94420 + Interview Questions in Chemistry in Class 12 Page 569 InterviewSolution

28401.	The mass of AgCl precipitated when a solution containing 11.70 g of NaCl is added to a solution containing 3.4g of AgNO_(3), is [Atomie mass of Ag -108, Atomic mass of Na - 23]
Answer» 5.74 g 1.17 g 2.87 g 6.8 g SOLUTION :(c) : `NaCl + AgNO_(3) to AgCl+NaNO_(3)` No. of MOLES : `(11.70)/(58.5) (3.4)/(170)` = 0.2= 0.02= 0.02 Here, `AgNO_(3)` is limiting REAGENT. `:.` Mass of AgCl precipitated = `0.02 XX 143.5 = 2.87` g

Discussion

28402.	The molecular weight of benzoic acid in benzene asxc determined by depression in freezing point method corresponds to
Answer» IONIZATION ofbenzoic acid dimerization ofbenzoic acid trimerization of BENZOIC acid solvation of benzoic acid Solution :Benzoic acid FORMS a DIMER in BENZENE.

Discussion

28403.	The mass of a unit cell of CsCl corresponds to
Answer» `1 Cs^(+)` and 1 `Cl^(–)` 1 `Cs^(+)` and 6 `Cl^(–)` 4 `Cs^(+)` and 4 `Cl^(–)` 8 `Cs^(+)` and 1 `Cl^(–)` ANSWER :A

Discussion

28404.	The molecular weight of benzoic acid in benzene as determined by depressing in freezing point method corresponds to -
Answer» IONIZATION of BENZOIC acid dimerization of benzoic acid trimerization of benzoic acid solvation of benzoic acid Answer :B

Discussion

28405.	The mass of a unit cell of CsCl corresponds to-
Answer» `8Cs^(+)` and `1CL^(¯)` `1Cs^(+)` and `6Cl^(¯) `1Cs^(+)` and `1Cl^(¯)` `4CS^(+)` and `4Cl^(¯)` ANSWER :C

Discussion

28406.	The molecular weight of benzoic acid as determined by depression in freezing point method corresponds to :
Answer» IONISATION of benzoic ACID Dimerisation of benzoic acid Trimerisation of benzoic acid Solution of benzoic acid Answer :A::B::C::D

Discussion

28407.	The mass of a unit cell of CsCI corresponds to:
Answer» `8Cs^+` and `8Cl^_` `1Cs^+` and `6Cl^_` `1Cs^+` and `1CL^_` `4CS^+` and` 4Cl^_` ANSWER :C

Discussion

28408.	The molecular weight of a tribasic acid is M. What will be its equivalent weight
Answer» `(M)/(2)` `M` `(M)/(3)` `SQRT((M)/(3))` ANSWER :C

Discussion

28409.	The mass of a piece of paper is 0.02 g and the mass of a solid substance and the piece of paper is 20.036 g. If the volume of the solid is 2.16 cm^3, calculate its density to the proper number of significant digits.
Answer» `9.27 G CM^(-3)` `9.3 g cm^(-3)` `9.267 g cm^(-3)` `43.24 g cm^(-3)` Solution :`{:(20.036),(-0.02),(OVERLINE(20.016) ):}` DENSITY ` = (20.02)/(2.16) = 9.268 " or" 9.27 g cm^(-3)`

Discussion

28410.	The molecular weight of a gas is 44. The volume occupied at STP by 2.2 g of the gas would be
Answer» 1.12 litre 2.24 litre 11.2 litre 22.4 litre Answer :A

Discussion

28411.	The molecular weight of a gas which diffuses through a porous plug of 1/6th of the speed of hydrogen under identical conditions is :
Answer» 27 72 36 48 Answer :B

Discussion

28412.	The molecular weight of O_(2) and SO_(2) are 32 and 64 respectively. If one litre of O_(2) at 15^(@)C and 750 mm contains N molecules, the number of molecules in two litres of SO_(2) under the same conditions of temperature and pressure will be
Answer» N/2 N 2N 4N Solution :EQUAL volume of DIFFERENT GASSES under same conditions of TEMP. and PRESSURE contain equal number of molecules.

