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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1051. |
The weight of `1 xx 10^(22)` molecules of `CuSO_(4). 5H_(2)O` is …….. |
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Answer» Correct Answer - A::D The weight of `1 x10^(22)` molecules of `CuSO_(4).5H_(2)O` is `4.14 g`. |
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| 1052. |
The weight of `1 xx 10^(22)` molecules of `CuSO_(4). 5H_(2)O` isA. 41.59gB. 415.9gC. 4.159gD. 0.4159g |
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Answer» Correct Answer - C |
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| 1053. |
Match parameters involved in column-I with those in column-II. |
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Answer» Correct Answer - `(a rarrp,r)(b rarr t);(c rarr q,t)(d rarr s, t)` |
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| 1054. |
6 gm of nitrogen on successive reaction with different compounds gets finally converted into 30 gm `[Cr(NH_(3))_(x)Br_(2)].` Value of x is: [Atomic mass of Cr=52] |
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Answer» Correct Answer - 4 |
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| 1055. |
Match the column-I (Reaction) with column-II (maximum yield of the product) |
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Answer» Correct Answer - `(a rarr r);(b rarr s);(c rarr p);(c rarr s);(d rarr q)` |
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| 1056. |
5 moles of A, 6 molesof Z are mixed with suffifcient amount of C to finally produce F. Then find the maximum molesof F which can be produced. Assuming that the product formed can also be reused. Reactions are : `A + 2Z rightarrow B` `B + C rightarrow Z + F`A. 3 molesB. 4.5 molesC. 5 molesD. 6 moles |
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Answer» Correct Answer - C |
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| 1057. |
`H_(2)SO_(4)` can be prepared in following two step process : [% yield of each step is indicated] `(P) S(g) + O_(2)(g) overset(50% yield)to SO_(2)(g)` (Q) `NO_(2)(g) + SO_(2)(g) + H_(2)O(l) overset(80% yield)to H_(2)SO_(4)(aq) + NO(g)` What mass of sulphur is required to prepare 1960 gm `H_(2)SO_(4)` ?A. 640gmB. 1450 gmC. 1600gmD. 1280gm |
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Answer» Correct Answer - C |
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| 1058. |
A mixture containing 3 moles each of `C_4H_8` and `C_6H_6` undergoes complete combustion with `O_2` to form `CO_2 and H_2O` Calculate total mass of `CO_2` produced :A. 1320 gmB. 610 gmC. 528 gmD. 792 gm |
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Answer» Correct Answer - A |
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| 1059. |
27 g C and 48 g `O_2` are allowed to react completely to form CO and `CO_2`.The weight ratio of CO and `CO_2` formed, is :A. `7:11`B. `3:4`C. `14:11`D. `9:8` |
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Answer» Correct Answer - C |
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| 1060. |
Two substance C and `O_(2)` are allowed to react completely to form CO and `CO_(2)` mixture , leaving none of the reactants . Its is known that when I mole of `CO_(2)` ,100 Kcal of energy is released and when 1 mole of carbon reacts with 0.5 mole of `O_(2)` to give of CO,25 Kcal is liberated . Using this information match column I and column II. |
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Answer» Correct Answer - `(a rarrq );(b rarr s);(c rarr p);(d rarr r)` |
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| 1061. |
