178 + Interview Questions in MOLE-AND-EQUIVALENT-CONCEPT IN GENERAL AWARENESS

1.	What volume of water is required to make `0.20N` solution from `1600 mL` of `0.2050N` solution?
Answer» Correct Answer - `40 mL`;

Discussion

2.	(i) Butyric acid contains only `C, H` and `O.A4.24 mg` sample of butyric acid is completely burned. It gives `8.45 mg` of carbon dioxide `(CO_(2))` and `3.46 mg` of water. What is the mass percentage of each element in butyric acid? (ii) If the elemental composition of butyric acid is found to be `54.2% C, 9.2% H` and `36.6%O`, determine the empirical formula. (iii) The molecular mass of butyric acid was determined of experiment to be `88`. What is the moleculare formula ?
Answer» (i) `%C(12xx8.45)/(44xx4.24)xx100= 54.35%` `%H=(2xx3.46)/(18xx4.24)xx100= 9.06%` `%O=[100-54.35-9.06]=36.59%` (ii) `\|{:(%,%at.wt.="Factor","factor/Lower number"),(C=54.2,(54.2)/(12)=4.52,(4.52)/(2.29)=2),(H=9.2,(9.2)/(1)=9.2,(9.2)/(2.29)=4),(O=36.6,(36.6)/(16)=2.29,(2.29)/(2.29)=1):}:\|` `:.` Empirical forumula of butyric acid is `C_(2)H_(4)O` (iii) Empirical formula mass `=44` Molecular mass `=88` `("Molecular mass")/("Empirical mass")=(88)/(44)=2` `:. "Molecular formula"="Empirical forumula"xxn` `=C_(4)H_(8)O_(2)`

Discussion

3.	If half mole of oxygen combine with `Al` to form `Al_(2)O_(3)` the weight of `Al` used in the reaction is:A. `27 g`B. `40.5 g`C. `54 g`D. `18 g`
Answer» Correct Answer - D `4Al+3O_(2)rarr2Al_(2)O_(3)`

Discussion

4.	A saturated solution is prepared at `70^(circ)C` containing `32.0g CusO_(4).5H_(2)O per 100 g` solution. A `335 g` sample of this solution is then cooled to `0^(circ)C` so that. `CuSO_(4).5H_(2)O` crystallises out. If the concentration of a saturated solution at `0^(circ)C` is `12.5g CuSO_(4).5H_(2)O per 100.0 g` solution, how much of `CuSO_(4).5H_(2)O` is crystallised?
Answer» Correct Answer - `74.47 g`;

Discussion

5.	What volume of `0.20M H_(2)SO_(4)` is required to produce `34.0 g` of `H_(2)S` by the reaction? `8KI+5H_(2)SO_(4)rarr4K_(2)SO_(4)+4I_(2)+H_(2)S+4H_(2)O`
Answer» Correct Answer - `25 "litre"`;

Discussion

6.	The weight of `1 xx 10^(22)` molecules of `CuSO_(4). 5H_(2)O` isA. `4.144 g`B. `5.144 g`C. `6.144 g`D. None of these
Answer» Correct Answer - A Mol.wt.of `CuSO_(4).5H_(2)O=249.6` `:.` Mass of `1xx10^(22)` molecules `=(249.6xx110^(22))/(6.023xx10^(23))=4.144 g`

Discussion

7.	Chlorophyll, a green colouring matter contains `2.68% Mg`. The number of atoms of `Mg` present in `1 g` chlorophyll are :A. `6.72 xx 10^(20)`B. `6.72xx10^(21)`C. `6.72 xx 10^(22)`D. `6.72 xx 10^(23)`
Answer» Correct Answer - A `100 g "chlorophyll"-=2.68 g Mg` `1 g "chlorophyll"-=2.68xx10^(-2) g Mg` `=2.68xx10^(-2)xx(6.023xx10^(23))/(24)"atoms" Mg` `=6.72xx10^(20) "atoms" Mg`