Discussion

28413.	The mass of a non-volatile solute of molar mass 40 g mol^(-1) that should be dissolved in 114 g of octane to lower its vapour pressure by 20% is
Answer» 10 g 11.4 g 9.8 g 12.8 g Solution :Using Raoult's Law, `(P_(o)-P_(s))/(P_(s))=(w)/(m)xx(M)/(W)` If `P_(o)=100 mm`, then `P_(s)=80mm` `(100-80)/(80)=(w)/(114)xx(114)/(40) "" THEREFORE "" w=10 g`

Discussion

28414.	The molecular weight and equivalent weight of which one of the following is the same ?
Answer» `H_(2)SO_(4)` `KMnO_(4)` `H_(2)C_(2)O_(4)` `NAOH` SOLUTION :Both the molecular weight and equivalent weight for NaOH is 40. Molecular weight of NaOH=23+16+1=40 Equivalent weight of NaOH `=("Molecular weight of NaOH")/("ACIDITY of NaOH")=(40)/(1)=40`

Discussion

28415.	The molecular velocities of two gases at same temperature are u_(1) and u_(2) and their masses are m_(1) and m_(2) respectively. Which of the following expressions is correct ?
Answer» `( m_(1))/( u_(1)^(2)) = ( m_(2))/( u_(2)^(2))` `m_(1)u_(1) = m_(2) u_(2)` `( m_(1))/( u_(1)^(2)) = ( m_(2))/( u_(2)^(2))` `m_(1) u_(1)^(2) = m_(2) u_(2)^(2)` ANSWER :D

Discussion

28416.	The molecular size of Icl and Br_(2) is approximately same, but b.p. if Icl is about 40^(@)C higher than that of Br_(2). It is because :
Answer» ICL bond is stronger than Br-Br bond IE of IODINE `lt` IE of bromine Icl is polar while `Br_(2)` is nonpolar I has larger size than Br Answer :C

Discussion

28417.	The mass of a non-volatile solute of molar mass 40 "g mol"^(-1) that should be dissolved in 114 g of octane to lower its vapour pressure by 20% is :
Answer» <P>10 g 11.4 g 9.8 g 12.8 g Solution :`(p_(A)^(@)-p_(A))/(p_(A)^(@))=(W_(B)M_(A))/(W_(A)M_(B))` Lowering in VAPOUR pressure is 20% `:.(p_(A)^(@)-p_(A))/(p_(A)^(@))=(20)/(100)=0.2` `W_(B)=?, W_(A)=114g` `M_(B)=40g, M_(A)=114g` `:.0.2=(W_(B)xx114)/(20xx114)` or `W_(B)=8G`.

Discussion

28418.	The molecular velocity of any gas is :
Answer» inversely PROPORTIONAL to absolute temperature DIRECTLY proportional to square of temperature directly proportional to square root of temperature inversely proportional to the square root of temperature. SOLUTION :MOLECULAR velocity `prop sqrt( T )`.

Discussion

28419.	The mass of a neutron is of the order of:
Answer» `10^(-23) KG` `10^(-24) kg `10^(-26) kg` `10^(-27) kg` Answer :D

Discussion

28420.	The molecular orbital with the lowest eenrgy is filled first. Thus sigma(1s)is filled first where as 'sigma^(*)(2p) isfilled in thelast, also the maximum number of electron in bonding and antibonding molecular orbitals are according to Pauli and Hund's rule. As an electron in an antibonding molecular orbital cancels out the stability introduced by the electron in a bonding molecular orbital, it means that in order for bonding of atoms to occur there should be an excess of bonding electron sover antibonding electrons. In case where the number of bonding and antibonding electrons are equal, no bond will be formed between the atoms. With the help of above discussion, we can define easily bond order, relative bond length, relative stability and magnetic properties for a molecule. Which of the following are paramagnetic in nature ?
Answer» `B_(2), N_(2)` and `C_(2)` `O_(2), B_(2) , N_(2)^(-)` and `O_(2)^(-)` `O_(2)^(-), N_(2)^(+)` and CO `B_(2) ,C_(2)` and `F_(2)` Answer :B

