982 + Interview Questions in SOME-BASIC-CONCEPTS-OF-CHEMISTRY IN GENERAL AWARENESS Page 11 InterviewSolution

501.	In which reaction there is a change in valencyA. `2NO_(2) rarr N_(2)O_(4)`B. `2NO_(2)+H_(2)O rarr HNO_(2)+HNO_(3)`C. `NH_(4)OH rarr NH_(4)^(+)+OH^(-)`D. `CaCO_(3) rarr CaO+CO_(2)`
Answer» Correct Answer - B `2overset(+4)(N)O_(2)+H_(2)O rarr HNO_(2)+Hoverset(+5)(N)O_(3)`. In this reaction oxidation state changes.

Discussion

502.	A sample of `AIF_(3)` contains `3.0 xx 10^(24)` `F^(-)` ions. The number of formula units of the sample areA. `90xx10^(24)`B. `30xx10^(24)`C. `075xx10^(24)`D. `10xx10^(24)`
Answer» Correct Answer - D `3F^(-) -= 1` Formula unit `(AIF_(3))` `3*0xx10^(24)F^(-) = 1xx10^(24)` Formula unit `(AIF_(3))`

Discussion

503.	The compound `Yba_2Cu_3O_7` which shows super conductivity has copper in oxidation state_____. Assume that the rare earth element yttrium is in its usual `+3` oxidation state.A. `3//7`B. `7//3`C. 3D. 7
Answer» Correct Answer - B `Yba_(2) overset(**)(C)u_(3) O_(7)` `3 + 2 xx 2 + 3x - (2xx 7) = 0` `3 + 4 + 3x - 14 = 0` `3x = 7` `x = (7)/(3)`

Discussion

504.	The mass of 70% `H_(2)SO_(4)` required for neutralization of one mole of NaOH is:A. 49 gB. 98 gC. 70 gD. `34*3 g`
Answer» Correct Answer - C `2NaOH + H_(2)SO_(4) to Na_(2)SO_(4)+ H_(2)O` Pure `H_(2)SO_(4)` required for 2 mole of NaOH `=(49xx100)/(70) = 70g`

Discussion

505.	0.126 g of acid required 20 mL of 0.1 N NaOH for complete neutralisation. The equivalent mass of an acid isA. `45`B. `53`C. `40`D. `63`
Answer» Correct Answer - D Let equivalent mass be E `:.` g-equivalent of acid `= (0.126)/(E)` G-equivalents of acid = G equi. Of base or `(0.126)/(E) = 20xx10^(-3)xx0.1` or `E = (0.126xx10^(3))/(20xx0.1)=63`

Discussion

506.	Molarity of liquid HCl with density equal to `1.17 g//mL` is:A. (a)`36.5`B. (b)`18.25`C. (c )`32.05`D. (d)`4.65`
Answer» Correct Answer - C Liquid HCl is `100%` pure `:. M=(100xx1.17xx1000)/(36.5xx100)=32.05`

Discussion

507.	Assertion If `10 mL` of a `H_(2)O_(2)` solution required `8.00 mL` of `0.02 M` acidified `KMnO_(4)` solution for complete oxidation, `12.50 mL` of same `H_(2)O_(2)` will oxidise completely to `5.00 mL` of `0.10 M Na_(2)C_(2)O_(4)` solution. Reason `H_(2)O_(2)` act as both oxidising as well as reducing agent.A. Both assertion and reason are correct and reason is the correct explanation of the assertion,B. Both assertion and reason are correct but reason is not the correct explanation of assertion.C. Assertion is correct but reason is incorrect.D. Assertion is incorrect but reason is correct.
Answer» Correct Answer - B

Discussion

508.	Assertion If certain volume of a basic solution require `x mL` of `HCl, 2x mL` of `H_(2)SO_(4)` of same molarity would be required. Reason `HCl` is a monobasic acid while `H_(2)SO_(4)` is a diabasic acid.A. Both assertion and reason are correct and reason is the correct explanation of the assertion,B. Both assertion and reason are correct but reason is not the correct explanation of assertion.C. Assertion is correct but reason is incorrect.D. Assertion is incorrect but reason is correct.
Answer» Correct Answer - D

Discussion

509.	With increase of temperature, which of these changes?A. MolalityB. Weight fraction of soluteC. Fraction of solute present in 1L of waterD. Mole fraction
Answer» Correct Answer - C Volume increases with rise in temperature. Further some water molecules may get evaporated at high temperature.