Discussion

8.	Copper forms two oxides. For the same amount of copper, twice as much oxygen was used to form first oxide than to form second one. What is the ratio of the valencies of copper in first and second oxides?
Answer» Correct Answer - `2:1`;

Discussion

9.	A metal oxide has `40%` oxygen. The equivalent weight of the metal is:A. `12`B. `16`C. `24`D. `48`
Answer» Correct Answer - A Meq. of metal = Meq. of oxygen `(60)/(E )=(40)/(8)`, `E=12`

Discussion

10.	Chemical reactions involve interation of atoms and molecules. A large number of atoms `//` molecules `(` approximately `6.023xx10^(23))` are present in a few grams of any chemical compound varying with their atomic `//` molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry, and radiochemistry. The following example illustrates a typical case, involving chemical `//` electrochemical reaction, which requires a clear understanding of the mole concept. A `4.0M` aqueous solution of `NaCl` is prepared and `500mL` of this solution is electrolyzed. This leads to the evolution of chlorine gas at one of the electrodes `(` atomic mass of `Na ` is 23 and `Hg` is `200)(1F=96500C)`. If the cathode is an `Hg` electrode, the maximum weight `(in g)` of amalgam formed from this solution isA. `200`B. `225`C. `400`D. `446`
Answer» Correct Answer - D `Na-Hg` (amalgam) formed `=2` moles at cathode. `2` mole of amalgam `=23xx2+2xx200` `=446`

Discussion

11.	Chemical reactions involve interation of atoms and molecules. A large number of atoms `//` molecules `(` approximately `6.023xx10^(23))` are present in a few grams of any chemical compound varying with their atomic `//` molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry, and radiochemistry. The following example illustrates a typical case, involving chemical `//` electrochemical reaction, which requires a clear understanding of the mole concept. A `4.0M` aqueous solution of `NaCl` is prepared and `500mL` of this solution is electrolyzed. This leads to the evolution of chlorine gas at one of the electrodes `(` atomic mass of `Na ` is 23 and `Hg` is `200)(1F=96500C)`. The total charge `(` coulomb `)` required for complete electrolysis isA. `24125`B. `48250`C. `96500`D. `193000`
Answer» Correct Answer - D `2 "moles"` of electrons ( `2` Faraday) are required. `because 1 F= 96500 "coulombs"` `therefore 2F=193000"coulombus"`

Discussion

12.	A mixture of ethane `(C_(2)H_(6))` and ethene `(C_(2)H_(4))` occupies `40 L` at `1.00 atm` and at `400 K`. The mixture reacts completely with `130 g` of `O_(2)` to produce `CO_(2)` and `H_(2)O`. Assuming ideal gas behaviour, calculate the mole fractions of `C_(2)H_(4)` and `C_(2)H_(6)` in the mixture.
Answer» For a gaseous mixture of `C_(2)H_(6)` and `C_(2)H_(14)` `PV=nRT` `:. 1xx40=nxx0.082xx400` `:.` Total mole of `(C_(2)H_(6)+C_(2)H_(4))=1.2195` Let mole `C_(2)H_(6)` and `C_(2)H_(4)` be `a` and `b` respectively. `a+b=1.2195`….(1) `C_(2)H_(6)+(7//2)O_(2)rarr2CO_(2)+3H_(2)O` `C_(2)H_(4)+3O_(2)rarr2CO_(2)+2H_(2)O` `:.` Mole of `O_(2)` needed for complete reaction of mixture `=7a//2+3b` `:. (7a)/(2)+3b=(130)/(32)`....(2) By eqs.(1) and (2), `a=0.808, b=0.4115` `:.` Mole fraction of `C_(2)H_(6)-0.808//1.2195=0.066` and Mole fraction of `C_(2)H_(4)=0.34`