Discussion

28421.	The molecular orbital with the lowest eenrgy is filled first. Thus sigma(1s)is filled first where as 'sigma^(*)(2p) isfilled in thelast, also the maximum number of electron in bonding and antibonding molecular orbitals are according to Pauli and Hund's rule. As an electron in an antibonding molecular orbital cancels out the stability introduced by the electron in a bonding molecular orbital, it means that in order for bonding of atoms to occur there should be an excess of bonding electron sover antibonding electrons. In case where the number of bonding and antibonding electrons are equal, no bond will be formed between the atoms. With the help of above discussion, we can define easily bond order, relative bond length, relative stability and magnetic properties for a molecule. According to MOT which statement is correct about Boron molecule ?
Answer» It is DIAMAGNETIC in nature It is paramagnetic in nature having MAGNETIC MOMENT 2.8 B.M. by using SPIN only formula It is paramagnetic but having magnetic moment 1.7 B.M. Its bond order is 2 Answer :B

Discussion

28422.	The mass of a litre of oxygen at standard conditions of temperature and presssure is 1.43 g and the of a litre of SO_(2) is 2.857 g. (i) How many molecules of each gas are there in this volume? (ii) What is the mass in grams of a single molecules of each gas ? (iii) What are the molecular masses of SO_(2) and O_(2) respectively?
Answer» SOLUTION :`1LO_(2)orSO_(2)=(1)/(22.4)mol=(1)/(22.4)xx6.02xx10^(23)" molecules"=2.688xx10^(22)" molecules"` `"Mass of 1 molecule of "O_(2)=(1.43)/(2.688xx10^(2))=1.0629xx10^(-23)g` Gram MOLECULAR mass of `O_(2)="Mass of 22.4 L at STP"=1.43xx22.4=32.032g` `"Molecular mass of "O_(2)=32.032u` Similarly, molecular mass of `SO_(2)` can be calculated.

Discussion

28423.	The mass of 2.24xx10^(-3)m^(3) of a gas is 4.4 g at 273.15 K and 101.325 kPa pressure. The gas may be
Answer» NO `NO_(2)` `C_(3)H_(8)` `NH_(3)` SOLUTION :`2.24xx10^(-3)m^(3)` of a gas = 2.24 L at STP Thus, 2.24 L of the gas at STP WEIGHT = 4.4 g `therefore 22.4` L of the gas at STP will weigh = 44 g But 22.4 L of a gas at sTP = Molar mass of the gas `therefore " Molar mass of the gas = 44 g"`

Discussion

28424.	The molecular shapes of SF_(4), CF_(4) and XeF_(4)are
Answer» the same with 2, 0 and 1 LONE PAIR of ELECTRONSON the central ATOM respectively the same with 1, 1 and 1 lone pair of electrons on the central atom respectively the same with 0, 1 and lone 2 pair of electrons on the central atom respectively the same with 1, 0 and 2 lone pair of electrons on the central atom respectively Answer :D

Discussion

28425.	The mass of 2.24 xx 10^(-3) m^(3) of a gas is 4.4 g at 273.15K and 101.325 kPa pressure. The gas may be
Answer» NO `NO_(2)` `C_(3) H_(8)` `NH_(3)` Solution :m = 4.4 g , V = `2.24 xx 10^(-3) = 2.24 L` T = 273.15 K p= 101. 325kPa = 101325Pa = 1 atm pV = NRT or pV `= ( m )/( M ) RT ` or `M = ( m RT )/( pV )` ` = ( 4.4 xx 0.0821 xx 273.15)/( 1 xx 2.24 ) = 44.05 ` `:.` The gas is `C_(3) H_(8)`