Many a time the reaction are carried out when the reactants are not present in the amounts required by a balanced chemical reaction. In such situations, one reactant is in excess over other. The reactant which is present in th lesser amount gets conserved after sometime and after that no further reaction takes place whatever be the amount of other reactant is present .Hence, the reactant which gets consumed , limits the amount of products formed and is therefore called limiting reagent. To determine the limiting reagent find the value of `phi` which is the ratio of (given) mole of a substance to the stoichiometric coefficient of that substance . The limiting reagent is the reagent which has minimum value of `phi`. In the above question , how much reactant is remaining at the completion of reaction and which one?A. NothingB. 5.9 kg `Cr_(2)O_(3)`C. 14.1 kg `Cr_(2)O_(3)`D. 14.1 kg Al |
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Answer» Correct Answer - B |
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| 1062. |
Many a time the reaction are carried out when the reactants are not present in the amounts required by a balanced chemical reaction. In such situations, one reactant is in excess over other. The reactant which is present in th lesser amount gets conserved after sometime and after that no further reaction takes place whatever be the amount of other reactant is present .Hence, the reactant which gets consumed , limits the amount of products formed and is therefore called limiting reagent. To determine the limiting reagent find the value of `phi` which is the ratio of (given) mole of a substance to the stoichiometric coefficient of that substance . The limiting reagent is the reagent which has minimum value of `phi`. The reduction of `Cr_(2)O+_(3)` by Al proceeds quantitatively on ignition of a suitable fuse. The reaction is : `2Al +Cr_(2)O_(3) rarr Al_(2)O_(3) +2Cr` How much metallic chromium can be made by bringing to reaction temperature a mixture of 5 kg Al and 20 kg `Cr_(2)O_(3)` ?A. 9.6 kgB. 14.4 kgC. 5.9 kgD. 1.41 kg |
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Answer» Correct Answer - A |
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| 1063. |
Estimation of phosphorous: Second method: A known mass of compound is heated with fuming `HNO_(3)` or sodium peroxide `(Na_(2)O_(2)` in Carius tube which converts phosphorous to `H_(3)PO_(4)`. Magnesia mixture `(MgCl_(2)+NH_(4)Cl)` is then added, which gives the percipate of magnesium ammonium phosphate `(MgNH_(4).PO_(4))` which on heating gives magnesium pyrophosphate `(Mg_(2)P_(2)O_(7))` , which is weighted. Percentage of P`=("Atomic mass of P")/("Molecular mass of " Mg_(2)P_(2)O_(7)) xx ("Molecular mass of "Mg_(2)P_(2)O_(7) xx 100)/("Mass of compound")=(62)/(222) xx("Mass of " Mg_(2)P_(2)O_(7)xx 100)/("Mass of compound")` An organic compound has 6.2% of phosphorus .In the reaction sequence , all phosphorous is converted to `Mg_(2)P_(2)O_(7)` . Find wt. of `Mg_(2)P_(2)O_(7)` formedA. 2.22B. 10.2C. 15D. 20 |
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Answer» Correct Answer - A |
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| 1064. |
Estimation of phosphorous: Second method: A known mass of compound is heated with fuming `HNO_(3)` or sodium peroxide `(Na_(2)O_(2)` in Carius tube which converts phosphorous to `H_(3)PO_(4)`. Magnesia mixture `(MgCl_(2)+NH_(4)Cl)` is then added, which gives the percipate of magnesium ammonium phosphate `(MgNH_(4).PO_(4))` which on heating gives magnesium pyrophosphate `(Mg_(2)P_(2)O_(7))` , which is weighted. Percentage of P`=("Atomic mass of P")/("Molecular mass of " Mg_(2)P_(2)O_(7)) xx ("Molecular mass of "Mg_(2)P_(2)O_(7) xx 100)/("Mass of compound")=(62)/(222) xx("Mass of " Mg_(2)P_(2)O_(7)xx 100)/("Mass of compound")` 0.12 gm of and organic compound containing phosphorus gave 0.22 gm of `Mg_(2)P_(2)O_(7)` by the usual analysis. Calculate the percentage of phosphorus in the compound.A. 25B. 9.25C. 801D. `51.20` |
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Answer» Correct Answer - D |
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| 1065. |
Estimation of halogens: Carius method: A known mass of compound is heated with conc. `HNO_(3)` in the pressure of `AgNO_(3)` contained in a hard glass tube known as Carius tube in a furance . C and H are oxidised to `CO_(2)` and `H_(2)O` . The halogen forms the corresponding AgX. It si filtered m dried and weighted . Extimation of sulphur: A known mass of compound is heated with fuming `HNO_(3)` or sodium peroxide `(Na_(2)O_(2))` in the presence of `BaCl_(2)` solution in Carius tube. Sulphur is oxidised to `H_(2)SO_(4)` and percipipated as `BaSO_(4)` .It is filtered , dried and weighed. Percentage of S `=("Atomic mass of S")/("Molecular mas of" BaSO_(4)) xx ("Mass of" BaSO_(4) xx 100)/("Mass of compound")` 0.15 gm of an organic compound gave 0.12 gm of silver bromide by the Carius method. Find the percentage of bromine in the compound.A. 34B. 40C. 17D. 68 |
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Answer» Correct Answer - A |
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| 1066. |
Estimation of halogens: Carius method: A known mass of compound is heated with conc. `HNO_(3)` in the pressure of `AgNO_(3)` contained in a hard glass tube known as Carius tube in a furance . C and H are oxidised to `CO_(2)` and `H_(2)O` . The halogen forms the corresponding AgX. It si filtered m dried and weighted . Extimation of sulphur: A known mass of compound is heated with fuming `HNO_(3)` or sodium peroxide `(Na_(2)O_(2))` in the presence of `BaCl_(2)` solution in Carius tube. Sulphur is oxidised to `H_(2)SO_(4)` and percipipated as `BaSO_(4)` .It is filtered , dried and weighed. Percentage of S `=("Atomic mass of S")/("Molecular mas of" BaSO_(4)) xx ("Mass of" BaSO_(4) xx 100)/("Mass of compound")` 0.2595 gm of an organic substance when treated by Carius method gave 0.35 gm of `BaSo_(4)` . Calculate the percentage of sulphur in the compound.A. 9B. 30.4C. 18.52D. 40.52 |
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Answer» Correct Answer - C |
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| 1067. |
(d) Estimation of phosphorous: A known mass of compound is heated with fuming `HNO_(3)` or sodium peroxide `(Na_(2)O_(2))` in Carius tube which converts phosphorrous to `H_(3)PO_(4)` . Magnesia mixture `(MgCl_(2) + NH_(4)Cl)` is then added, which gives the precipitate of magnesium ammonium phosphate `(MgNH_(4).PO_(4))` which on heating gives magnesium pyrophosphate `(Mg_(2)P_(2)O_(7))` , which is weighed. 0.124 gm of an organic compound containing phosphorus gave 0.222 gm of `Mg_(2)P_(2)O_(7)` by the usual analysis . Calculate the percentage of phosphorous in the compound (Mg=24, P=31)A. 25B. 75C. 62D. 50 |
| Answer» Correct Answer - D | |
| 1068. |
Estimation of halogens: Carisu Method: A known mass of compound is heated with conc. `HNO_(3)` in the presence of `AgNO_(3)` contained in a hard glass tube known as carius tube in a furnce. C and H are oxidised to `CO_(2)` and `H_(2)O` . The halogen forms the corresponding AgX. It is filtered , dried , and weighed. Estimation of sulphur: A known mass of compound is heated with fuming `HNO_(3)` or sodium peroxide `(Na_(2)O_(2))` in the presence of `BaCl_(2)` solution in Carius tube. Sulphur is oxidised to `H_(2)SO_(4)` and precipitated as `BaSO_(4)` . It is filerted, dried and weighed. 0.32 gm of an organic substance when treated by Carius method gave 0.466 gm of `BaSO_(4)` . Calculate the percentage of sulphur in the compound (Ba=137)A. `10.0`B. `34.0`C. `20.0`D. `30.0` |
| Answer» Correct Answer - C | |
| 1069. |
Estimation of halogens: Carisu Method: A known mass of compound is heated with conc. `HNO_(3)` in the presence of `AgNO_(3)` contained in a hard glass tube known as carius tube in a furnce. C and H are oxidised to `CO_(2)` and `H_(2)O` . The halogen forms the corresponding AgX. It is filtered , dried , and weighed. Estimation of sulphur: A known mass of compound is heated with fuming `HNO_(3)` or sodium peroxide `(Na_(2)O_(2))` in the presence of `BaCl_(2)` solution in Carius tube. Sulphur is oxidised to `H_(2)SO_(4)` and precipitated as `BaSO_(4)` . It is filerted, dried and weighed. 0.15 gm of an organic compound gave 0.12 gm of silver bromide by the Carius method. Find the percentage of bromine in the compound. (Ag=108, Br=80)A. `34.0`B. `46.0`C. `80.0`D. `50.0` |
| Answer» Correct Answer - A | |
| 1070. |
No. of moles in 100 mg of heptane is …………than those in 10 mg of propyneA. 4 times greaterB. 4 times lesserC. 2.5 times greaterD. 2.5 times lesser |
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Answer» Correct Answer - A No. of moles in 100 mg of heptane `(C_(7)H_(16))` `= ((100xx10^(-3)g))/(("100g mol"^(-1)))=1xx10^(-3)` No. of moles in 10 mg of propyne `(C_(3)H_(4))` `= ((10xx10^(-3)g))/(("40 g mol"^(-1)))=(1)/(4)xx10^(-3)` No. of moles of heptane are four times the no. of moles of propyne. |
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| 1071. |
The numbers of significant figures in 0.0500 areA. OneB. ThreeC. TwoD. Four |
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Answer» Correct Answer - B The significant figures in 0.0500 are three because zeros to the left of the first non-zero digit are not significant while zeroes at the end but to the right of the decimal point are significant. |
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| 1072. |
One molal solution contains 1 mole of a solute in :A. 1000 g of the solventB. one litre of the solventC. one litre of the solutionD. 22.4 litres of the solution |
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Answer» Correct Answer - A Definition of molality. |
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| 1073. |
A molal solution is one that contains 1 mol of a solute inA. `1000 g` of solventB. `1 L` of solvenC. `1 L` of solutionD. `22.4 L` of solution |
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Answer» Correct Answer - A A molal solution is one that contains 1 mol of a solute in `1000 g` or `kg` of the solvent. Molality `(m) = ("Moles of solute")/("Mass of solvent in kg")` |
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| 1074. |
Which of following relations are correct?A. `1 eV = 9.11 xx 10^(-4) J`B. `1 L = 1 dm^(3)`C. `1 J = 1.98 "cal"`D. 1 atm `= 1.01325` bar |
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Answer» Correct Answer - A::B::C::D All correct |
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| 1075. |
Which of the following statements are correct?A. French chemist `A`. Lavoisier is called th father of chemistry and proposed the law of conservation of mass.B. French chemist joseph proust proposed the law of definite proportionsC. Dalton proposed the law of multiple proportion.D. Richter proposed the law of reciprocal proportions. |
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Answer» Correct Answer - A::B::C::D All correct |
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| 1076. |
In which mode of expression, the concentration of a solution remains independent of temperature?A. MolarityB. NormalityC. FormalityD. Molality |
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Answer» Correct Answer - D Molaity, normality, and formality are calaculated against the volume of the solution. The volume of the solution changes with change in temperature, therefore these quanities do not reamain constant with temperature. Molality `(m) = ("Moles of solute")/("Mass of solvent in kg")` The molality of a solution remains independent of temperature because it involes only mass, which is independent of temperature. |
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| 1077. |
In which mode of expression, the concentration of the solution is independent of the temperature ?A. MolarityB. NormalityC. FormalityD. Molality |
| Answer» Correct Answer - D | |
| 1078. |
`Na_(2)SO_(4).xH_(2)O` has `50% H_(2)O`. Henxe, `x` is :A. `4`B. `5`C. `6`D. `8` |
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Answer» Correct Answer - D `%` by `wt`. Of `H_(2)O` `=(wt. "of" H_(2)O)/("Total wt".) xx100` `50=(18x)/(142+18x)xx100` `71+9x=18x` `x=71//9=7.88~~8` |
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| 1079. |
`13.4 g` of a sample of unstable hydrated salt `Na_(2)SO_(4).XH_(2)O` was found to contains `6.3 g` of `H_(2) O`. The number of molecular of water of crystalistion isA. 5B. 7C. 2D. 10 |
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Answer» Correct Answer - B Weight of salt `= 13.4 g` weight of `H_(2) O = 6.3 g` Weight of anhydrous salt `= 13.4 - 6.3 = 7.1 g` Moles of anhydrous salt `= (7.1)/(842) = 0.05` Moles of `H_(2) O = (6.3)/(18) = 0.35 "mol"` `0.05 "mol"` of anhydrous salt `implies 0.35 "mol of" H_(2)O` 1 mole of anhydrous salt `implies (0.35)/(0.05) = 7 "mol"` |
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| 1080. |
A mixture of ethylene and excess of `H_(2)` had a pressure of `600 mmHg` the mixture was passed over nickel catalyst to convert ethylene to ethane.The pressure of the resultant mixture at the similar conditions of temperature and volume dropped to `400 mm Hg` The fraction of `C_(2) H_(4)` by volume in the original mixture isA. `1//3` rd of the total volumeB. `1//4` th of the total volumeC. `2//3` rd of the total volumeD. `1//2` nd of the total volume |
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Answer» Correct Answer - A Let `n` mol of `(C_(2) H_(3) + H_(2))` `X "mol of" C_(2)H_(4)` `H_(2) =(n - x)` mole `underset(x)(C_(2))H_(4) + underset(x)(H_(2)) rarr underset(x "mol")(C_(2) H_(6))` After reaction `(C_(2) H_(6) + H_(2)` left) `x + n - x - x = n - x` Total `H_(2) = (n - x), H_(2)` reacted `= x]` `H_(2)` left `= (n - x - x)` `n = 600, n - x = 400` `(n)/(n - x) = (600)/(400), x = (4)/(4)` volume of `C_(2) H_(4)` `= (1)/(3)` rd of total volume |
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| 1081. |
The weight of `1 L` of ozonised oxygen at `STP` was found to be `1.5 g`. When `100 mL` of this mixture at `STP` was treated with turpentine oil, the volume was reduced to `90 mL`. The molecular weight of ozone isA. 49B. 47C. 46D. 47.9 |
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Answer» Correct Answer - D Volume of `O_(3)` in `100 mL` of ozonised `O_(2)` `= 100 - 90 = 10 mL` (dissolved in turpentine) Volume of `O_(3)` in `1 L` of ozonised `O_(2)` `= (10 x 1000)/(100) = 100 mL` Volume of `O_(2)` in `1 L = 1000 - 100 = 900 mL` Weight of `900 mL` of `O_(2)` at `STP = (900 xx 32)/(22400) = 1.286 g` Weight of `100 mL "of" O_(3)` at `STP = 1.5 - 1.286` `= 0.214 g` Now `100 mL` of`O_(3)` at `STP = 0.214 g` `22400 mL` of `O_(3)` at `STP` Weighs `= (0.214 xx 22400)/(100)` `= 47.94 g` Molecular weight of `O_(3) = 47.94` |
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| 1082. |
`H_(2)SO_(4)` solution `(80% "by weight and specific gravity" 1.75 g//ml)` is used to prepare `2 litre` of `0.25 MH_(2)SO_(4)` (aq). The volume of `H_(2)SO_(4)` solution (original) which must be used is :A. `107.18ml`B. `43.75ml`C. `35ml`D. None of these |
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Answer» Correct Answer - C Let `x ml` of `H_(2)SO_(4)` solution `(n_(H_(2)SO_(4)))_(i) = (n_(H_(2)SO_(4)))_(i)` `1.75x xx (80)/(100) = 2xx0.25` `x = 35 mL` |
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| 1083. |
The empirical formula of a commercial ion-exchange resin is `C_(8) H_(70 SO_(3) Na`. The resin is used to soften water as follows: `Ca^(2+) + 2C_(8)H_(7)SO_(3)Na rarr (C_(8) H_(7) SO_(3))_(2) Ca + Na^(o+)` expressed in `mol//g` resin? |
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Answer» `(1 mol Ca^(2+))/((2 "mol resin")(206 g "resin"//mol. "reson"))` `= 0.00246 mol Ca^(2+)//g "resin"` |
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| 1084. |
For the following reaction, the mass of water produced from 445 g of `C_(57)H_(110)O_(6)` is : `2C_(57)H_(110)O_(6)(s) + 163O_(2)(g) to 114CO_(2)(g) + 110 H_(2)O(l)`A. `495 g`B. `490 g`C. `890 g`D. `445 g` |
| Answer» Correct Answer - A | |
| 1085. |
An unknown chlorohydrocarbon has 3.55% of chlorine. If each molecule of the hydrocarbon has one chlorine atom only, chlorine atoms present in 1 g of chlorohydrocarbon are : (Atomic wt. of C1 = 35.5 u, Avogador constant `= 6.023xx10^(23) mol^(-1)) `A. `6.023xx10^(21)`B. `6.023xx10^(23)`C. `6.023xx10^(20)`D. `6.023xx10^(9)` |
| Answer» Correct Answer - C | |
| 1086. |
A 10 mg effervescent tablet containing sodium bicarbonate and oxalic acid releases 0.25 ml of `CO_(2)` at T = 298.15 K and p = 1 bar. If molar volume of `CO_(2)` is 25.0 L under such condition, what is the percentage of sodium bicarbonate in each tablet ? [Molar mass of `NaHCO_(3) = 84 g mol^(-1)`]A. `16.8`B. `8.4`C. `0.84`D. `33.6` |
| Answer» Correct Answer - B | |
| 1087. |
How many grams of `H_(2)SO_(4)` are present in `500ml` of `0.2M H_(2)SO_(4)` solution ? . |
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Answer» `M = ("moles")/("vol") rArr` moles of `H_(2)SO_(4) = M xx V = 0.2 xx (500)/(1000) L = 0.1` Mass of `H_(2)SO_(4) = 0.1 xx 98 = 9.8g` . |
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| 1088. |
A compound contains `42.3913 % K,15.2173 % Fe,19.5652 % C and 22.8260 % N`. The molecular mass of the compound is 368 u. Find the molecular formula of the compound. |
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Answer» Step I : Determination of empirical formula of the compound `{("Element","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole. no ratio"),("K",42.3913,39,(42.3913)/(39)=1.087,(1.087)/(0.2717)=4,4),("Fe",15.2173,56,(15.2173)/(56)=0.2717,(0.2717)/(0.2717)=1,1),("C",19.5652,12,(19.5652)/(12)=1.630,(1.630)/(0.2717)=6,6),("N",22.8260,14,(22.8260)/(14)=1.630,(1.630)/(0.2717)=6,6):}` The empirical formula of the compound `= K_(4)FeC_(6)N_(6)` Step II. Determination of molecular formula of the compound Empirical formula mass `= 4xx39 +56 + 6 xx 12 + 6 xx 14` `= 156 + 56 + 72 + 84 = 368` u Molecular mass = 368 u (Given) `:. n=("Molecular mass")/("Empirical formula mass")=(368)/(368)=1` Molecular formula `= n xx` Empirical formula `= 1 xx (K_(4) FeC_(6)N_(6)) or K_(4)[Fe(CN)_(6)]` The name of the compounbd is Potassium ferrocyanide. |
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| 1089. |
Comprehension # 9 A `10ml` mixture of `N_(2)`, a alkane `& O_(2)` undergo combustion in Eudiometry tube. There was contraction of `2ml`, when residual gases are passed are passed through `KOH`. To the remaining mixture comprising of only one gas excess `H_(2)` was added `&` after combustion the gas product is absorbed by water. causing a reduction in volume of `8ml`. Gas produced after introduction of `H_(2)` in the mixture? Identify the hydrocarbon.A. `CH_(4)`B. `C_(2)H_(6)`C. `C_(3)H_(8)`D. `C_(4)H_(10)` |
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Answer» Correct Answer - A `C_(n)H_(2n+2)+O_(2)rarrCO_(2)+H_(2)O` `{:(n=7),(CH_(4)):}` |
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| 1090. |
Comprehension # 9 A `10ml` mixture of `N_(2)`, a alkane `& O_(2)` undergo combustion in Eudiometry tube. There was contraction of `2ml`, when residual gases are passed are passed through `KOH`. To the remaining mixture comprising of only one gas excess `H_(2)` was added `&` after combustion the gas product is absorbed by water. causing a reduction in volume of `8ml`. Gas produced after introduction of `H_(2)` in the mixture? Volume of `N_(2)` present in the mixture?A. `2ml`B. `4ml`C. `6ml`D. `8ml` |
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Answer» Correct Answer - B `NH_(3) = 8 ml` `{:(N_(2)+3H_(2)rarr2NH_(3)),(4ml" "8):}` |
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| 1091. |
The expression relating mole fraction of solute `(chi_(2))` and molarity `(M)` of the solution is: (where `d` is the density of the solution in `g L^(-1)` and `Mw_(1)` and `Mw_(2)` are the molar masses of solvent and solute, respectivelyA. `x_(2) = (M xx Mw_(1))/(M (Mw_(1) xx Mw_(2)) + 1000d)`B. `x_(2) = (M xx Mw_(1))/(M (Mw_(1) xx Mw_(2)) + d)`C. `x_(2) = (M xx Mw_(1))/(M (Mw_(1) xx Mw_(2)) - 1000d)`D. `x_(2) = (M xx Mw_(1))/(M (Mw_(1) xx Mw_(2)) - d)` |
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Answer» Correct Answer - B Let the volume of solution `= 1 L` Weight of soulition `= 1 xx d = dg` Number of moles of solute `(n_(2))` in `1 L` solution `= M` `:. (W_(2))/(Mw_(2)) = M` `W_(2)` (weight of solute) `= M xx Mw_(2)` `W_(1)` (weight of solvent) = Weight of solution - Weight of solute `= (d - M xx Mw_(2))` Number of moles of solvent `(n_(1))` `= (W_(1))/(Mw_(1)) = ((d - M xx Mw_(2))/(Mw_(1)))` `chi_(2) = (n_(2))/(n_(1) + n_(2)) = (M)/(((d - M x Mw_(2))/(Mw_(1))) + M)` `= (M xx Mw_(1))/(M (Mw_(1) - Mw_(2)) + d)` |
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| 1092. |
Comprehension # 6 The percentage labelling of oleum is a unique process by means of which, the percentage composition of `H_(2)SO_(4), SO_(3)` (free) and `SO_(3)` (combined) is calculated. Oleum is nothing but it is a mixture of `H_(2)SO_(4)` and `SO_(3)` i.e., `H_(2)S_(2)O_(7)`, which is obtained by passing. `SO_(3)` in solution of `H_(2)SO_(4)`. In order of dissolve free `SO_(3)` in oleum, dilution of oleum is done, in which oleum converts into pure `H_(2)SO_(4)`. It is shown by the reaction as under : `H_(2)SO_(4)+SO_(3)+H_(2)Orarr2H_(2)SO_(4)("pure")` or " "`SO_(3)+H_(2)OrarrH_(2)SO_(4)("pure")` When `100g` sample of oleum is diluted with desired weight of `H_(2)O("in" g)`, then the total mass of pure `H_(2)SO_(4)` obtained after dilution is known as percentage labelling in oleum. For example, if the oleum sample is labelled as `""109%H_(2)SO_(4)"` it means that `100 g` of oleum on dilution with `9m` of `H_(2)O` provides `109g` pure `H_(2)SO_(4)`, in which all free `SO_(2)` in `100g` of oleum is dissolved. For `"118% H_(2)SO_(4)` labelled oleum, if the number of moles of free `SO_(3)`, number of moles of `H_(2)SO_(4)` and number of moles of `H_(2)O` be `x`, `y` and `z` respectively, then what will be value `x+y+z` ?A. `3.2`B. `3.2`C. `4.2`D. `2.2` |
| Answer» Correct Answer - D | |
| 1093. |
Comprehension # 9 A `10ml` mixture of `N_(2)`, a alkane `& O_(2)` undergo combustion in Eudiometry tube. There was contraction of `2ml`, when residual gases are passed are passed through `KOH`. To the remaining mixture comprising of only one gas excess `H_(2)` was added `&` after combustion the gas product is absorbed by water. causing a reduction in volume of `8ml`. Gas produced after introduction of `H_(2)` in the mixture? Volume of `O_(2)` remained after the first combustion?A. `4ml`B. `2ml`C. `0`D. `8ml` |
| Answer» Correct Answer - C | |
| 1094. |
0.01 mole of iodoform `(CHI_(3))` reacts with Ag to produce a gas whose volume at NTP is `2CHI_(3)+6Ag to6Agl(s)+C_(2)H_(2)(g)`A. 224 mLB. 112 mLC. 336 mLD. None of these |
| Answer» Correct Answer - 2 | |
| 1095. |
Comprehension # 1 Potash is any potassium mineral that is used for its potassium content. Most of the potash produced in the United States goes into fertilizer. The major sources of potash are potassium choride `(KCI)` and potassium sulphate `(K_(2)SO_(4))`. Potash production is often reported as the potassium oxide `(K_(2)O)` equivalent or the amount of `K_(2)O` that could be made from a given mineral. `KCI` cost `Rs 50 per kg` What mass `("in kg")` of `K_(2)O` contains the same number of moles of `K` atoms as `1.00 kg KCI` ?A. `0.158 kg`B. `0.315 kg`C. `1.262 kg`D. `0.631 kg` |
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Answer» Correct Answer - D mole of K in `KCl = (1000)/(74.5) = 13.42` mole of `K_(2)O` form `13.42` mole of `K = (13.42)/(2)=6.71` mass of `K_(2)O` `=6.71xx94=630.8 gm = 0.631 kg` |
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| 1096. |
Comprehension # 1 Potash is any potassium mineral that is used for its potassium content. Most of the potash produced in the United States goes into fertilizer. The major sources of potash are potassium choride `(KCI)` and potassium sulphate `(K_(2)SO_(4))`. Potash production is often reported as the potassium oxide `(K_(2)O)` equivalent or the amount of `K_(2)O` that could be made from a given mineral. `KCI` cost `Rs 50 per kg` For what price must `K_(2)SO_(4)` be sold in order to supply the same amount of potassium as in `KCI` ?A. `"Rs". 58.40 kg^(-1)`B. `"Rs". 50.00 kg^(-1)`C. `"Rs". 42.82 kg^(-1)`D. `"Rs". 25.00 kg^(-1)` |
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Answer» Correct Answer - C `"the price of" K_(2)SO_(4)` `=(50)/(174)xx74.5xx2 implies Rs. `42.82 kg^(-1)` |
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| 1097. |
6 litre of mixture of methane and propane on complete combustion gives 7 litre of `CO_(2)`. Find out the composition of the mixture ?A. `5.5 `lit, `0.5` lit.B. `4.5` lit, `1.5 ` litC. `3.5` lit, `1.5 ` litD. None of these |
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Answer» Correct Answer - A Let the volume of `CH_(4)` in the mixture `=x ` litre. Then the volume of `C_(3)H_(5)` in the mixture `=(6-x)` litre `{:(CH_(4),+,2O_(2),rarr,CO_(2),+,2H_(2)O),(1 litre,,,,1 litre,,),(x litre,,,,x litre,,):}` `{:(C_(3)H_(8),+,5O_(2),rarr,3CO_(2),+,4H_(2)O),(1 litre,,,,3 litre,,),(6-x,,,,3(6-x),,):}` `x+3(6-x)=7` `x+18-3x=7` `-2x=-11` `x=5.5litre`. |
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| 1098. |
A 3 L gas mixture of propane `(C_(3)H_(8))` and butane `(C_(4)H_(10))` on complete combustion at 25 C produced 10 L `CO_(2.)` Assuming constant P and T conditions what was volume of butane present in initial mixture? |
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Answer» Correct Answer - 1 |
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| 1099. |
A gaseous mixture of 3 L of propane `(C_(3)H_(8))` and butane `(C_(4)H_(10))` on complete combustion at `25^(@)C` produced 10 L of `CO_(2)`. Find out the composition of the gaseous mixture. |
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Answer» Correct Answer - 2 L, 1L Let the volume of propane in the mixture = xL `:.` The volume of butane in the mixture `= (3-x)L` Now let us calculate the volume of `CO_(2)` evolved with the help of chemical equations Step I. Calculation of volume of `CO_(2)` from xL of propane The combustion equation for propane is : `underset(1L)(C_(3)H_(8))+5O_(2)rarrunderset(3L)(3CO_(2))+4H_(2)O` 1 L of propane `(C_(3)H_(8))` from `CO_(2)=3L` x L of propane `(C_(3)H_(8))` from `CO_(2)=3xL` Step II. Calculation of volume of `CO_(2)` from (3-x)L of butane The combustion equation for butane is : `underset(1L)(C_(4)H_(10))+(13)/(2)O_(2)rarrunderset(4L)(4CO_(2))+5H_(2)O` 1L of butance `(C_(4)H_(10))` from `CO_(2)=4L` (3-x)L of butane `(C_(4)H_(10))` from `CO_(2) = 4 xx (3-x)L` Step III. Calculation of composition of the mixture Total volume of `CO_(2)` formed in the step I and step II = [3x + 4(3-x)]L But the volume of `CO_(2)` actually formed = 10 L `:. 3x +4(3-x)=10 , 3x +12 - 4x =10 - x = -2 or x = 2L` `:.` Volume of propane = 2L Volume of butane `= (3-2)=1L`. |
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| 1100. |
20 mL of a solution of sulphuric acid neutrlise 21.2 mL of 30 % solution of sodium carbonate. How much water should be added to 100 mL of this solution to bring down its strength to decinormal ? |
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Answer» Step1. Calculation of normality of sulphuric acid solution Strength of `Na_(3)CO_(3)` solution `= 30 g//100 mL = 300 gL^(-1)` Normality of `Na_(2)CO_(3)` solution `= ("Strength")/("Equivalent mass")=((300 gL^(-1)))/((53"g equiv"^(-1)))` `=5.66 "equiv. L"^(-1) =5.66N` `N_(1)xxV_(1)-=N_(2)xxV_(2)or N_(1)=(N_(2)xxV_(2))/(V_(1))` `H_(2)SO_(4))" "(Na_(2)CO_(3))` `N_(1)=((5.66N)xx(21.2"mL"))/((20.0mL))=6.0N` Step II. Calculation of volume of water to be added Normality of `H_(2)SO_(4)` solution `= 6.0 N` Volume of `H_(2)SO_(4)` solution `= 100 mL` Normality of diluted `H_(2)SO_(4)` solution `= 0.1 N` Volume of diluted `H_(2)SO_(4)` solution may be calculated as : `underset(("Diluted"H_(2)SO_(4)))(N_(1)xxV_(1))=underset(("Concentrated" H_(2)SO_(4)))(N_(2)xxV_(2))` `V_(1)=(N_(2)xxV_(2))/(N_(1))=((6.0N)xx(100mL))/((0.1N))=6000mL` Volume of water to be added `= (6000-100)=5900` mL. |
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