Discussion

13.	Calculate the concentration of a solution obtained by mixing `300 g` of `25%` by weight solution of `NH_(4)Cl` and `150 g` of `40%` by weight solution of `NH_(4)Cl`.
Answer» Weight of solution I and II `=300+150=450 g` Weight of `NH_(4)Cl` in II solution `=(40xx150)/(100)=60 g` `:.` Total weight of `NH_(4)Cl= 75+60= 135 g` `:. %` by wt. of mixed solution `=(135)/(450)xx100= 30%`

Discussion

14.	The `%` composition of `NH_(3), H_(2)O` and `N_(2)O_(3)` is as given below: `NH_(3) rarr 82.35% N` and `17.65%H` `H_(2)O rarr 88.90% ` and `11.10% H` `N_(2)O_(3) rarr 63.15%O` and `36.85%N` On the basis of above data prove law of reciprocal proportions.
Answer» (i) For `NH_(3)`: `1` part hydrogen reacts with `=(82.35)/(17.65)= 4/67` Part `N` (ii) For `H_(2)O` `1` part hydrogen reacts with `=(88.9)/(11.10)= 8.01` part `O` Thus the ratio `N:O:4.67:8.01implies0.58` (iii) For `N_(2)O_(3)`: In which `N` and `O` reacts with each other and ratio `N:O::36.85:63.15~~0.58` beacuse the two ratio are saem, thus law of reciprocal proportions is correct.

Discussion

15.	A mixture of `NH_(3(g))` and `N_(2)H_(4_((g)))` is placed in a sealed container at `300 K`. The total pressure is `0.5 atm`. The container is heated to `1200 K`, at which time both substances decompose completely according to the equations: `2NH_(3(g))rarrN_(2(g))+3H_(2(g))` `N_(2)H_(4_((g)))rarrN_(2(g))+2H_(2(g))` After decomposition is complete, the total pressure at `1200 K` is found to be `4.5 atm`. Find the amount (mole) per cent of `N_(2)H_(4(g))` in the original mixture.
Answer» Correct Answer - `25%`;

Discussion

16.	Calculate the `%` of free `SO_(3)` in oleum ( a solution of `SO_(3)` in `H_(2)SO_(4)`) that is labelled `109% H_(2)SO_(4)` by weight.
Answer» Correct Answer - `40%`;

Discussion

17.	When a metal is burnt, its weight is increased by `24%`. The equivalent weight of the metal wil be:A. `25`B. `24`C. `33.3`D. `76`
Answer» Correct Answer - C Eq. of metal `=`Eq.of example or `(100)/(E )=(24)/(8)` `:. E= 33.3`

Discussion

18.	`0.71 g` of chlorine combines with certain weight of a metal giving `1.11 g` of its chloride. The eq.wt. of the metal is:A. `40`B. `20`C. `80`D. none of these
Answer» Correct Answer - B Let metal chloride be `MCl_(n)`, `:.` Eq. of metal `=` Eq. of choride or `or (1.11-0.71)/(E )=(0.71)/(35.5)` `:. E=20`

Discussion

19.	Eq.wt. of an acid salt `NaHSO_(4)` is:A. `M//1`B. `M//2`C. `M//3`D. none of these
Answer» Correct Answer - A It is an acid salt, Only `1 H` its replaceable.

Discussion

20.	`A g` of a metal displaces `VmL of H_(2)` at `NTP` Eq.wt. of metal, `E` is (are):A. `E=(Axx1.008xx22400)/("Vol.of" H_(2)"displaced"xx2)`B. `E=(AxxEq."mass .of"H)/("mass of"H_(2)"displaced")`C. `E=(Axx1.008)/("Vol.of displaced"xx0.000897)`D. None of these
Answer» Correct Answer - A::B::C

Discussion

21.	Assuming `100%` ionisation, the solution having highest normality is:A. `1 M H_(2)SO_(4)`B. `1 M H_(3)PO_(3)`C. `1 M H_(3)PO_(4)`D. `1 M HNO_(3)`
Answer» Correct Answer - C `H_(3)PO_(4)` is tribasic, thus `1 MH_(3)PO_(4)-=3NH_(3)PO_(4) at 100% "ionisation"`

Discussion

22.	Insulin contains `3.4%` sulphur. Calculate minimum mol.wt. of insulin.A. `941.176`B. `944`C. `945.27`D. none of these
Answer» Correct Answer - A `3.4 gS=100 g "insulin"` `:. 32 g S=(100xx32)/(3.4)=941.176` Insulin must contain at least one atom of `S` in its one molecule.