Discussion

510.	How many moles of lead (II) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g of HCl?A. `0.333`B. `0.011`C. `0.029`D. `0.044`
Answer» Correct Answer - C `PbO+2HCLtoPbCl_(2)+H_(2)O` Moles of PbO `= (6.5)/(224)=0.029` Moles of HCl `=(3.2)/(36.5)=0.088` Thus PbO is limiting reatant `:. PbCl_(2)` formed = 0.029 mol

Discussion

511.	The number of moles of `KMnO_(4)` that will be needed to react with one mole of sulphite ion in acidic solution isA. `4//5`B. `2//5`C. `1`D. `3//5`
Answer» Correct Answer - B Ionic equation for the reaction is `2MnO_(4)^(-)+6H^(+)+5SO_(3)^(2-)to2Mn^(2-)+5SO_(4)^(2-)+3H_(2)O` 5 moles of `SO_(3)^(2-)` ions are oxidized by `MnO_(4)^(-)` ions = 2 mol `:.` 1 mole of `SO_(3)^(2-)` ions is oxidized by `MnO_(4)^(-)` ions = 2/5 mol.

Discussion

512.	How many molecules are there in `3.46g` sample of hydrogen chloride `(HCl)`?A. `3.72xx10^(22)`B. `4.87xx10^(22)`C. `1.95xx10^(22)`D. `3.786xx10^(22)g`
Answer» Correct Answer - D `underset(HCl)("gram of")rarr underset(HCl)("moles of")rarr underset(HCl)("molecules of ")` `3.46g HCl xx (1 mol CHl)/(36.5g HCl) xx (6.022xx10^(23) HCl "molecules")/(1 mol HCl)` `= 5.72xx10^(22) HCl` molecules

Discussion

513.	One atom mass unit (amu) is equivalent toA. `3.786xx10^(-24)g`B. `1.661xx10^(-24)g`C. `2.687xx10^(-24)g`D. `5.099xx10^(-24)g`
Answer» Correct Answer - B `= (1)/(12) xx ((12g mol^(-1))/(6.022xx10^(23) "atoms mol"^(-1)))` `= 1.661xx10^(-24) g`

Discussion

514.	If the density of water is 1 g `cm^(-3)` then the volume occupied by one molecule of water is approximatelyA. (a)`6.023xx10^(-3) cm^(3)`B. (b)`3.0xx10^(-3) cm^(3)`C. (c )`5.5xx10^(-23) cm^(3)`D. (d)`9.0xx10^(-23) cm^(3)`
Answer» Correct Answer - b `1000 g H_(2)O=1000 cm^(3) H_(2)O` `1000/18 "mole" H_(2)O=1000 cm^(3) H_(2)O` `1000/18xx6.023xx10^(23)` molecule of `H_(2)O` `=1000 cm^(3) H_(2)O` `implies ` Volume of `1 "molecule"=1000/1000xx18/(6.023xx10^(23))` `=3.0xx10^(-23) cm^(3)`.

Discussion

515.	An electric discharge is passed through a mixture containing 50 c.c. of `O_(2)` and 50 c.c. of` H_(2)`. The volume of the gases formed (i) at room temperature and (ii) at `110^(@)`C will beA. (a)(i) 25 c.c. (ii) 50 c.c.B. (b)(i) 50 c.c. (ii) 75 c.c.C. (c )(i) 25 c.c. (ii) 75 c.c.D. (d)(i) 75 c.c. (ii) 75 c.c.
Answer» Correct Answer - C At room temperature `2H_(2(g))+O_(2(g))rarr 2H_(2)O_((l))` `" "t=0 " " 50 ml" " 50 "ml"" "` 0 `" "t=t" " 50-2x" " 50-x" " 2x` `" " =0" "25 gases (50)` liquid In this case `H_(2)` is limiting reagent `x=25 ml` At `110^(@)C 2H_(2(g))+O_(2(g))rarr 2H_(2)O_((g)) V_(gas)=75 ml` `t=t" 0" " 25ml " "50ml`

Discussion

516.	A mixture of gases contains `H_(2)` and `O_(2)` gases in the ratio of `1:4 (w//w)`. What is the molar ratio of the two gases in the mixture?A. (a)`16:1`B. (b)`2:1`C. (c )`1:4`D. (d)`4:1`
Answer» Correct Answer - b Number of moles of `H_(2)=1/2` Number of moles of `O_(2)=4/32` Hence, molar ratio `=1/2:4/32=4:1`

Discussion

517.	`100 g CaCO_(3)` reacts with `1 litre 1 N HCl`. On completion of reaction how much weight of `CO_(2)` will be obtainA. (a)5.5 gB. (b)11 gC. (c )22 gD. (d)33 g
Answer» Correct Answer - c `underset(100 g)(CaCO_(3))+underset(2N)(2HCl) rarr CaCl_(2)+underset(44 g)(CO_(2))+H_(2)O` `100 g CaCO_(3) " with " 2 N " HCl gives " 44 g CO_(2)` `100 g CaCO_(3) " with 1 N HCl gives " 22 g CO_(2)`

