1153 + Interview Questions in MOLE-CONCEPT IN GENERAL AWARENESS Page 15 InterviewSolution

701.	The atomic weights of two alements A and B are 40 and 80 reapectively. If x g of A contains y atoms, how many atoms are present in 2x g of B?A. `y/2`B. `y/4`C. yD. 2y
Answer» Correct Answer - C

Discussion

702.	How may millilitres of a `9 N H_(2)SO_(4)` solution will be required to neutralize completely `20mL` of a `3.6 N NaOH` solution ?A. `18.0mL`B. `8. 0 Ml`C. `16.0mL`D. `80.0 mL`
Answer» Correct Answer - 2 `N_(NaOH)=3.6" "," "V_(NaOH)=20" "," "N_(H_(2)SO_(4))=9" "," "V_(H_(2)SO_(4))=?` `V_(H_(2)SO_(4))=(3.6xx20)/(9)=8mL`

Discussion

703.	Calculate percentage change in `M_(avg)` of the mixture, if `PCl_(5)` undergo `50%` decomposition. `PCl_(5) rArr PCl_(3) + Cl_(2)`A. 0.5B. 0.6666C. 0.3333D. zero
Answer» Correct Answer - C

Discussion

704.	When 22.4 litres of `H_(2)(g)` is mixied with 11.2 litres of `Cl_(2)(g)`, each at S.T.P, the moles of HCl (g) formed is equal to :A. 0.5 mol of HCl (g)B. 1.5 mol of HCl (g)C. 1.0 mol of HCl (g)D. 2.0 mol of HCl (g)
Answer» Correct Answer - C No. of moles in 22.4 L of `H_(2)(g)` at S.T.P = 1 mol No. of moles in 11.2 L of `Cl_(2)(g)` at S.T.P = 0.5 mol `underset("1 mol")(H_(2)(g))+underset("1 mol")(Cl_(2)(g))rarrunderset("2 mol")(2HCl(g))` `Cl_(2)(g)` is the limiting reactant since it has only 0.5 mole available for chemical reaction `:.` No. of moles of HCl (g) formed = 1mol.

Discussion

705.	The litres of `CO_(2)` represented by 4.4 g of `CO_(2)` at S.T.P are :A. 2.4 litresB. 2.24 litresC. 44 litresD. 22.4 litres
Answer» Correct Answer - B 44 g (molar mass) of `CO_(2)` represent = 22.4 L at S.T.P 4.4 of `CO_(2)` represent `= ((22.4L))/((44g))xx(4.4g)=2.24L`.

Discussion

706.	The average atomic mass of a mixture containing 79 mol % of `""^(24)Mg` and remaining 21 mole % of `""^(25)Mg` and `""^(26)Mg`, is 24.31,% mole of `(26)Mg` isA. 5B. 20C. 10D. 15
Answer» Correct Answer - C

Discussion

707.	How many litres of oxygen at 1 atm & and 273K will be required to burn completely 2.2 of propene `(C_(3)H_(8))`A. 11.2 LB. 22.4 LC. 5.6 LD. 44.8 L
Answer» Correct Answer - C

Discussion

708.	Calculate the number of molecules and number of atoms present in 11.2 litres of oxygen `(O_(2))` at N.T.P
Answer» Correct Answer - `3.01 xx 10^(23)` molecules, `6.02 xx 10^(23)` atoms. 22.4 litres of `O_(2)` at N.T.P contain `= 6.022xx10^(23)` molecules 11.2 litres of `O_(2)` at N.T.P contain `= (11.2)/(22.4)xx6.022 xx 10^(23)` molecules `= 3.01 xx 10^(23)` molecules `= 6.02 xx 10^(23)` atoms.

Discussion

709.	We know that balancing of a chemical equation is entirely bases on law of conservation of mass. However the concept of Principle of Atom Conservation (POAC) can also be related to law of consevation of mass in a chemical reaction. So, POAC can also act as a technique for balancing a chemical equation. For example, for a reaction: `ABC_(3) rarr AB+C_(2)` On applying POAC for A , B and C and related the equations , we get : `(n_(ABC_(3)))/(2) =(n_(AB))/(2) =(n_(C_(2)))/(3)` (`n_(x)` : number of moles of X) Thus , the cofficients of `ABC_(3)` , AB and `C_(2)` in the balanced chemical equation will be 2,2 and 3 respectively and the balanced chemical equation can be represented as , `2ABC_(3) rarr 2AB + 3C_(2)` If in the above question, the atomic masses of X and Y are 10 and 30 respectively, then the mass of `XY_(3)` formed when 120 g of `Y_(2)` reacts completely with X is:A. 133.3 gB. 200gC. 266.6gD. 400g
Answer» Correct Answer - A