Discussion

23.	How many moles of `NaOH` are contained in `27 mL` of `0.15 M NaOH`?
Answer» Milli-moles of `NaOH=MxxV_("in"mL)=0.15xx27` `:.` Moles of `NaOH=(0.15xx27)/(1000)= 4.05xx10^(-3)`

Discussion

24.	How many atoms are contained in a mole of `Ca(OH)_(2)`?A. `30xx6.02xx10^(23)` atoms/molB. `5xx6.02xx10^(23)`atoms/molC. `6xx6.02xx10^(23)`atoms/molD. none of these
Answer» Correct Answer - B `1 "molecule of" Ca(OH)_(2)` contains `5 "atoms"`, `:. 1 "mole of" Ca(OH)_(2)` contains `5 N` atoms.

Discussion

25.	`0.5` mole of `H_(2)SO_(4)` is mixed with `0.2` mole of `Ca(OH)_(2)`. The maximum number of mole of `CaSO_(4)` formed is:A. `0.2`B. `0.5`C. `0.4`D. `1.5`
Answer» Correct Answer - A Eq. of `H_(2)SO_(4)= 0.5xx2=1.0`, Eq. of `Ca(OH)_(2)=0.2xx2=0.4`, Thus, Eq. of `CaSO_(4)` formed `=0.4` `:. "Mole of" CaSO_(4) "formed" =(0.4)/(2)=0.2`

Discussion

26.	Statement `1 "mole" O_(3)=N "molecule"O_(3)=3N` atoms of `O=48 g` Explanation A mole is the amount of matter that contains as many as objects as the amount of atoms exactly in `12 g C^(12)`.A. `S` is correct but `E` is wrong.B. `S` is wrong but `E` is wrong.C. Both `S` and `E` are correct and `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S.
Answer» Correct Answer - C Explanation is correct reason for statements.

Discussion

27.	Chlloride samples are prepared for analysis by using NaCl, KCl, `NH_4Cl` separately or as mixtures. What minimum volume of a `5.0%` by weight `AgNO_3` Solution (Density`=1.04`) must be added to a sample wehging 0.3 g in order to ensure complete precipitation of choride in every possible cases?
Answer» Correct Answer - `18.33 mL`;

Discussion

28.	Weight of 1 litre milk is `1.032 kg`. It contains butter fat (density `865 kg m^(-3))` to the extent of `4.0%` by wt`//`volume. Calculate the density of the fat -free skimmed milk.
Answer» Let `100 m^(3)` milk contains `4m^(3)` fat. `:.` Weight of butter fat in `1 m^(3)` milk `=(4)/(100)xx865= 35 kg` Also Weight of `10^(3)cm^(3)` milk `=1.032 kg` `:.` Weight volume of skimmed milk `= 1.0-0.04=0.96m^(3)` `:.` Density of fat free skimmed milk`=(997)/(0.96)` `=10 38.5 kg//m^(3)`

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

What volume of water is required to make `0.20N` solution from `1600 mL` of `0.2050N` solution?

