 2.292 kJ 1.292 kJ 0.292 kJ 3.392 kJ Solution :Mili-equivalent of HCL = `500xx0.1` =50 milli-moles = 0.05 moles Milli-equivalent of NaOH= `200xx0.2` = 40 milli-moles = 0.04 moles When 1 MOLE of `H^(+)` ions and 1 mole of `OH^(-)` ions are neutralized, 1 mole of water is formed and 57.1 kJ of energy is released. 0.04 moles of `H^(+)` ions react with 0.04 moles of `OH^(-)` ions to form 0.04 moles of water molecules. Heat EVOLVED = `0.04xx57.1=2.29 kJ`.

Discussion

42.	The amount of heat evolved when one mole of H_2SO_4 reacts with two mole of NaOH is:
Answer» 57.3 KJ 2x57.3 KJ 57.3/2 KJ None Answer :C

Discussion

43.	The amount of heat evolved when 500 cm^(3) of 0.1 M HCl is mixed with 200 cm^(3) of 0.2 M NaOH is ……….
Answer» 2.292 kJ 1.292 kJ 0.292 kJ 3.392 kJ Solution :`underset(underset(0.1 M)(500 cm^(3)))(HCl) + underset(underset(0.2 M)(200 cm^(3)))NaOH rarr NACL + H_(2)O` Moles of HCl `= (0.1)/(1000) XX 500 = 0.05 moles` Moles of NaOH `= (0.2)/(1000) xx 200 = 0.04 moles` Moles of `H_(2)O` formed = 0.04 moles. Heat released when 1 MOL `H_(2)O` is formed = 57.3 kJ Heat released when 0.04 mol `H_(2)O` is fromed `= 0.04 xx 57.3 = 2.292 kJ`

Discussion

44.	The amount of Glucose to be dissolved in 500g of water so as to produce the same lowering in vapour pressure as that of 0.2 molal aqueous urea solution
Answer» 9 G 18 g 36 g 1.8 g ANSWER :B

Discussion

45.	The amount of gas adsorbed physically on charcoal increases with :
Answer» TEMPERATURE and pressure Temperature and DECREASES with pressure Pressure and decreases with temperature None Answer :C

Discussion

46.	The amount of energy required to remove the most loosely bound electron from an isolated gaseous atom is called as first ionization energy (IE_1).Similarly the amount of energies required to knock out second third etc. electrons from the isolated gaseous cation are called successive ionization energies and IE_3gtIE_2gtIE_1 (i)Nuclear charge (ii)Atomic size (iii)penetration effect of the electrons (iv)shielding effect of the inner electrons and (v)electronic configurations (exactly half filled & completely filled configurations are considered extra stable) affect the ionisation energies. On the other hand, the amount of energy released when a neutral isolated gaseous atom accepts an extra electron to form gaseous anion is called electron affinity. O(g)+e^(-) overset("Exothermic")toO^(-)(g),DeltaH_(eg)=-141 kJ "mol"^(-1)...(a) O^(-)(g)+e^(-)overset("Endothermic")to O^(2-)(g) , DeltaH_(eg)=+780 kJ "mol"^(-1)...(b) In (b) the energy has to be supplied for the addition of second electron due to electrostatic repulsion between an anion and extra electron (same charged species).The electron affinity of an element depends upon (i) atomic size (ii)nuclear charge & (iii)electronic configuration.In general, ionisation energy and electron affinity increases as the atomic radii decreases and nuclear charge increases across a period.In general in a group, ionisation energy and electron affinity decrease as the atomic size increases. The members of third period have some higher (e.g. S and Cl) electron affinity values than the members of second period (e.g. O and F) because second period elements have very small atomic size.Hence there is a tendency of electron - electron repulsion, which results in less evolution of energy in the formation of corresponding anion. Which one of the following statements is incorrect in relation to ionisation enthalpy ?
Answer» Ionization enthalpy increases for each successive valence shell electron The GREATEST INCREASE in ionization enthalpy is EXPERIENCED on removal of electron from core of noble gas configuration End of valence ELECTRONS is marked by a big jump in ionization enthalpy Removal of electron from orbitals bearing lower n value is easier than from orbital having HIGHER n value Solution :Orbitals bearing lower value of n will be more closer to the nucleus and thus electrons will experience greater attraction from nucleus and so its removal will be difficult not easier.