Discussion

28426.	The molecular orbital with the lowest eenrgy is filled first. Thus sigma(1s)is filled first where as 'sigma^(*)(2p) isfilled in thelast, also the maximum number of electron in bonding and antibonding molecular orbitals are according to Pauli and Hund's rule. As an electron in an antibonding molecular orbital cancels out the stability introduced by the electron in a bonding molecular orbital, it means that in order for bonding of atoms to occur there should be an excess of bonding electron sover antibonding electrons. In case where the number of bonding and antibonding electrons are equal, no bond will be formed between the atoms. With the help of above discussion, we can define easily bond order, relative bond length, relative stability and magnetic properties for a molecule. In an antibonding molecular orbital, there is a line between the two probability contours of hydrogen atoms. This is called
Answer» Antinode Nodal plane A plane where ELECTRON CHARGE DENSITY is maximum A and C both are CORRECT Answer :A

Discussion

28427.	The molecular mass of NaCl determined by studying freezing point depression of it's 0.5% aqueous solution is 30. The apparent degree of dissociantion of NaCl is :
Answer» 0.95 `0.5` `0.6` `0.3` SOLUTION :`i=1+ALPHA =("cal.m.wt.")/("exp.m.wt.")= 58.5/30` `:. 1 + alpha = 1.95` `:. alpha =0.05`

Discussion

28428.	The mass of 1xx10^(22) molecules of blue vitriol (CuSO_(4).xH_(2)O) is 4.144g. The values of 'x' will be:
Answer» ANSWER :5

Discussion

28429.	The molecular mass of CO_(2) is 44 amu and Avogadro's number is 6.02xx10^(23). Therefore, the mass of one molecule of CO_(2) is:
Answer» `7.31xx10^(-23)` `3.65xx10^(-23)` `1.01xx10^(-23)` `2.01xx10^(-23)` ANSWER :A

Discussion

28430.	The mass of 11.2 L of ammonia gas at S.T.P. is
Answer» `8.5 g` `85 g` `17 g` `1.7 g` SOLUTION :MOLAR mass of `NH_(3)=14+3xx1=17g MOL^(-1)` `:.` Mass of 22.4 L of `NH_(3)` at S.T.P. = 17 g `:.` Mass of 11.2L of `NH_(3)` at S.T.P. = 8.5 g

Discussion

28431.	The molecular mass of chloride MCl, is 74.5 . The equivalent mass of the metal M will be
Answer» 39 74.5 110 35.5 Solution :Mol. MASS =At. Mass +35.5=Eq. mass `xx` VALENCY +35.5. valency of M from the formula MCL is1

Discussion

28432.	The mass of 112 cm^(3) " of " CH_(4) gas at STP is
Answer» 0.08g 1.6g 0.16G 0.8g Solution :`PV = (WRT)/(M) rArr W = (MPV)/(RT)` `= (16g MOL^(-1) xx 1"ATM" xx 0.1120"lit")/(0.082 "atm lit" k^(-1) mol^(-1) xx 273K), W= 0.089`

Discussion

28433.	The molecular mass of an organic compound which contains only one nitrogen atom can be
Answer» 41 76 146 152 Solution :All compounds containing an odd number of nitrogen atoms have odd molar MASS and with EVEN number OFNITROGEN atoms have even molar mass. Therefore, option(a) is correct . Itis called nitrogen RULE (b), (c ) and (d) are RULED out.

Discussion

28434.	The mass of 112cm^(3) " of " NH_(3) gas at STP is
Answer» 0.085 g 0.850 g 8.500 g 80.500 g Solution :Mass of 22,400 `cm^3` of `NH_3` at STP = 17g mass of `112 cm^3` of `NH_3` at STP `= (17)/(22400) XX 112 = 0.085 g `

Discussion

28435.	The molecular mass of acetic acid dissolved in water is 60 and when dissolved in benzene it is 120. This difference in behaviour of CH_(3)COOH is because
Answer» Water prevents ASSOCIATION of acetic ACID Acetic acid does not fully dissolve in water Acetic acid fully DISSOLVES in benzene Acetic acid does not IONIZE in benzene Answer :B

Discussion

28436.	The atomic mass of thorium is 232 and its atomic number is 90. During the course of its radioactive disintegration 6 alpha and 4 beta particles are emitted. What is the atomic mass and atomic number of the atom?
Answer» ANSWER :208,82