Discussion

710.	In which of the following pairs, `10 g` of each have an equal number of molecules?A. `N_(2)O` and `CO`B. `N_(2)` and `C_(3)O_(2)`C. `N_(2)` and `CO`D. `N_(2)O` and `CO_(2)`
Answer» Correct Answer - C::D Equal number of moleucles are present when moles are same. For the same mass the molarcular weight has tobe same. Hence, `Mw "of" N_(2) = Mw "of" CO = 28 g` `Mw "of" N_(2) O = Mw "of" CO_(2) = 44 g`

Discussion

711.	The density of 2.0 M solution of a solute is `1.2 g mL^(-1)`. If the molecular mass of the solute is `100 g mol^(-1)`, then the molality of the solution isA. 2.0 mB. 1.2 mC. 1.0 mD. 0.6 m
Answer» Correct Answer - A 2 M solution has 2 moles of solute dissolved in one litre (1000 mL) of the solution Mass of the solution `= V xx d = 1000 mL xx 1.2 "g mL"^(-1) = 1200 g` Mass of solvent `= (1200 - 200)g = 1000g = 1kg` Molality of solution (m) `= ("No. of moles of solute")/("Mass of solvent in Kg")` `("2 mol")/("1 kg")=2 "mol Kg"^(-1)=2`m.

Discussion

712.	Which ofhte following will have maximum number of C atoms ?A. 5.8 g of glyoxal `(C_(2)H_(2)O_(2))`B. 3.1g of acetone` (C_(3)H_(6)O)`C. 11.6 g of fumaric (acid) `(C_(4)H_(4)O_(4))`D. 12 g of urea `(CON_(2)H_(4))`
Answer» Correct Answer - C

Discussion

713.	`6g` of a hydrocarbon on combustion in excess of oxygen produces `17.6g` of `CO_(2)` and `10.8g` of `H_(2)O`. The data illustrates the law of `:`A. conservation of massB. multiple proportionsC. constant proportionsD. reciprocal proportions
Answer» Correct Answer - 1 `{:(C_(X)H_(Y)+,,O_(2),rarr,CO_(2),+,H_(2)O),(6g,,excess,,,,):}` mass of Hydrocarbon `=6g` mass of carbon `=(12)/(44)xx` mass of `CO_(2)=(12)/(44)xx17.6=4.8g` mass of Hydrogen `=(2)/(18)xx` mass of `H_(2)O=(2)/(18)xx10.8=1.2g` Total mass of carbon & Hydrogen `=4.8+1.2=6g` Mass of Hydrogen `=` Total mass of `(C+H)=6g`. Law of conservation of mass.

Discussion

714.	What is the basic difference between empirical and molecular formulae ?
Answer» Empirical formula gives the simplest ratio of the atoms of different elements in the molecule of a substance but the molecular formula gives their actual ratio.

Discussion

715.	The chemical compounds are represented by both Empirical and Molecular formulae. Whereas the former is only theoritical, the latter is the actual formula of a compound. In some cases, the two may be even same. A very popular organic compound is a major constituent of alcoholic drinks. (i) What is the empirical of the compound ? (ii) What is its molecular formula ? (iii) Give the chemical name of the compound (iv) Find the percentage composition of the compound ?
Answer» (i) The empirical formula is `C_(2)H_(6)O` (ii) The molecular formula is also the same i.e., `C_(2)H_(6)O or C_(2)H_(5)OH` (iii) The compound is known as ethanol or ethyl alcohol (iv) Molecular mass of `C_(2)H_(5)OH=2xx` Atomic mass of C + Atomic mass of `O + 6 xx` Atomic mass of H `= 2xx12+16 +6 xx1=46` u Percentage of carbon `(C)=("Mass of C")/("Molecular mass of "C_(2)H_(5)OH)xx100=((24u))/((46u))xx100=52.2` Percentage of oxygen `(O)=("Mass of O")/("Molecular mass of "C_(2)H_(5)OH)xx((16u))/((46u))xx100=34.8` Percentage of hydrogen `(H)=("Mass of H")/("Molecular mass of "C_(2)H_(5)OH)xx100=((6u))/((46u))xx100=13.0`.

Discussion

716.	3.0 litre of water are added to 2.0 litre to 2.0 litre of 5 M HCl. What is the molarity of HCl in the resultant solution?
Answer» Correct Answer - 2

Discussion

717.	How much water should be added to `2M HCl` solution to form `1"litre"` of `0.5M HCl` ? .
Answer» Let `V` be initial volume Then mol of `HCl` = constant `2 xx V = 1 xx 0.5 " "rArr" " V =0.25 L` Volume of water added `=1 - 0.25 = 0.75 L` .