Discussion

47.	The amount of energy required to remove the most loosely bound electron from an isolated gaseous atom is called as first ionization energy (IE_1).Similarly the amount of energies required to knock out second third etc. electrons from the isolated gaseous cation are called successive ionization energies and IE_3gtIE_2gtIE_1 (i)Nuclear charge (ii)Atomic size (iii)penetration effect of the electrons (iv)shielding effect of the inner electrons and (v)electronic configurations (exactly half filled & completely filled configurations are considered extra stable) affect the ionisation energies. On the other hand, the amount of energy released when a neutral isolated gaseous atom accepts an extra electron to form gaseous anion is called electron affinity. O(g)+e^(-) overset("Exothermic")toO^(-)(g),DeltaH_(eg)=-141 kJ "mol"^(-1)...(a) O^(-)(g)+e^(-)overset("Endothermic")to O^(2-)(g) , DeltaH_(eg)=+780 kJ "mol"^(-1)...(b) In (b) the energy has to be supplied for the addition of second electron due to electrostatic repulsion between an anion and extra electron (same charged species).The electron affinity of an element depends upon (i) atomic size (ii)nuclear charge & (iii)electronic configuration.In general, ionisation energy and electron affinity increases as the atomic radii decreases and nuclear charge increases across a period.In general in a group, ionisation energy and electron affinity decrease as the atomic size increases. The members of third period have some higher (e.g. S and Cl) electron affinity values than the members of second period (e.g. O and F) because second period elements have very small atomic size.Hence there is a tendency of electron - electron repulsion, which results in less evolution of energy in the formation of corresponding anion. Which one of the following statements is correct ?
Answer» The elements LIKE F, CL , Br etc having HIGH values of electron affinity act as strong oxidising agent The element having low values of ionisation energies act as strong reducing agent The formation of `Be^(-)`(g) from Be (g) is an endothermic process All of these Solution :All statements are correct.

Discussion

48.	The amount of energy, which is required to separate the nucleons from a nucleus is called
Answer» BINDING ENERGY Lattice energy Kinetic energy None of these Solution :DEFINITION of binding energy

Discussion

49.	The amount of energy required to remove the electron from a Li^(2+) ion in its ground state is how many times greater than the amount of energy required to remove the electron from an H atom in its ground state:
Answer» 9 2 3 5 Answer :A

Discussion

50.	The amount of energy required to remove an electron from the last orbit of an isolated (free) atom in gaseous state is known as ionization potential or first ionization potential of the element. Similarly, the energy required for the removal of the electron from the unipositive ion (produced above) is referred to as second ionization potential and thus the third, fourth etc. lonization potentials may be defined in general as the number of ionization potentials of an element may be as much as its number of electrons. Which of the following statemetn is correct ?
Answer» IE of elemnts increase along the period IE of the `3^(rd)` GROUP elements is less than that of the SECOND group elements. IE of group 16 element is less than that of group 15 elements. All of the above Answer :D

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Which of the following is an anhydride of HClO_(4)?

The anhydride of HClO_4 is-

The anhydride of carbonic acid H_2CO_3 is :

The angular wave function depends upon quantum numbers.

The angular momentum of electron of H atom is proportional to:

The angular momentum of electron in nth orbit is given by:

The angular momentum of an electron in a particular orbit of Li^(2+) ion is 5.2728xx10^(-34) kg m^(2)//"sec". Calculate the frequency of the spectral line when electron falls from this level to the level where angular momentum of electron will be 3.1636xx10^(-34)kgm^(2)//"sec".

The angular momentum of an electron in a particular orbit of Li^(2+) ion is 5.2728 xx 10^(-34) kg m^(2) // sec. Calculate the frequency of the spectral line when electron falls from this level to the level where angular momentum of electron will be 3.1636xx 10^(-34) kg-m^(2)// sec.