Discussion

28437.	The molecular mass of a solute cannot be calculated by which of the following?
Answer» `M_(B) =(W_(B)xxRT)/(PIV)` `M_(B) =(p^(@)W_(B) M_(A))/((p^(@) -p) W_(A))` `M_(B) =(Delta T_(b) W_(B) XX 1000)/(K_(b) W_(A))` `M _(B) =(K_(b) W_(B) xx1000)/(Delta T_(b) xxW_(A))` Solution :`M_(B) =(Delta T _(b) xx W_(B)xx1000)/(K _(b) xx W_(A))` is WRONG. The correct from is `M _(B) =(K_(b)xx W_(B)xx 1000)/(Delta T_(b) xx W_(A))`

Discussion

28438.	The mass of 1 curie of U-234 is :
Answer» `3.7xx10^(10)G` `2.348xx10^(-23)g` `3.7xx10^(-10)g` `1.438xx10^(-11)g` Solution :1 Curie of a SUBSTANCE is the amount of substance which gives `3.7xx10^(10)` DISINTEGRATIONS per SECOND. NUMBER of disintegrations is proportional to the number of atoms so that `6.02xx10^(23)` atoms of U-234 has mass =234g `3.7xx10^(10)` atoms of U-234 has mass `=(234)/(6.02xx10^(23))xx3.7xx10^(10)` `=1.438xx10^(-11)g`

Discussion

28439.	The molecular mass of a compound is 75 (i) Calculate the mass of 100 molecules, in amu. (ii) Calculate the mass of 5000 molecules , in gm. (iii) What is the mass of 6.022xx10^(20) molecules , in gm (iv) How many molecules are in its 2.5 mg
Answer» Solution :(i) mass of 1 molecules =75 amu `therefore` mass of 100 molecules=7500 amu (ii) Mass of 5000 molecules =`5000xx75` amu `=5000xx75xx1.67xx10^(-24)=6.26 25xx10^(-19)` gm `because 6.022xx10^(20)` molecules weights `(75)/(6.022xx10^(23))xx6.022xx10^(20)=0.075` gm (iv) `therefore `75 gm compound CONTAINS `6.022xx10^(23)` molecules `because 2.5xx10^(-3)` gm will contain `(6.022xx10^(23))/(75)xx2.5xx10^(-3)=2.007xx10^(19)` molecules.

Discussion

28440.	The mass number and the atomic number of uranium are 238 and 92 respectively. If on nuclear disintegration 6 alpha- and 4 beta-particles are emitted, find the mass number and atomic number of the atom formed.
Answer» Solution :APPLYING group DISPLACEMENT law, EMISSION of `6 alpha`-PARTICLES will cause a decrease in mass number 24 `(6 xx4)` units and in atomic number by `12(6 xx 2)` units. Now with the emission of `4 beta`-particles, mass number will not change but atomic number will increase by 4 units. Thus the mass number and the atomic number of the atom formed will be 214 (i.e. 238-24) and 84 (i.e., `92-12+4`) respectively.

Discussion

28441.	The mass defect of the nuclear reaction ._(5)B^(8) rarr ._(4)Be^(8) + ._(1)e^(0) is Delta m, the wrong expression "is"//"are"
Answer» `Delta m =` atomic mass of `(._(4)Be^(8) - ._(5)B^(8))` `Delta m =` atomic mass of `(._(4)Be^(8) - ._(5)B^(8)) +` mass of one ELECTRON `Delta m =` atomic mass of `(._(4)Be^(8) - ._(5)B^(8)) +` mass of one positron `Delta m =` atomic mass of `(._(4)Be^(8) - ._(5)B^(8)) +` mass of TWO electrons Answer :A::B::C

Discussion

28442.	The molecular formula of Urotropine is ..............
Answer» `(CH_2)_6N_4` `(CH_2)_4N_6` `(CH_2)_2N_2` `(CH_2)_6N_6` SOLUTION :`(CH_2)_6N_4`

Discussion

28443.	The mass deposited at an electrode is directly proportional to ……………. .
Answer» ATOMIC WEIGHT equivalent weight MOLECULAR weight atomic number Answer :B

Discussion

28444.	The molecular formula of Wilkinson catalyst, used in hydrogenation of alkenes is
Answer» <P>`CO(CO)_(3)` `(Ph_(3)P)_(3)RhCl` `[Pt(NH_(3))_(2)Cl_(2)]` `K[Ag(CN)_(2)]` SOLUTION :Wilkinston catalyst has molecular formula `(Ph_(3))P_(3)` RhCl and is used in HYDROGENATION of alkenes.

Discussion

28445.	The mass average molecular mass & number average molecular mass of a polymer are 40,000 and 30,000 respectively. The polydispersity index of polymer will be
Answer» lt1 gt1 1 0 Solution :AVERAGE number MOLECULES weight `BAR(M)_(N) =30,000` Average mass molecular weight `bar(M)_(w)=40,000` polydispersity index (PDI) `= (bar(M)_(w))/(bar(M)_(n))=40,000/30,000=1.33`

Discussion

28446.	The molecular formulae for phosgene and tear gas are......and.....respectively.
Answer» `SOCl_(2)` and `C Cl_(2)NO_(2)` `COCl_(2)` and `C Cl_(2) NO_(2)` `COCl_(2)` and `C Cl_(3)NO_(2)` `SOCl_(2)` and `C Cl_(3) NO_(2)` ANSWER :D

Discussion

28447.	The mass aof 350cm^(3) of a diatomic gas at 273 K at 2 atmospheres pressure is one gram. Calculate the mass of one atom of the gas.
Answer» Solution :(i) Find the volume of the gas at STP which on CALCULATION comes out to be `700cm^(3)` (ii) Find the mass of 1 mole, i.e, 22.4 L of gas at STP, i.e., `(22400)/(700)=32g` (III) Calculate the mass of one molecule, i.e., `(32)/(6.022xx10^(23))G` and divide it by 2 to get the mass of one atom of the gas.

Discussion

28448.	The mass and charge of 1 mole electrons will be:
Answer» 1kg,96500C 0.55mg: 96500C 1.55mg:96500C 5.5mg 96500C Answer :B

Discussion

28449.	The Markovnikov rule is best applicable to the reaction between
Answer» `C_2H_4+HCl` `C_3H_6+Br_2` `C_3H_6+HBr` `C_3H_8+Cl_2` SOLUTION :Markownikoff's rule states that addition of halides to unsymmetrical alkenes takes place in such a MANNER that the negative part of the adduct attches itself to that carbon ATOM which CONTAINS LESSER number of hydrogen atoms. Here `C_3H_6(CH_3-CH=CH_2)` is an unsymmetrical alkene and the reagent is HBr, hence Markownikoff's rule is best applicable for this reaction

Discussion

28450.	The molecular formula of starch is ………………
Answer» `[C_(6)(H_(2)O)_(6)]_(N)` `[C_(6)(H_(2)O)_(5)]_(n)` `C_(6)H_(12)O_(5)` `C_(6)H_(12)O_(6)` Solution :`[C_(6)(H_(2)O)_(5)]_(n)`

Discussion

Explore topic-wise InterviewSolutions in .

The mass of AgCl precipitated when a solution containing 11.70 g of NaCl is added to a solution containing 3.4g of AgNO_(3), is [Atomie mass of Ag -108, Atomic mass of Na - 23]

The molecular weight of benzoic acid in benzene asxc determined by depression in freezing point method corresponds to

The mass of a unit cell of CsCl corresponds to

The molecular weight of benzoic acid in benzene as determined by depressing in freezing point method corresponds to -

The mass of a unit cell of CsCl corresponds to-

The molecular weight of benzoic acid as determined by depression in freezing point method corresponds to :

The mass of a unit cell of CsCI corresponds to:

The molecular weight of a tribasic acid is M. What will be its equivalent weight

The mass of a piece of paper is 0.02 g and the mass of a solid substance and the piece of paper is 20.036 g. If the volume of the solid is 2.16 cm^3, calculate its density to the proper number of significant digits.

The molecular weight of a gas is 44. The volume occupied at STP by 2.2 g of the gas would be

The molecular weight of a gas which diffuses through a porous plug of 1/6th of the speed of hydrogen under identical conditions is :

The molecular weight of O_(2) and SO_(2) are 32 and 64 respectively. If one litre of O_(2) at 15^(@)C and 750 mm contains N molecules, the number of molecules in two litres of SO_(2) under the same conditions of temperature and pressure will be

The mass of a non-volatile solute of molar mass 40 g mol^(-1) that should be dissolved in 114 g of octane to lower its vapour pressure by 20% is

The molecular weight and equivalent weight of which one of the following is the same ?

The molecular velocities of two gases at same temperature are u_(1) and u_(2) and their masses are m_(1) and m_(2) respectively. Which of the following expressions is correct ?

The molecular size of Icl and Br_(2) is approximately same, but b.p. if Icl is about 40^(@)C higher than that of Br_(2). It is because :

The mass of a non-volatile solute of molar mass 40 "g mol"^(-1) that should be dissolved in 114 g of octane to lower its vapour pressure by 20% is :

The molecular velocity of any gas is :

The mass of a neutron is of the order of:

The mass of 2.24xx10^(-3)m^(3) of a gas is 4.4 g at 273.15 K and 101.325 kPa pressure. The gas may be

The molecular shapes of SF_(4), CF_(4) and XeF_(4)are

The mass of 2.24 xx 10^(-3) m^(3) of a gas is 4.4 g at 273.15K and 101.325 kPa pressure. The gas may be

The molecular mass of NaCl determined by studying freezing point depression of it's 0.5% aqueous solution is 30. The apparent degree of dissociantion of NaCl is :

The mass of 1xx10^(22) molecules of blue vitriol (CuSO_(4).xH_(2)O) is 4.144g. The values of 'x' will be:

The molecular mass of CO_(2) is 44 amu and Avogadro's number is 6.02xx10^(23). Therefore, the mass of one molecule of CO_(2) is:

The mass of 11.2 L of ammonia gas at S.T.P. is

The molecular mass of chloride MCl, is 74.5 . The equivalent mass of the metal M will be

The mass of 112 cm^(3) " of " CH_(4) gas at STP is

The molecular mass of an organic compound which contains only one nitrogen atom can be

The mass of 112cm^(3) " of " NH_(3) gas at STP is

The molecular mass of acetic acid dissolved in water is 60 and when dissolved in benzene it is 120. This difference in behaviour of CH_(3)COOH is because

The atomic mass of thorium is 232 and its atomic number is 90. During the course of its radioactive disintegration 6 alpha and 4 beta particles are emitted. What is the atomic mass and atomic number of the atom?

The molecular mass of a solute cannot be calculated by which of the following?

The mass of 1 curie of U-234 is :

The molecular mass of a compound is 75 (i) Calculate the mass of 100 molecules, in amu. (ii) Calculate the mass of 5000 molecules , in gm. (iii) What is the mass of 6.022xx10^(20) molecules , in gm (iv) How many molecules are in its 2.5 mg

The mass number and the atomic number of uranium are 238 and 92 respectively. If on nuclear disintegration 6 alpha- and 4 beta-particles are emitted, find the mass number and atomic number of the atom formed.

The mass defect of the nuclear reaction ._(5)B^(8) rarr ._(4)Be^(8) + ._(1)e^(0) is Delta m, the wrong expression "is"//"are"

The molecular formula of Urotropine is ..............

The mass deposited at an electrode is directly proportional to ……………. .

The molecular formula of Wilkinson catalyst, used in hydrogenation of alkenes is

The mass average molecular mass & number average molecular mass of a polymer are 40,000 and 30,000 respectively. The polydispersity index of polymer will be

The molecular formulae for phosgene and tear gas are......and.....respectively.

The mass aof 350cm^(3) of a diatomic gas at 273 K at 2 atmospheres pressure is one gram. Calculate the mass of one atom of the gas.

The mass and charge of 1 mole electrons will be:

The Markovnikov rule is best applicable to the reaction between

The molecular formula of starch is ………………