Discussion

518.	What is the equivalent mass of ` IO_4^(-)` when it is converted into `I_2` in acid medium ?A. `M//6`B. `M//7`C. `M//5`D. `M//4`
Answer» Correct Answer - B `underset((O.N. of I = +7))(IO_(4)^(-))+8H^(+)tounderset((O.N. of I=0))(I_(2))+4H_(2)O-7e^(-)` Eq. mass of `O_(4)^(-) = ("Mol mass")/(7) = (M)/(7)`

Discussion

519.	How many atoms are contained in one mole of sucrose `(C_(12)H_(22)O_(11))`?A. `45xx6.02xx10^(23)` atoms/moleB. `5xx6.62xx10^(23)`atoms/moleC. `5xx6.02xx10^(23)` atoms/moleD. None of these
Answer» Correct Answer - A Atoms in 1 molecule of `C_(12)H_(22)O_(11)` `= 12+22+11=45` `:.` Atom in 1 mole `= 45xxN_(A)` `= 45xx6.02xx10^(23)` atoms/mole

Discussion

520.	The weight of one molecule of a compound `C_(60)H_(12)` isA. `1.3xx10^(-20)`gB. `5.01xx10^(-21)`gC. `3.72xx10^(23)`gD. `1.4xx10^(-21)`g
Answer» Correct Answer - D Wt. of one molecule `= ("Mol mass of" C_(60)H_(122))/(6.023xx10^(23))` `= ((12xx60)+122)/(6.023xx10^(23)) = (842)/(6.023xx10^(-23))` `=1.4xx10^(-21)g`

Discussion

521.	Define second.
Answer» The second is the duration of 9192631770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium - 133 atom.

Discussion

522.	A piece of metal is 3 inch (represented by in) long. What is its length in cm?
Answer» We know that 1 in = 2.54 cm From this equivalence, we can write `(1"in ")/(2.54 cm ) =1=(2.54 cm )/(1 "in ")` thus ,`(1 "in")/(2.54 cm) `equals 1 and `(2.54 cm) /(1 "in")` also equals 1. Both of these are called unit factors. If some number is multiplied by these unit factors (i.e., 1), it will not be affected otherwise. Say, the 3 in given above is multiplied by the unit factor. So, 3 in 3 in `xx(2.54 cm) /(1 "in")= 3xx 2.54 cm = 7.62 cm` Now , the unit factor by which multiplication to is to done is that unit factor `((2.54 cm )/(1"in")` in the above case) which gives the desired units i.e., the numerator should have that part which is required in the desired result. It should also be noted in the above example that units can be handled just like other numerical part. It can be cancelled, divided, multiplied, squared, etc. Let us study one more example.

Discussion

523.	A jug contains 2L of milk. Calcualte the volume of the milk in `m^(3)`
Answer» Since 1L = `1000 cm ^(3)` and 1m = 100 cm , which gives `(1m)/(100 cm) =1 = (1 m^(3))/(1 m) ` TO get `m^(3) ` from the above unit factors , the first unit factor is taken ant it is cubed `((1m)/(100 cm ))^(3) implies (1 m^(3))/(10^(6) cm^(3))=(1)^(3) =1 ` Now 2 L `= 2xx1000 cm^(3)` the above is multiplied by the unit factor `2xx1000 cm^(3) xx(1 m^(3))/(10 ^(6)cm^(3)) =(2 m^(3))/(10^(3)) = 2xx10^(-3) m^(3)`

Discussion

524.	Number of atoms of oxygen present in 10.6 g of `Na_(2)CO_(3)` will beA. `6.02xx10^(22)`B. `12.04xx10^(22)`C. `1.806xx10^(23)`D. `31.80xx10^(28)`
Answer» Correct Answer - C Molar mass of `Na_(2)CO_(3) = 2xx23+1216xx3` `=46+12+48` `=106gmol^(-1)` `:.` 10.6 g of `Na_(2)CO_(3)-=0.1` mol of `Na_(2)CO_(3)` `=0.3` mol of oxygen atoms `=0.3xx6.02xx10^(-23)` oxygen atom `=1.806xx10^(23)` oxygen atom

Discussion

525.	4.4 g of `CO_(2)` contains how many litres of `CO_(2)` at S.T.PA. `2.4 L`B. `2.24L`C. `44L`D. `22.4L`
Answer» Correct Answer - B 44g of `CO_(2)` at S.T.P. = 22.4 L `:.` 4.4 g of `CO_(2)` at S.T.P. = 2.24 L

Discussion

526.	The volume of 1.0 g of hydrogen in litres at N.T.P. isA. `2.24`B. `22.4`C. `1.12`D. `11.2`
Answer» Correct Answer - D Vol. of 2 g `H_(2)` at S.T.P. = 22.4 litres `:.` Vol. of 1 g `H_(2) = (22.4)/(2) = 11.2L`