The angular momentum of an electron in 2p orbitals is

The angle strain in cyclopropane is :

The angle strain in cyclobutanne is

The angle strain in cyclobutane is

The angle corresponding to maximum diffra ction of x-rays on solid crystal is determined by electrometre reading in

The angle between the bonding orbitals of a molecules AX_(3) with zero dipole moment is

The analysis of a uranium reveals that ratio of mole of .^(206)Pband .^(238) U in sample is 0.2 . If effective decay constant of process .^(238)U rarr .^(206) Pb is lambda then age of rock is

The anaesthetic which is administrated by injection is-

The amountphosphoric acid, (H_(3)PO_(4)) required to prepare 550 mL of 0.40 N solution (i) assuming complete neutralization (ii) assumingreduction to HPO_(3)^(2-)are respectively :

The amount of urea to be dissolved in 500 ml of water (K = 18.6 K "mole"^(-1) in 100 g solvent) to produce a depression of 0.186^(@)C in freezing point is

The amount of urea to be dissolved in 500mL of water (K=18.6 K mol^-1 100 g solvent) to produce a depression of 0.186^@C in freezing point is:

The amount of substance deposited by the passage of 1 amp of current for 1 second is equal to

The amount of solute (molar mass 60 g mol^(-1)) that must be added to 180 g of water so that the vapour pressure of water is lowered by 10% is

The amount of substance which gives 3.7xx10^(7) dps is :

The amount of solute required to prepare 10 L of a decimolar solution is :

The amount of solute ( molar mass 60 g "mol"^(-1) ) that must be added to 180 g of water so that the vapour pressure of water is lowered by 10% is

The amount of solute (molar mass "60 g mol"^(-1)) that must be added to 180 g of water so that the vapour pressure of water is lowered by 10% is

The amount of sodium deposited by 5 ampere current for 10 minute from fused NaCl is

The amount of silver deposited on passing 2 faraday of charge though an aqueous solution of AgNO_3 is :

The amount of sodium deposited by 5 ampere current for 10 minute from fused NaCI is :

The amountof silver deposited on passing 2 F of electricty throughtaqueous solutionof AgNO_(3) is

The amount of silver (at. mass=108) deposited from a solution of silver nitrate, when a current of 9650 coulombs was passed is:

The amount of lime, Ca(OH)_(2) required to remove the hardness in 60 L of pond water containing 1.62 mg of calcium bicarbonate per 100 ml of water, will be :

The amount of KMnO_4 required to prepare 100 ml of 0.1 N solution in alkaline medium

The amount of ice that will separate out from a solution containing 25 g of ethylene glycol in 100 g of water that is cooled to -10^(@)C, will be [Given : K_(f) "for" H_(2)O = 1.86K mol^(-1) kg]

The amount of ice that will separate out on cooling the solution containing 50g of ethylene glycol (CH_2OH)_2 in 200 g of water at -9.3^@C is

The amount of ice that will separate out on cooling a solution containing 50 g of ethylene glycol in 200 g water to -9.3^@C is: (K_f = 1.86 K molality^-1)

The amount of hydrazine (N_(2)H_(4)) oxidized to N_(2)by 19.4 gK_(2)CrO_(4) which itself reduces to Cr(OH)_(4)^(-) is :

The amount of heat required to raise the temperature of 75 kg of ice at 0^oC to water at 10^oC is (latent heat of fusion of ice is 80 cal/g, specific heat of water is 1 cal/g^oC)

The amount of heat measured for a reaction in a bomb calorimeter is

The amount of heat evolved when 500cm^(3) of 0.1M HCl is mixed with 200cm^(3) of 0.2 M NaOH is

The amount of heat evolved when one mole of H_2SO_4 reacts with two mole of NaOH is:

The amount of heat evolved when 500 cm^(3) of 0.1 M HCl is mixed with 200 cm^(3) of 0.2 M NaOH is ……….

The amount of Glucose to be dissolved in 500g of water so as to produce the same lowering in vapour pressure as that of 0.2 molal aqueous urea solution

The amount of gas adsorbed physically on charcoal increases with :

The amount of energy, which is required to separate the nucleons from a nucleus is called

The amount of energy required to remove the electron from a Li^(2+) ion in its ground state is how many times greater than the amount of energy required to remove the electron from an H atom in its ground state: