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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
One litre flask containing vapour of methyl alcohol (Mol mass 32) at a pressure of 1 atm. And 25^(@)Cwas evacuated till the final pressure was 10^(-3)mm. How many molecules of methyl alcohol were left in the flask ? |
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Answer» For converting this volume to volume at S.T.P. , apply `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)), i.e., (10^(-3)xx1000)/(298)=(760xxV_(2))/(273)"or" V_(2)=1.205xx10^(-3)cm^(3)` `22400 cm^(3)` at S.T.P.`-=6.02xx10^(23)` molecules `:. 1.205xx10^(-3)cm^(3)` at S.T.P. `-=(6.02xx10^(23))/(22400)xx1.025xx10^(-3)` molecules `=3.24xx10^(16)` molecules |
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| 2. |
One litre each of 1M Al_(2)(SO_(4))_(3) and 1M BaCl_(2) are mixed. What is the molarity of sulphate ions in the resultant solution? |
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Answer» eq. `BaCl_(2)=1xx1xx2=2eq`. `THEREFORE eq [SO_(4)^(2-)]` left `=6-2=4` eq `therefore [SO_(4)^(2-)]=(4)/(2)=2N=1M` |
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| 4. |
One litre 2M Na_(3)PO_(4) and 1 litre of 1M BaCl_(2) solutions are mixed. What is the normality of phosphate ions in the filtrate? |
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Answer» eq. `BaCl_(2)=1xx1xx2=2` `therefore` eq. `PO_(4)^(3-)` ppted =2 `therefore` eq. in FILTRATE = `6-2=4` `therefore [PO_(4)^(3-)]=(4)/(1+1)=2` |
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| 5. |
One lit. of aqueous solution of H_(2)SO_(4) contain 4.9 g of the acid . 100 ml of this solution is taken in a 1 lit. flask and diluted with water upto the mark . The pH of the dilute solution is |
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Answer» `2.0` |
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| 6. |
One lit of SO_(3)was placed in a two litre vessels of a certain temperature. The following equilibrium was established in the vessel 2SO_(3(g)) hArr2SO_(2(g)) +O_(2(g)) the equilibrium mixture reacted with 0.2 mole KMnO, in acidic medium. Kc value is 1.25 x10^(-x) then the value of x is: |
Answer»  Only `SO_(2)` is will oxidised So, no. of EQUIVALENT of `SO_(2)` = equivalent of `KMnO_(4)` `2x xx 2=0.2 xx 5 implies 2x=0.5` `Kc=(((0.5)/(2))^(2)((0.25)/(2)))/(((0.5)/(2))^(2))=0.125` `implies 1.25 xx 10^(-1)=1.25 xx 10^(-x) implies x=1` |
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| 7. |
One lakh atoms of gold weigh 3.271 xx 10^(-17) g. What is the atomic mass of gold? |
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| 8. |
One liquid contain nonvolatile impurity. What technique will be applied for purification of it? |
| Answer» SOLUTION :DISTILLATION | |
| 9. |
One kilogram sample of hard water contains 4.44 mg of CaCl_(2) and 1.9 mg of NaCl. The total hardness in tems of ppm of CaCO_(3) is : |
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| 10. |
One kg of graphite is burnt in a closed vessel. The same amount of the same sample is burnt in an open vessel. Will the heat evolved in the two cases be same ? If not, in which case itwould be greater? |
| Answer» SOLUTION :Same in both CASES because `Deltan_(G) = 0`. | |
| 11. |
One litre of a solution contains 18.9 gm of HNO_(3) and one lire of another solution contains 3.2 gm of NaOH. In what volume ratio must these solutions be mixed to obtain a neutral solution? |
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Answer» Solution : X.is a buffer solution ` PH = pk_a+ log ([" Salt"])/(["acid"])` .Y. is a solution CONTAINING acid HCl and base `CH_3 COOK `which give 1 mole of `CH_3 COOH,` a weak base. ` pH =(pK_a - logC)/(2) = (pK_a)/(2)` The pH RATIO of solution .X. and solution .Y. `= PK_a ,( pk_a)/(2) =2:1` |
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| 12. |
To convert 1 litre of 1.123N solution of an acid into 1N solution, how much volume of water should be added ? |
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| 13. |
One kilogram of sea water contains 6 mg of dissolved O_(2). The concentration of O_(2) in the sample in ppm is : |
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Answer» 0.6 `10^(6)` mg water contains 6 mg `O_(2)` `THEREFORE` CONCENTRATION of `O_(2)` is 6 PPM. |
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| 14. |
One important source of energy of volacans erruption is: |
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Answer» Hot mealn STEAM TRAPPED in earth |
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| 15. |