Discussion

718.	Gypsum is a hydrate of calcium sulphate 1.0 g of the sample contains 0.791 g of `CaSO_(4)`. How many of `CaSO_(4)` are there in the sample ? Assuming that the rest of sample is water, how many moles of water are there in the sample ? Show that the result is in consistent with the formula `CaSO_(4).2H_(2)O`
Answer» Mass of `CaSO_(4)=40 +32 + 64 = 136 "g mol"^(-1)` Molar mass of `CaSO_(4)=40+32+64=136 "g mol"^(-1)` Mass of water present in 1.0 g of the sample `= 1.0 - 0.791 = 0.209 g` Moles of `CaSO_(4)=("Mass of "CaSO_(4))/("Molar mass")=((0.791g))/((136"g mol"^(-1)))=5.816xx10^(-3)mol` Moles of water `= ("Mass of water")/("Molar mass")=((0.209g))/((18 gmol^(-1)))=11.6xx10^(-3)mol` Ratio of `CaSO_(4) : H_(2)O = 5.816 xx 10^(-3) : 11.6 xx 10^(-3) = 1:2` `:.` Molecular formula of hydrated salt `= CaSO_(4).2H_(2)O`.

Discussion

719.	100 ml of 5 M `NaOH` solution (density 1.2 g/ml) added to 200 mL of another `NaOH` solution which has a density of 1.5 g/ml and contains 20 mass percent of `NaOH.` What will be the volume of the gas (at STP) in litres liberated when aluminium reacts with this (final) solution? [Give answer excluding decimal places] The reaction is `Al+NaOH+H_(2)OrarrNaAlO_(2)+H_(2)` `(At. wt. Na=23, 0=16, H=1)`
Answer» Correct Answer - 68

Discussion

720.	1.0 g of metal nitrate gave 0.86 g of metal sulphate. Calculate the equivalent weight of metal.
Answer» `M (NO_(3))_(n) rarr M_(2) (SO_(4))_(n)` `:. Eq. of M (NO_(3))_(n) = Eq. of M_(2) (SO_(4))_(n)` `("Weight of M nitrate")/(Ew of M + Ew of NO_(3)^(ɵ)) = ("Weight. Of sulphate")/(Ew of M + Ew of SO_(4)^(2-))` `(1.0)/(E + (62)/(1)) = (0.86)/(E + (96)/(2))` `(1)/(E + 62) = (0.86)/(E + 48)` `E = 38`

Discussion

721.	34 gm of ammonia on decomposition gives some hydrogen gas along with `N_(2)` gas. Hydrogen underwent combustion with oxygen gas and water wa formed. Calculate number of drops of water formed if each drop contain 6 ml of water.
Answer» Correct Answer - 9

Discussion

722.	A mixture pf `H_(2) , N_(2)` and `O_(2)` occupying 100 ml underwent reaction so as to form `H_(2)O_(2)(l)` and `N_(2)H_(2)(g)` as the only products , causing the volume to contract by 60 ml. The remaining mixture was passed through pyrogallol causing a contraction of 10 ml . to the remaining mixture excess `H_(2)` was added and the above reaction was repeaped , causing a reduction in volume of 10 ml .(No other products are formed) What is the volume of `N_(20H_(2)(g)` formed in this reaction after adding excess of `H_(2)(g)`?A. 20 mlB. 30 mlC. 10 mlD. 40 ml
Answer» Correct Answer - C correct answer will be option B as answered in the following link https://www.sarthaks.com/1468419/mixture-and-occupying-100-underwent-reaction-as-to-form-h-o-and-n-h-as-the-only-products

722.

A mixture pf `H_(2) , N_(2)` and `O_(2)` occupying 100 ml underwent reaction so as to form `H_(2)O_(2)(l)` and `N_(2)H_(2)(g)` as the only products , causing the volume to contract by 60 ml. The remaining mixture was passed through pyrogallol causing a contraction of 10 ml . to the remaining mixture excess `H_(2)` was added and the above reaction was repeaped , causing a reduction in volume of 10 ml .(No other products are formed) What is the volume of `N_(20H_(2)(g)` formed in this reaction after adding excess of `H_(2)(g)`?A. 20 mlB. 30 mlC. 10 mlD. 40 ml

Answer» Correct Answer - C

correct answer will be option B

as answered in the following link

https://www.sarthaks.com/1468419/mixture-and-occupying-100-underwent-reaction-as-to-form-h-o-and-n-h-as-the-only-products

Discussion

723.	A mixture pf `H_(2) , N_(2)` and `O_(2)` occupying 100 ml underwent reaction so as to form `H_(2)O_(2)(l)` and `N_(2)H_(2)(g)` as the only products , causing the volume to contract by 60 ml. The remaining mixture was passed through pyrogallol causing a contraction of 10 ml . to the remaining mixture excess `H_(2)` was added and the above reaction was repeaped , causing a reduction in volume of 10 ml .(No other products are formed) What is the volume of `N_(20H_(2)(g)` formed in this reaction after adding excess of `H_(2)(g)`?
Answer» Correct Answer - 4 After first step reaction there exists excess O₂as evident by subsequent absorption by pyrogallol .. So H2 is not present there in excess after first step. After absorption of O2 the gas mixture should contain N2H2(g) and excess N2. This excess N2 is reacted with H2 added in excess to carry out 2nd step reaction according to the following equation N₂(g). + H₂(g) -> N₂H₂ (g) Where it is obvious that 1ml N₂ reacts with 1ml H₂ to produce 1ml N₂H₂(g) and contraction in volume will be 1ml. As the given contraction in volume is 10 ml the volume of N₂H₂ formed in 2nd step should be 10ml and the volume of N₂ consumed will be 10 ml also. Now we will proceed to estimate the volume of N₂H₂ (g) produced in first phase. To do this let us consider x ml H₂ goes to react with O₂ and y ml with N2 . x ml H₂gas will combine with 0.5x ml O₂ (g) to produce 0 ml H₂O(l) and the contraction due to this reaction will be 1.5x ml Again following the equation N₂(g). + H₂(g) -> N₂H₂ (g) we can say, y ml H₂will require y ml N₂ to react and will produce y ml N₂H₂ (g). The contraction in volume due to this is y ml So volume of N₂ spent in first step will be y ml. Summing up we can say ,the composition of the mixture at first is as follows volume of H₂=> (x+y)ml , volume of N₂ => (y+10) ml and volume of O₂=> (0.5x+10)ml So by the question we get (x+y)ml +(y+10) ml +(0.5x+10)ml = 100 ml =>1.5x +2y =80 ................(1) considering the total contraction we get 1.5x +y =60 ................(2) subtracting (2) from (1) we get y = 20 ml Hence total volume of N₂H₂ produced in two steps will be (10+20) ml = 30 ml

723.

A mixture pf `H_(2) , N_(2)` and `O_(2)` occupying 100 ml underwent reaction so as to form `H_(2)O_(2)(l)` and `N_(2)H_(2)(g)` as the only products , causing the volume to contract by 60 ml. The remaining mixture was passed through pyrogallol causing a contraction of 10 ml . to the remaining mixture excess `H_(2)` was added and the above reaction was repeaped , causing a reduction in volume of 10 ml .(No other products are formed) What is the volume of `N_(20H_(2)(g)` formed in this reaction after adding excess of `H_(2)(g)`?

Answer» Correct Answer - 4

After first step reaction there exists excess O₂as evident by subsequent absorption by pyrogallol .. So H2 is not present there in excess after first step. After absorption of O2 the gas mixture should contain N2H2(g) and excess N2. This excess N2 is reacted with H2 added in excess to carry out 2nd step reaction according to the following equation

N₂(g). + H₂(g) -> N₂H₂ (g)

Where it is obvious that 1ml N₂ reacts with 1ml H₂ to produce 1ml N₂H₂(g) and contraction in volume will be 1ml. As the given contraction in volume is 10 ml the volume of N₂H₂ formed in 2nd step should be 10ml and the volume of N₂ consumed will be 10 ml also.

Now we will proceed to estimate the volume of N₂H₂ (g) produced in first phase. To do this let us consider x ml H₂ goes to react with O₂ and y ml with N2 .

x ml H₂gas will combine with 0.5x ml O₂ (g) to produce 0 ml H₂O(l) and the contraction due to this reaction will be 1.5x ml

Again following the equation N₂(g). + H₂(g) -> N₂H₂ (g) we can say, y ml H₂will require y ml N₂ to react and will produce y ml N₂H₂ (g). The contraction in volume due to this is y ml So volume of N₂ spent in first step will be y ml.

Summing up we can say ,the composition of the mixture at first is as follows

volume of H₂=> (x+y)ml , volume of N₂ => (y+10) ml and volume of O₂=> (0.5x+10)ml

So by the question we get

(x+y)ml +(y+10) ml +(0.5x+10)ml = 100 ml

=>1.5x +2y =80 ................(1)

considering the total contraction we get

1.5x +y =60 ................(2)

subtracting (2) from (1) we get y = 20 ml

Hence total volume of N₂H₂ produced in two steps will be (10+20) ml = 30 ml

Discussion

724.	The volume of 1 mol of a gas at standard temperature and pressure is .A. 11.2 litreB. 22.4 litresC. 100 litreD. none of these
Answer» Correct Answer - B At `STP` or `NTP` volume of any gas = 22.4 L

Discussion

725.	What is the mass of the precipitate formed when 50 mL of 16.9% solution of `AgNO_(3)` is mixed with 50 mL of 5.8% NaCl solution?
Answer» 100 mL of `AgNO_(3)` solution have mass = 16.9 g 50 mL of `AgNO_(3)` solution have mass = 8.45 g Number of moles of `AgNO_(3)` present `= ("Mass of" AgNO_(3))/("Molar mass")=((8.45g))/((170"g mol"^(-1)))=0.048` mol 100 mL of NaCl solution have mass = 5.8 g 50 mL of NaCl solution have mass = 2.9 g Number of moles of NaCl present `= ("Mass of NaCl")/("Molar mass")=((2.9g))/((58.5"g mol"^(-1)))=0.049` mol The chemical equation for the reaction may be written as : `{:(,AgNO_(3)+,NaClrarr,AgCl("ppt")+,NaNO_(3)),("Initial moles","0.049 mol","0.049 mol",0,0),("Final moles",0,0,"0.049 mol",0.049 mol"):}` Number of moles of AgCl formed `= 0.049` mol Mass of AgCl formed `= (0.049 mol xx 143.5 g mol^(-1))` `= 7.03 ~~ 7.0 g`.

Discussion

726.	What is would be the molality of a solution obtained by mixing equal volumes of 30% by weight `H_(2) SO_(4) (d = 1.218 g mL^(-1))` and 70% by weight `H_(2) SO_(4) (d = 1.610 g mL^(-1))`? If the resulting solution has density `1.425 g mL^(-1)`, calculate its molality.
Answer» Let `V ml` of each are mixed. For solution I `H_(2) SO_(4)` is 30 % by wegith. `:.` Weight of `H_(2) SO_(4) = 30 g` and weight of solution `= 100 g` `:.` Volume of solution `= (100)/(1.218) mL` contains `30 g H_(2) SO_(4)` `:. V mL` contains `(30 xx V xx 1.218)/(100) g H_(2) SO_(4)`. For solution II `H_(2) SO_(4)` is 70% by weight. `:.` weight of `H_(2) SO_(4) = 70 g` Weight of solution `= 100 g` `:.` Volume of solution `= (100)/(1.610) mL` i.e., `(100)/(1.610) mL` contains `70 g H_(2) SO_(4)` `:. V mL` contains `= (70 xx V xx 1.610)/(100) H_(2) SO_(4)` `= [ (30 xx 1.218)/(100) + (70 xx 1.610)/(100)] V g = 1.492 V g` ltbRgt Total volume of solution `= 2 V mL` `:.` Molarity `(M)` of solution `= (1.4924 V)/(98 xx (2 V)/(1000)) = 7.61` Now, Weight of total solution `= 2 V xx 1.425 g = 2.85 V g` `:.` Weight of water `= (2.85 V - 1.4924 V) g` `= 1.376 V g` `:.` Molality `(m)` of solution `= (1.7924 V)/(98 xx (1.3576 V)/(1000)) = 11.22` Alternatively : (use the formula) `d_(sol) = M ((Mw_(2))/(1000) + (1)/(m))` `1.425 = 7.61 ((98)/(1000) + (1)/(m))` `:. M = 11.22`

Discussion

727.	Represent the following chemical changes in the form of chemical equations and balance them. (a) A copper coin is placed in a solution of corrosive sublimate, mercuric chloride. The products obtained are cupric chloride and mercury. (b) A piece of (a) sulphur, (b) charcoal burns vigorously when dropped in molten potassium nitrate, because potassium nitrate decomposes to form potassium nitrite and oxygen, and this oxygen helps to burn charcoal and sulphur giving out carbon dioxide and sulphur dioxide, respectively. (c ) Aqueous ammonium hydroxide solution is made to react with aqueous copper sulphate solution and a bluish white precipitate of cupric hydroxide, and ammonium sulphate are formed.
Answer» (a) The copper displaces the mercury from mercuric chloride and forms cupric chloride and mercury. `(Cu+HgCl_(2) to CuCl_(2) + Hg)` (b)` 2 KNO_(3) to 2 KNO_(2) + O_(2)` `S + O_(2) to SO_(2)` `C + O_(2) to CO_(2)` (c ) `2NH_(4)OH + CuSO_(4) to Cu(OH)_(2) + (NH_(4))_(2) SO_(4)`

Discussion

728.	The percentage of metal in two metal oxides, A and B is `74.2% and 59%`, respectively. Show that the law of multiple proportions is obeyed.
Answer» `{:("Metal Oxides",,"Percentage of Metal",,"Percentage of Oxygen"),(" A",," 74.2",," 25.8"),(" B",," 59",," 41"):}` In A, 74.2 g of metal combines with 25.8 g of oxygen ? combines with 1 g of oxygen ` = (74.2 g)/(25.8 g) = 2.87` g In B, 59 g of metal combines with 41 g of oxygen ? combines with 1 g of oxygen ` =59/41 g = 1.43` g `:.` The ratio by weight of metal combining with fixed mass of oxygen in A and B ` = 2.87 : 1.43 = 2 : 1` Hence, law of multiple proportions is obeyed.

Discussion

729.	The molarity of a solution obtained by mixing 750 mL of 0.5 M HCl with 250 mL of 2 M HCl will beA. `0.875 M`B. `1.00 M`C. `1.75 M`D. `0.975 M`
Answer» Correct Answer - A

Discussion

730.	It is found that with an increase in temperature by `40%`, the volume decreases by `20%` with change in pressure. Find the percentage change in pressure.A. `40%` decreaseB. `60%` decreaseC. `75%` increaseD. `80%` increase
Answer» Correct Answer - C

Discussion

731.	Calculate the molarity of a solution obtained by mixing 250 mL of `0.5` M HCl with 750 mL of 2 M HCl.A. `1.8`B. `2.0`C. `1.6`D. `0.8`
Answer» `N_("Mixture")=(N_(1)V_(1)+N_(2)V_(2))/(V_(1)+V_(2))` ` = ((0.5)250+(2xx750))/(250+750) = 1.625`N `:. ` Molarity = 1.625 N(as HCl is monobasic acid)

Discussion

732.	50 mL of a gaseous hydrocarbon is mixed with excess of oxygen and burnt and cooled to the laboratory temperature. The reduction in volume was found to be 150 mL. The gas is then passed into caustic potash, when there is a further reduction in the volume of 150 mL. Provide all the volumes are measured at the same conditions of temperature and pressure, find out the molecular formula of the hydrocarbon.
Answer» Chemical reaction equation to combustion reaction is `{:(C_(x)H_(y)+O_(2) to CO_(2)+H_(2)O),("50mL"" 150mL"" 150mL"):}` We know volume of gas has same number of moles, at same temperature and pressure `{:("50 moles"" 150 moles"" 150 moles"),(C_(x)H_(y)+(x+y/4)O_(2) toxCO_(2) + y/2 H_(2)O),("50 mole"" 150 mole"" 150 mole"):}` Given that reduction in volume, when the gases mixture is cooled to the laboratory temperature = 150 mL `:. y/2 (50) = 150 rArr y = 6`. and `x (50) = 150 rArr x = 3` `:.` Hydrocarbon is `C_(3)H_(6)`.

Discussion

733.	If 40 g of ethyl alcohol is dissolved in 50 mL of water, then calculate the weight/volume percentage of ethyl alcohol present in the solution. (density of ethyl alcohol = 0.8 g/mL)
Answer» Amount of ethyl alcohol = 40 g Volume of ethyl alcohol =` 40/0.8 = 50 `mL Volume of water = 50 mL Total volume of the solution is 100 mL `(W/v)%` ethyl alcohol is `40/100 xx 100 = 40%`.

Discussion

734.	The reduction of acidified solution of ferric ions by hydrogen gas takes gas takes place in the presence of zinc. Explain.
Answer» Reduction of acidified solution of ferric ions by hydrogen gas does not take place. If a small amount of Zn is added to reaction mixture, the hydrogen gas is converted into nascent hydrogen by Zn. The nascent hydrogen is highly reactive reactive and it reduces ferric ions to ferrous ions.

Discussion

735.	What is the molarity of a solution containing 15 g of NaOH dissolved in 500 mL of solution?
Answer» `"Molarity"="Weight"/"GMW "xx1000/V = 15/40 xx1000/500 =0.75`M.

Discussion

736.	At STP, a certain amount of hydrogen is produced by the reaction of 550 g of impure zinc with excess amount of HCl. The same volume of hydrogen at STP reduces `Fe_(3)O_(4)` and produces 336 g of iron. Find the percentage of zinc. (Atomic mass of Zn is 65)
Answer» `Zn+2HCl to ZnCl_(2) + H_(2) , 4H_(2) + Fe_(3)O_(4) to 4H_(2)O + 3Fe` Amount of iron produced ` = 336 g rArr` number of moles of iron ` = (336)/56 = 6` To produce 6 moles of iron, it requires 8 moles of hydrogen which further requires 8 moles of Zn. Weight of Zn = `8 xx 65.5 = 524` g, % purity =`(524)/(550) xx 100 = 95.27 %`.

Discussion

737.	Brass is an alloy of copper and zinc. A sample of brass weighing `5.793` g, when treated with excess of dil. `H_(2)SO_(4)` gave 336 mL of hydrogen gas at ` 136.5` K and 785 mm pressure. If the hydrogen gas is collected over water and pressure of water vapour under the given conditions is 25 mm, find out the percentage weight of copper in the alloy.
Answer» Among Cu and Zn, only Zn displaces `H_(2)" from "H_(2)SO_(4)` `Zn+H_(2)SO_(4) to ZnSO_(4) + H_(2)` Pressure = 785 - 25 = 760 mm Temperature = 136.5 K, volume = 336 mL `(P_(1)V_(1))/T_(1) = (P_(2)V_(2))/T_(2)` `(760 xx 336)/(136.5) = (760 xx V_(2))/(273)` ` V_(2) = 672` mL `(V_(2))` Vol. of `H_(2)` liberated at STP = 672 mL Weight of zinc required = `(65.5 xx 672)/(22400) = 1.965` g Weight of copper in brass = `5.793 - 1.965 = 3.828` g % of Cu in the alloy = `(3.828)/(5.793) xx 100 = 66%`

Discussion

738.	Calculate the weight of zinc required for the liberation of 10 g of hydrogen gas on reaction with `H_(2)SO_(4)`.
Answer» `{:(Zn+H_(2)SO_(4) toZnSO_(4) +H_(2)),(" 1 mole" " 1 mole"):}` Atomic weight of zinc is ` 65.5` 1 mole of hydrogen = 2 g of hydrogen 10 g of hydrogen = `10/2 = 5` moles 1 mole of `H_(2)` gas is produced from 1 mole of zinc 5 moles of `H_(2)` gas are produced from 5 moles of zinc. 1 mole of zinc = 65.5 g 5 mole of zinc = ` 65.5 xx 5 = 327. 5 ` g

Discussion

739.	What is the percentage by weight of sulphuric acid, if 13 g of `H_(2)SO_(4)` is dissolved to make 78 g of solution?
Answer» Weight of `H_(2)SO_(4) = 13` g Weight of solution = 78 `(W/W)%" of "H_(2)SO_(4)=13/78 xx100=16.6%`.

Discussion

740.	The ratio by mass of sulphur and oxygen in `SO_(2)` isA. `1 : 2`B. ` 2 : 1`C. ` 1 : 1`D. ` 1 : 4`
Answer» Correct Answer - C

Discussion

741.	Critical temperatures of the gases A, B, C and D are (a) `5.2 `K (b) `33.2 ` K (c )` 126.3` K (d) ` 191.1` K Arrange them in the ascending order of intermolecular forces of attraction.A. C,B,D,AB. A,B, C,DC. D,C,B,AD. A,C,B,D
Answer» Correct Answer - B

Discussion

742.	1 g atom of nitrogen contains `6.023 xx 10^(23)` atoms of nitrogen.
Answer» Correct Answer - 1

Discussion

743.	Which of the following pairs of gases corresponds to the ratio of the rates of diffusion as `sqrt2 : 1`?A. `H_(2) and He`B. `He and CH_(4)`C. `H_(2) and CH_(4)`D. `CH_(4) and SO_(2)`
Answer» Correct Answer - A

Discussion

744.	A vessel contains equal masses of hydrogen, helium and methane. Find out the fractions of the partial pressures in the mixture.
Answer» (i) fraction of number of moles of `CH_(4), H_(2) ` and He in the mixture (ii) relation between mole fraction and total pressure (iii) `(8P)/13,(4P)/13, P/13`

Discussion

745.	A vessel contains equal number of moles of helium and methane. Due to a hole in the vessel, half of the gaseous mixture effused out. What is the ratio of the number of moles of helium and methane remaining in the vessel?
Answer» (i) relation between molecular weight and rate of effusion (ii) determination of the ratio of the number of molecules of helium and methane which are effused out (iii) calculation of the ratio of the remaining molecules (iv) `(2x)/3` moles

Discussion

746.	One mole of any substance contains `6.022 xx 10^(23)` atoms/molecules. Number of molecules of `H_(2)SO_(4)` present in 100 mL of 0.02 M `H_(2)SO_(4)` solution is :A. `12.044 xx 10^(20)` moleculesB. `6.022 xx 10^(23)` moleculesC. `1 xx 10^(23)` moleculesD. `12.044 xx 10^(23)` molecules.
Answer» Correct Answer - A Step I : Moles of `H_(2)SO_(4)(x)` `(0.02 mol_(L)^(-1))=(x)/((100//1000L))` `x=0.02xx0.1=0.002mol` Step II : Molecules of `H_(2)SO_(4)` `= 6.022 xx 10^(23) xx 0.002` `=12.044 xx 10^(20)` molecules.