If half mole of oxygen combine with `Al` to form `Al_(2)O_(3)` the weight of `Al` used in the reaction is:A. `27 g`B. `40.5 g`C. `54 g`D. `18 g`

What volume of `0.20M H_(2)SO_(4)` is required to produce `34.0 g` of `H_(2)S` by the reaction? `8KI+5H_(2)SO_(4)rarr4K_(2)SO_(4)+4I_(2)+H_(2)S+4H_(2)O`

The weight of `1 xx 10^(22)` molecules of `CuSO_(4). 5H_(2)O` isA. `4.144 g`B. `5.144 g`C. `6.144 g`D. None of these

Chlorophyll, a green colouring matter contains `2.68% Mg`. The number of atoms of `Mg` present in `1 g` chlorophyll are :A. `6.72 xx 10^(20)`B. `6.72xx10^(21)`C. `6.72 xx 10^(22)`D. `6.72 xx 10^(23)`

Copper forms two oxides. For the same amount of copper, twice as much oxygen was used to form first oxide than to form second one. What is the ratio of the valencies of copper in first and second oxides?

A metal oxide has `40%` oxygen. The equivalent weight of the metal is:A. `12`B. `16`C. `24`D. `48`

Calculate the concentration of a solution obtained by mixing `300 g` of `25%` by weight solution of `NH_(4)Cl` and `150 g` of `40%` by weight solution of `NH_(4)Cl`.

The `%` composition of `NH_(3), H_(2)O` and `N_(2)O_(3)` is as given below: `NH_(3) rarr 82.35% N` and `17.65%H` `H_(2)O rarr 88.90% ` and `11.10% H` `N_(2)O_(3) rarr 63.15%O` and `36.85%N` On the basis of above data prove law of reciprocal proportions.

Calculate the `%` of free `SO_(3)` in oleum ( a solution of `SO_(3)` in `H_(2)SO_(4)`) that is labelled `109% H_(2)SO_(4)` by weight.

When a metal is burnt, its weight is increased by `24%`. The equivalent weight of the metal wil be:A. `25`B. `24`C. `33.3`D. `76`

`0.71 g` of chlorine combines with certain weight of a metal giving `1.11 g` of its chloride. The eq.wt. of the metal is:A. `40`B. `20`C. `80`D. none of these

Eq.wt. of an acid salt `NaHSO_(4)` is:A. `M//1`B. `M//2`C. `M//3`D. none of these

`A g` of a metal displaces `VmL of H_(2)` at `NTP` Eq.wt. of metal, `E` is (are):A. `E=(Axx1.008xx22400)/("Vol.of" H_(2)"displaced"xx2)`B. `E=(AxxEq."mass .of"H)/("mass of"H_(2)"displaced")`C. `E=(Axx1.008)/("Vol.of displaced"xx0.000897)`D. None of these

Assuming `100%` ionisation, the solution having highest normality is:A. `1 M H_(2)SO_(4)`B. `1 M H_(3)PO_(3)`C. `1 M H_(3)PO_(4)`D. `1 M HNO_(3)`

Insulin contains `3.4%` sulphur. Calculate minimum mol.wt. of insulin.A. `941.176`B. `944`C. `945.27`D. none of these

How many moles of `NaOH` are contained in `27 mL` of `0.15 M NaOH`?

How many atoms are contained in a mole of `Ca(OH)_(2)`?A. `30xx6.02xx10^(23)` atoms/molB. `5xx6.02xx10^(23)`atoms/molC. `6xx6.02xx10^(23)`atoms/molD. none of these

`0.5` mole of `H_(2)SO_(4)` is mixed with `0.2` mole of `Ca(OH)_(2)`. The maximum number of mole of `CaSO_(4)` formed is:A. `0.2`B. `0.5`C. `0.4`D. `1.5`

Chlloride samples are prepared for analysis by using NaCl, KCl, `NH_4Cl` separately or as mixtures. What minimum volume of a `5.0%` by weight `AgNO_3` Solution (Density`=1.04`) must be added to a sample wehging 0.3 g in order to ensure complete precipitation of choride in every possible cases?

Weight of 1 litre milk is `1.032 kg`. It contains butter fat (density `865 kg m^(-3))` to the extent of `4.0%` by wt`//`volume. Calculate the density of the fat -free skimmed milk.