Discussion

527.	Volume of 0.1 M `K_(2)Cr_(2)O_(7)` required to oxidise 35 mL of 0.5 M Fe `SO_(4)` solution isA. 29.2 mLB. 17.5 mLC. 175 mLD. 145 mL
Answer» Correct Answer - A `(({:(K_(2)Cr_(2)O_(7)+4H_(2)SO_(4)toK_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+4H_(2)O+3[0]),(2FeSO_(4)+H_(2)SO_(4)+[O]toFe_(2)(SO_(4))_(3)+H_(2)Oxx3):})/(K_(2)Cr_(2)O_(7)+7H_(2)SO_(4)+6FeSO_(4)toK_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+7H_(2)O+3Fe_(2)(SO_(4))_(3)))/` `(M_(1)V_(1))/(n_(1) = (M_(2)V_(2))/(n_(2))` `(K_(2)Cr_(2)O_(7) = (FeSO_(4))` `(0.1xxV_(1))/(1)=(0.5xx35)/(6xx0.1)` `V_(1) = (0.5xx35)/(6xx0.1)=29.2 mL`

Discussion

528.	`6.0xx10^(20)` molecules of urea are present in 100 L of his solution. The concentration of urea solution isA. 0.001 MB. 0.1 MC. 0.02 MD. 0.01 M
Answer» Correct Answer - D No. of moles of urea `= (6.02xx10^(20))/(6.02xx10^(23))=10^(-3)` mol `:. 10^(-3)` mol of urea are present in 100 mL `:.` Moles of urea present in 1L (1000 mL) `=10^(-3)xx10=10^(-2)` `:.` Molarity of solution `= 10^(-2) M = 0.01 M`

Discussion

529.	The answer to the following problem in standard exponential form is :`(2.0xx10^(13)) + (1.5xx10^(14))`A. `3.5xx10^(13)`B. `3.5xx10^(14)`C. `2.15xx10^(13)`D. `1.7xx10^(14)`
Answer» Correct Answer - D `2.0xx10^(13)+1.5xx10^(14) = 2.0xx10^(13)+15xx10^(13)` `= 17xx10^(13) = 1.7xx10^(14)`

Discussion

530.	To differentiat between C-12, C-13 and C-14 the instrument that you would use isA. infra-red spectrometerB. atomic absorption spectrometerC. mass spectrometerD. ultraviolet spectrometer
Answer» Correct Answer - C Mass spectrometer can easily differentiate between different isotopes of an element

Discussion

531.	Accurate determination of atomic masses is done with the instrument called asA. spectrophotometerB. mass spectrometerC. atomic absorption spectrometerD. calorimeter
Answer» Correct Answer - B Accurate determination of atomic masses is done with mass spectrometer.

Discussion

532.	Which among the following is the heaviest?A. one mole of oxygenB. one molecule of sulphur dioxideC. 100 a.m.u. of uraniumD. ten moles of hydrogen
Answer» Correct Answer - B (A) 1 mole of `O_(2) = 32g` (B) 1 molecule of `SO_(2) = (64)/(6.02xx10^(23)g)` `1.06xx10^(-22)g` (C) 100 a.m.u. of U `= (1)/(6.02xx10^(23))xx100g` `=1.66xx10^(-22)g` (D) 10 moles of `H_(2) = 10xx2 = 20g` (E) 44 g of `CO_(2) = 44g` Clearly (E) is maximum.

Discussion

533.	What is the simplest formula of the compound which has the following percentage composition - Carbon 80%. Hydrogen 20% ? If the molecular mass is 30, calculate its molecular formula.
Answer» Calculation of empirical formula. `\|{:("Element","% age","Atomic","Constent","No of","Sample","Simplest whole"),(,," mass","in sample","moles"," ratio"," no. ratio"),(" C"," 80"," 12"," 80 g",(80)/(12)=6.66,(6.66)/(6.66)=1," 1"),(" H"," 20"," 1"," 20 g",(20)/(1)=20,(20)/(6.66)=3," 3"):}\|` `therefore` Empirical formula is `CH_(3)` Calculation of molecular formula Empirical formula mass `=12xx1+1xx3=15` `n=("Molecular mass")/("Empirical formula mass")=(30)/(15)=2` Molecular formula = Empirical formula `xx2` `= CH_(3)xx2=C_(2)H_(6)` .

Discussion

534.	When pentane , `C_(5) H_(12)` is burned in excess oxygen , the products of the reaction are `CO_(2)(g)` and `H_(2)O (l)` . The balanced equation for this combustion is `C_(5) H_(12)(g) + xO_(2)(g) rarr 5CO_(2)(g) + 6 H_(2) O (l)` The coefficient (x) of oxygen should beA. 16B. 12C. 11D. 8
Answer» Correct Answer - D

Discussion

535.	Which one of the following statements is false ?A. An element of a substance contains only one kind of atomsB. Milk is a homogenous mixtureC. A compound can be decomposed into its constituentsD. All homogenous mixtures are called solutions
Answer» Correct Answer - B

Discussion

536.	The number of g-atom of oxygen in `6.02xx10^(24) CO` molecule isA. 1B. 0.5C. 5D. 10
Answer» Correct Answer - D One molecula of `CO` contains one oxygen atom. `:. 6.02xx10^(24)` molecules of `CO` contain `6.02xx10^(24)` oxygen atoms. `6.02xx10^(23)` atoms of oxygen `=1` g-atom of oxygen `:. 6.02xx10^(24)` atoms of oxygen `=10` g-atom of oxygen.

Discussion

537.	Sodium nitrate on reduction with Zn in presence of NaOH solution produces `NH_(3)`. Mass of sodium nitrate nitrate absorbing 1 mole of electron will beA. `7.750`B. `10.625`C. `8.000`D. `9.875`
Answer» Correct Answer - B Ammonia is formed by reduction of nitrates and nitrites with Zn and NaOH. Zn and caustic soda produce nascent hydrogen which reacts with nitrates to form ammonia. `NaNO_(3)+8H^(+)+8e overset(Zn//NaOH)(rarr) NaOH+NH_(3)+2 H_(2)O` From the equation Mass of `NaNO_(3)` absorbing 8 moles of electron `=85 g` `:.` Mass of `NaNO_(3)` absorbing 1 mole of electron `=85/8=10.625 g`.

Discussion

538.	0.078 g of a hydrocarbon occupy 22.414 mL of volume at S.T.P. The empirical formula of the hydrocarbon is CH. The molecular formula of hydrocarbon isA. `C_(2)H_(2)`B. `C_(6)H_(6)`C. `C_(8)H_(8)`D. `C_(4)H_(4)`
Answer» Correct Answer - B 22.414 mL of gaseous hydrocarbon at S.T.P. has mass = 0.078 g 22414 mL of gaseous hydrocarbon at S.T.P. will have mass = 78 g `:.` Molar mass of hydrocarbon `= 78 gmol^(-1)` Empirical formula mass of hydrocarbon (CH) `= 12+1=13` `:. n = ("Molecular mass")/("E.F. mass") = (78)/(13) =6` `:.` Molecular mass of hydrocarbon `= (CH)_(6)=C_(6)H_(6)`

Discussion

539.	The empirical formula of a compound is `CH_2O_2`. What could be its molecular formula?A. `C_2H_2O_2`B. `C_2H_2O_4`C. `C_2H_4O_4`D. `CH_4O_4`
Answer» Since empirical formula is multiplied by n to get molecular formula. `CH_2O_2 " will give only " C_2H_4O_4` as its molecular formula. `(CH_2O_2)_n` where n=1, 2, 3, …..etc.

Discussion

540.	A gas has molecular formula `(CH)_n`. If vapour density of the gas is 39, what should be the formula of the compound ?A. `C_3H_3`B. `C_4H_4`C. `C_2H_2`D. `C_6H_6`
Answer» Mol. Wt. `=2xxV.D. = 2xx39=78` `(CH)_n=(13)_n " or " 13xxn=78` n=78/13=6 `therefore " molecular formula " = C_6H_6`

Discussion

541.	Which one of the following is the lightestA. 0.2 mole of hydrogen gasB. `6.023xx10^(22)` molecules of nitrogenC. 0.1 g of silverD. 0.1 mole of oxygen gas
Answer» Correct Answer - C (a) Moles = given wt./molecular wt. Weight of `H_(2)=` moles `xx` molecular wt. `=0.2xx2=0.4 g` (b) `6.023xx10^(23)` represents 1 mole Thus `6.023xx10^(22)` will represent 0.1 mole Wight of `N_(2)=0.1xx28=2.8=2.8 g` (c) Weight of silver `=0.1 g` (d) Weight of oxygen `=32xx0.1=3.2 g` (e) Weight of water `=1 g` Thus, 0.1 g silver is the lightest.

Discussion

542.	250 ml of a sodium carbonate solution contains 2.65 grams of `Na_(2)CO_(3)`. If 10 ml of this solution is diluted to one litre, what is the concentration of the resultant solution (mol. Wt. of Na_(a)`CO_(3)=106`)A. 0.1 MB. 0.001 MC. 0.01 MD. `10^(-4)` M
Answer» Correct Answer - B Molarity `=(W(gm)xx1000)/("molecular wt."xxV(ml.))=(2.65xx1000)/(106xx250)=0.1 M`. 10 ml of this solution is diluted to 100 ml `M_(1)V_(1)= M_(2)V_(2)` `10xx0.1=1000xx x` `x=(0.1xx10)/1000=0.001 M`

Discussion

543.	250 ml of a sodium carbonate solution contains 2.65 grams of `Na_(2)CO_(3)`. If 10 ml of this solution is diluted to one litre, what is the concentration of the resultant solution (mol. Wt. of Na_(a)`CO_(3)=106`)A. (a)0.1 MB. (b)0.001 MC. (c )0.01 MD. (d)`10^(-4)M`
Answer» Correct Answer - B Molarity`=(W(g)xx1000)/("molecular wt". xxV(ml.))` `=(2.65xx1000)/(106xx250)=0.1 M` `10 ml` of this solution is diluted to `1000 ml N_(1)V_(1)=N_(2)V_(2)` `10xx0.1-1000xxx` `x=(0.1xx10)/1000=0.001 M`

Discussion

544.	Assertion : Molecular mass of A is `M/4` if the molecular mass of B is M. Reason : Vapour density of A four times that of B.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false.
Answer» Correct Answer - C Vapour density of `B=M/2` Vapour density of `A=M/4xx1/2 =M/8` Molecular mass of `A=M/4`.

Discussion

545.	At room temperature ratio of pressures of `CH_(4)` and `CO_(2)` kept in two separate containers of equal volume is `3:5`. Then two containers have equal number of-A. (a)molesB. (b)electronsC. (c )atomsD. (d)molecules
Answer» Correct Answer - C `P prop nimpliesn_(CH_(4))/(n_(CO_(2)))=3/5` `implies("no. of atoms in container of" CH_(4))/("no. of atoms in container of" CO_(2))=(3xx5)/(5xx3)=1` `implies` Both containers have equal no. of atoms

Discussion

546.	The valency factor of `I_(2)` when, (i) it is formed by the reaction of potassium iodide and potassium iodate in acid medium and (ii) when it reacts with hypo, are respectively:A. (a)`2, 2`B. (b)`5/3, 2`C. (c )`3/5, 2`D. (d)`5, 2`
Answer» Correct Answer - B `IO_(3)^(-)+I^(-)+H^(+) rarr I_(2)+H_(2)O` `I_(2)` can be said to have undergone disproportionation. :. V.f. of `I_(2)` for oxidation, `n_(1)=2(5-0)=10` and v.f. of `I_(2)` for reduction, `n_(2)=2(0-(-1))=2` `:. V.f. of I_(2) (overall) (n_(1)n_(2))/(n_(1)+N_(2))=(10xx2)/(10+2)=5/3` (ii) `I_(2)+2Na_(2)S_(2)O_(3) rarr Na_(a)S_(4)O_(6)+2Nal` `I_(2)+2e^(-) rarr 2I^(-) implies v.f.=2`

Discussion

547.	A metal M of equivalent mass E forms an oxide of molecular formula `M_(x)O_(y)`. The atomic mass of the metal is given by the correct equation.A. `2E(y//x)`B. `xyE`C. `E//y`D. `y//E`
Answer» Correct Answer - A

Discussion

548.	18 carat gold containsA. `60%` goldB. `75%` goldC. `18%` goldD. `60%` gold
Answer» Correct Answer - B `100%` gold is 24 carat. Thus, by unitary method, 18 carat gold is equivalent to `(18)/(24) xx 100% = 75%` gold.

Discussion

549.	The number of sodium atoms in 2 moles of sodium ferrocyanide isA. `2`B. `602xx10^(23)`C. `8xx602xx10^(23)`D. `4xx6*02xx10^(23)`
Answer» Correct Answer - C The formula of sodium ferrocyanide is `Na_(4)Fe[(CN)_(6)]`. The number of Na atoms in 2 mol of compound `= 2xx4` mol `= 8xx6*02xx10^(23)`

Discussion

550.	Which of the following will not have a mass of 10 g?A. `01` mol `CaCO_(3)`B. `151xx10^(23)` `Ca^(2+)` ionsC. `016` mol of `CO_(3)^(2)` ionsD. `7525xx10^(22)` Br atom.
Answer» Correct Answer - C `0.1` mole `CaCO_(3) = 01xx100 = 10g` `Ca^(2+) = (1.51xx10^(23))/(6023xx10^(23))xx40=10g` `CO_(3)^(2-) = 016xx60 = 96 g` `Br = (7525xx10^(22))/(602xx10^(23))xx80=10g`

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In which reaction there is a change in valencyA. `2NO_(2) rarr N_(2)O_(4)`B. `2NO_(2)+H_(2)O rarr HNO_(2)+HNO_(3)`C. `NH_(4)OH rarr NH_(4)^(+)+OH^(-)`D. `CaCO_(3) rarr CaO+CO_(2)`

A sample of `AIF_(3)` contains `3.0 xx 10^(24)` `F^(-)` ions. The number of formula units of the sample areA. `9*0xx10^(24)`B. `3*0xx10^(24)`C. `0*75xx10^(24)`D. `1*0xx10^(24)`

The compound `Yba_2Cu_3O_7` which shows super conductivity has copper in oxidation state_____. Assume that the rare earth element yttrium is in its usual `+3` oxidation state.A. `3//7`B. `7//3`C. 3D. 7

The mass of 70% `H_(2)SO_(4)` required for neutralization of one mole of NaOH is:A. 49 gB. 98 gC. 70 gD. `34*3 g`

0.126 g of acid required 20 mL of 0.1 N NaOH for complete neutralisation. The equivalent mass of an acid isA. `45`B. `53`C. `40`D. `63`

Molarity of liquid HCl with density equal to `1.17 g//mL` is:A. (a)`36.5`B. (b)`18.25`C. (c )`32.05`D. (d)`4.65`

With increase of temperature, which of these changes?A. MolalityB. Weight fraction of soluteC. Fraction of solute present in 1L of waterD. Mole fraction

How many moles of lead (II) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g of HCl?A. `0.333`B. `0.011`C. `0.029`D. `0.044`

The number of moles of `KMnO_(4)` that will be needed to react with one mole of sulphite ion in acidic solution isA. `4//5`B. `2//5`C. `1`D. `3//5`

How many molecules are there in `3.46g` sample of hydrogen chloride `(HCl)`?A. `3.72xx10^(22)`B. `4.87xx10^(22)`C. `1.95xx10^(22)`D. `3.786xx10^(22)g`

One atom mass unit (amu) is equivalent toA. `3.786xx10^(-24)g`B. `1.661xx10^(-24)g`C. `2.687xx10^(-24)g`D. `5.099xx10^(-24)g`

If the density of water is 1 g `cm^(-3)` then the volume occupied by one molecule of water is approximatelyA. (a)`6.023xx10^(-3) cm^(3)`B. (b)`3.0xx10^(-3) cm^(3)`C. (c )`5.5xx10^(-23) cm^(3)`D. (d)`9.0xx10^(-23) cm^(3)`

A mixture of gases contains `H_(2)` and `O_(2)` gases in the ratio of `1:4 (w//w)`. What is the molar ratio of the two gases in the mixture?A. (a)`16:1`B. (b)`2:1`C. (c )`1:4`D. (d)`4:1`

`100 g CaCO_(3)` reacts with `1 litre 1 N HCl`. On completion of reaction how much weight of `CO_(2)` will be obtainA. (a)5.5 gB. (b)11 gC. (c )22 gD. (d)33 g

What is the equivalent mass of ` IO_4^(-)` when it is converted into `I_2` in acid medium ?A. `M//6`B. `M//7`C. `M//5`D. `M//4`

How many atoms are contained in one mole of sucrose `(C_(12)H_(22)O_(11))`?A. `45xx6.02xx10^(23)` atoms/moleB. `5xx6.62xx10^(23)`atoms/moleC. `5xx6.02xx10^(23)` atoms/moleD. None of these

The weight of one molecule of a compound `C_(60)H_(12)` isA. `1.3xx10^(-20)`gB. `5.01xx10^(-21)`gC. `3.72xx10^(23)`gD. `1.4xx10^(-21)`g

Define second.

A piece of metal is 3 inch (represented by in) long. What is its length in cm?

A jug contains 2L of milk. Calcualte the volume of the milk in `m^(3)`

Number of atoms of oxygen present in 10.6 g of `Na_(2)CO_(3)` will beA. `6.02xx10^(22)`B. `12.04xx10^(22)`C. `1.806xx10^(23)`D. `31.80xx10^(28)`

4.4 g of `CO_(2)` contains how many litres of `CO_(2)` at S.T.PA. `2.4 L`B. `2.24L`C. `44L`D. `22.4L`

The volume of 1.0 g of hydrogen in litres at N.T.P. isA. `2.24`B. `22.4`C. `1.12`D. `11.2`

Volume of 0.1 M `K_(2)Cr_(2)O_(7)` required to oxidise 35 mL of 0.5 M Fe `SO_(4)` solution isA. 29.2 mLB. 17.5 mLC. 175 mLD. 145 mL

`6.0xx10^(20)` molecules of urea are present in 100 L of his solution. The concentration of urea solution isA. 0.001 MB. 0.1 MC. 0.02 MD. 0.01 M

The answer to the following problem in standard exponential form is :`(2.0xx10^(13)) + (1.5xx10^(14))`A. `3.5xx10^(13)`B. `3.5xx10^(14)`C. `2.15xx10^(13)`D. `1.7xx10^(14)`

To differentiat between C-12, C-13 and C-14 the instrument that you would use isA. infra-red spectrometerB. atomic absorption spectrometerC. mass spectrometerD. ultraviolet spectrometer

Accurate determination of atomic masses is done with the instrument called asA. spectrophotometerB. mass spectrometerC. atomic absorption spectrometerD. calorimeter

Which among the following is the heaviest?A. one mole of oxygenB. one molecule of sulphur dioxideC. 100 a.m.u. of uraniumD. ten moles of hydrogen

What is the simplest formula of the compound which has the following percentage composition - Carbon 80%. Hydrogen 20% ? If the molecular mass is 30, calculate its molecular formula.

When pentane , `C_(5) H_(12)` is burned in excess oxygen , the products of the reaction are `CO_(2)(g)` and `H_(2)O (l)` . The balanced equation for this combustion is `C_(5) H_(12)(g) + xO_(2)(g) rarr 5CO_(2)(g) + 6 H_(2) O (l)` The coefficient (x) of oxygen should beA. 16B. 12C. 11D. 8

Which one of the following statements is false ?A. An element of a substance contains only one kind of atomsB. Milk is a homogenous mixtureC. A compound can be decomposed into its constituentsD. All homogenous mixtures are called solutions

The number of g-atom of oxygen in `6.02xx10^(24) CO` molecule isA. 1B. 0.5C. 5D. 10

Sodium nitrate on reduction with Zn in presence of NaOH solution produces `NH_(3)`. Mass of sodium nitrate nitrate absorbing 1 mole of electron will beA. `7.750`B. `10.625`C. `8.000`D. `9.875`

0.078 g of a hydrocarbon occupy 22.414 mL of volume at S.T.P. The empirical formula of the hydrocarbon is CH. The molecular formula of hydrocarbon isA. `C_(2)H_(2)`B. `C_(6)H_(6)`C. `C_(8)H_(8)`D. `C_(4)H_(4)`

The empirical formula of a compound is `CH_2O_2`. What could be its molecular formula?A. `C_2H_2O_2`B. `C_2H_2O_4`C. `C_2H_4O_4`D. `CH_4O_4`

A gas has molecular formula `(CH)_n`. If vapour density of the gas is 39, what should be the formula of the compound ?A. `C_3H_3`B. `C_4H_4`C. `C_2H_2`D. `C_6H_6`

Which one of the following is the lightestA. 0.2 mole of hydrogen gasB. `6.023xx10^(22)` molecules of nitrogenC. 0.1 g of silverD. 0.1 mole of oxygen gas

250 ml of a sodium carbonate solution contains 2.65 grams of `Na_(2)CO_(3)`. If 10 ml of this solution is diluted to one litre, what is the concentration of the resultant solution (mol. Wt. of Na_(a)`CO_(3)=106`)A. 0.1 MB. 0.001 MC. 0.01 MD. `10^(-4)` M

250 ml of a sodium carbonate solution contains 2.65 grams of `Na_(2)CO_(3)`. If 10 ml of this solution is diluted to one litre, what is the concentration of the resultant solution (mol. Wt. of Na_(a)`CO_(3)=106`)A. (a)0.1 MB. (b)0.001 MC. (c )0.01 MD. (d)`10^(-4)M`

At room temperature ratio of pressures of `CH_(4)` and `CO_(2)` kept in two separate containers of equal volume is `3:5`. Then two containers have equal number of-A. (a)molesB. (b)electronsC. (c )atomsD. (d)molecules

The valency factor of `I_(2)` when, (i) it is formed by the reaction of potassium iodide and potassium iodate in acid medium and (ii) when it reacts with hypo, are respectively:A. (a)`2, 2`B. (b)`5/3, 2`C. (c )`3/5, 2`D. (d)`5, 2`

A metal M of equivalent mass E forms an oxide of molecular formula `M_(x)O_(y)`. The atomic mass of the metal is given by the correct equation.A. `2E(y//x)`B. `xyE`C. `E//y`D. `y//E`

18 carat gold containsA. `60%` goldB. `75%` goldC. `18%` goldD. `60%` gold

The number of sodium atoms in 2 moles of sodium ferrocyanide isA. `2`B. `6*02xx10^(23)`C. `8xx6*02xx10^(23)`D. `4xx6*02xx10^(23)`

Which of the following will not have a mass of 10 g?A. `0*1` mol `CaCO_(3)`B. `1*51xx10^(23)` `Ca^(2+)` ionsC. `0*16` mol of `CO_(3)^(2)` ionsD. `7*525xx10^(22)` Br atom.

A sample of `AIF_(3)` contains `3.0 xx 10^(24)` `F^(-)` ions. The number of formula units of the sample areA. `90xx10^(24)`B. `30xx10^(24)`C. `075xx10^(24)`D. `10xx10^(24)`

The number of sodium atoms in 2 moles of sodium ferrocyanide isA. `2`B. `602xx10^(23)`C. `8xx602xx10^(23)`D. `4xx6*02xx10^(23)`

Which of the following will not have a mass of 10 g?A. `01` mol `CaCO_(3)`B. `151xx10^(23)` `Ca^(2+)` ionsC. `016` mol of `CO_(3)^(2)` ionsD. `7525xx10^(22)` Br atom.