Discussion

747.	If Avogadro number `N_(A)`, is changed from `6.022 xx 10^(23) mol^(-1)` to `6.022 xx 10^(20) mol^(-1)`, this would changeA. the mass of one mole of carbonB. the ratioof chemical species to each other in a balanced equationC. the ratio of elements to each other in a compoundD. the definition of mass in units of grams
Answer» Correct Answer - A Mass of 1 mole `(6.022 xx 10^(23))` of carbon atoms = 12g If `N_(A)` is changed to `6.022 xx 10^(200)` then mass of 1 mole of carbon atoms `= ((12g))/((6.022xx10^(23)))xx(6.022xx10^(20))=1.2xx10^(-2)g`.

Discussion

748.	Calculate the density `("in gm L"^(-1))` of a 3.60 M sulphuric acid solution that is 29 % `H_(2)SO_(4)` by mass `("molar mass = 98 g mol"^(-1))`
Answer» By definition, 3.6 M or `3.6 xx 98 = 352.8 g` of `H_(2)SO_(4)` are present in 1L of the solution Now, 29g of `H_(2)SO_(4)` are present in the solution = 100 g 352.8 g of `H_(2)SO_(4)` are present in the solution `= ((100g)xx(352.8))/((29g))=1216g` Density of solution `= ("Mass of solution")/("Volume of solution")=((1216g))/((1000mL))=1.22"g mL"^(-1)`

Discussion

749.	Concentrated aqueous solution of sulphuric acid is `98 %` by mas and has density of `1.80 "g mL"^(-1)`. What is the volume of acid requird to male one litre `0.1 M H_(2)SO_(4)` solution ?
Answer» Volume of 98 % by mass of `H_(2)SO_(4)=("Mass")/("Density")=((100g))/((1.80"g mL"^(-1)))` `=55.55mL=0.0555L` Molarity of solution `(M)=("Mass of "H_(2)SO_(4)//"Molar mass")/("Volume of solution in litres")` `=((98.0g)//(98.0"g mol"^(-1)))/((0.0555L))=18.02 "molL"^(-1)=18.02` Volume of solution required to make 1L of `0.1 M H_(2)SO_(4)` solution `M_(1) V_(1) -= M_(2)V_(2)` `(18.02M)xxV_(1)=(0.1 M)xx(1.0 L)` `V_(1)=((0.1M)xx(1.0L))/((18.02M))=0.0055L=5.5mL`.

Discussion

750.	The correct sequence of deceasing mass of the following is : (i) `6.022 xx 10^(23)` atoms of oxygen (ii) `1.0 xx 10^(23)` molecules of `H_(2)S` (iii) `6.022 xx 10^(23)` molecules of oxygenA. (i) `gt` (ii) `gt` (iii)B. (i) `gt` (iii) `gt` (ii)C. (iii) `gt` (i) `gt` (ii)D. (ii) `gt` (i) `gt` (iii)
Answer» Correct Answer - C (i) `6.022 xx 10^(23)` atoms of oxygen have mass = 16.0 g (ii) `6.022xx 10^(23)` molecules of `H_(2)S` have mass = 34.0 g `1.0 xx 10^(23)` molecules of `H_(2)S` have mass `= (1xx10^(23))/(6.022xx10^(23))xx(34.0g)=5.645g` (iii) Mass of `6.022xx10^(23)` molecules of oxygen `(O_(2)) = 32 g`.

Discussion

Explore topic-wise InterviewSolutions in .

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A vessel contains equal number of moles of helium and methane. Due to a hole in the vessel, half of the gaseous mixture effused out. What is the ratio of the number of moles of helium and methane remaining in the vessel?

One mole of any substance contains `6.022 xx 10^(23)` atoms/molecules. Number of molecules of `H_(2)SO_(4)` present in 100 mL of 0.02 M `H_(2)SO_(4)` solution is :A. `12.044 xx 10^(20)` moleculesB. `6.022 xx 10^(23)` moleculesC. `1 xx 10^(23)` moleculesD. `12.044 xx 10^(23)` molecules.

Calculate the density `("in gm L"^(-1))` of a 3.60 M sulphuric acid solution that is 29 % `H_(2)SO_(4)` by mass `("molar mass = 98 g mol"^(-1))`

Concentrated aqueous solution of sulphuric acid is `98 %` by mas and has density of `1.80 "g mL"^(-1)`. What is the volume of acid requird to male one litre `0.1 M H_(2)SO_(4)` solution ?

The correct sequence of deceasing mass of the following is : (i) `6.022 xx 10^(23)` atoms of oxygen (ii) `1.0 xx 10^(23)` molecules of `H_(2)S` (iii) `6.022 xx 10^(23)` molecules of oxygenA. (i) `gt` (ii) `gt` (iii)B. (i) `gt` (iii) `gt` (ii)C. (iii) `gt` (i) `gt` (ii)D. (ii) `gt` (i) `gt` (iii)