One hydrocarbon X in presence of Pd, n-butane is formed on hydrogenation. X treated with acidic KMnO_(4), on oxidation carboxylic acid is formed of 2 carbon product. Give the structure of X and IUPAC name ? |
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Answer» Solution :2 carbon carboxylic ACID means `CH_(3)COOH` is formed. On hydrogenation of X n-Butane is formed. X is 4 carbon containing alkene, 2-butene is taken, reaction is as follows : structure of X : `CH_(3)CH=CHCH_(3)` 
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| 16. |
One hundred Litre of a vessel contains 14g of carbondioxide at 27^@C. What is the density ? |
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Answer» Solution :The idealgas equation, PV = nRT DENSITY of gas, `d = (PM)/(RT) or d = (W)/(V)` W = mass of gas = 14 g, V = volume = 100 L Substituting the values, the density of carbondioxide ` = (14)/(100) = 0.4 g L^(-1)` |
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| 17. |
One having high vapour pressure at temperature below its melting point is: |
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Answer» BENZOIC acid |
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| 18. |
On gram of the carbonate of a metal was dissolve din 35 CC 1N HCl.The resulting liquid required 50 CC N/10 caustic soda solution to neutralise it completely. The equivalent weight of metal carbonate is |
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Answer» 100 `=5C C 1NHCl` Volume of ACID left = `5C C 1NHCl` Volume of acid used by metal carbonate = `25.5 = 20C C 1 NHCl` No. of MeQ of `HCl=20xx1=20` No. of meQ of metal carbonate = `(1)/("GEW of M.C")xxoverset("1")(1000)` No. of mecq of HCl = No. of mecq of metal carbonate `20=(1)/("GSW of M.C.")XX100` GTW of M.C = 50 `therefore` EQ wt of M.C. = 50 |
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| 19. |
One gram of copper is dissolved in nitric acid and crystal hydrate Cu(NO_(3))_(2).3H_(2)O is obtained by evaporation . What is the weight of residue? |
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| 20. |
One gram of calcium is treated with excess dilute hydrochloric acid. Calculate the volume of hyrogen liberated at 700 torr 27^(@)C? |
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| 21. |
One gram of an alloy of Mg and Al, when treated with excess dilute hydrochloric acid gave 1.2L hydrogen at O""^(@)C and 0.92atm. Calculate the weight percentage of Mg in the alloy |
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| 22. |
One gram of anorganic compound on oxidation gave 2.2g CO_(2) and 0.9g of H_(2)O .Caculate the weight percentage of carbon and hydrogen. |
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| 23. |
One gram of an alloy of Al and Mg when treated with excess of dilute HCI forms MgCI_2, AICI_3 and hydrogen. The evolved hydrogen collected over Hg at 0°C has a volume of 1.20 litres at 0.92 atm pressure. Calculate the composition of the alloy. ("AI" = 27, "Mg" = 24) |
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Answer» Solution :Volume of `H_2` produced at NTP `= ( 1.2 xx 0.92)/( 273) xx (273)/(1)` `((p_1 V_1)/( T_1) = ( p_2 V_2)/( T_2) )` `= 1.104` LITRES. SINCE both `Al and Mg` produce `1.104` litres of `H_2`, equivalent of `Al+` equivalent of Mg `=`equivalent of hydrogen. `THEREFORE ("weight of Al")/( "eq. wt. of Al") + ("weight of Mg")/("eq. wt. of Mg")="eq. of hydrogen" = ("volume of hydrogen (NTP)")/("equivalent volume of hydrogen")`. Let the weight of `AI` be `x` g. `therefore (x)/( 27//3) + (1-x)/( 24//2) = (1.104)/( 11.2) ""(E_("AI")= (27)/(3) )` `(E_("Mg") = (24)/(2) )` ` x = 0.55` `(1-x) = 0.45`. Thus, weight of `"AI" = 0.55 g` and weight of `"Mg" = 0.45 g`. |
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| 24. |
One gram of a mixture of KMnO_(4) and K_(2)Cr_(2)O_(7) was treated with excess Kl in acid solution. The liberated iodine requred 200mL of 0.15N hypo for titration. Calculate the pracentage weight of potassium permanganate in the mixture. |
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| 25. |
One gram of a metallic oxide on reduction gives 0.68 g of the metal. The equivalent mass of the metal is : |
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Answer» 68 `= ("Mass of metal")/("Mass of OXYGEN")XX8=(0.68)/(0.32)xx8=17`. |
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| 26. |
One gram of a bromoalkane on heating with excess silver nitrate in Carius tube method gave 0.94g of yellow precipitate. What is the percent weight of halogen? |
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Answer» Solution :Weight of substance = 1g Yellow precipitate is AGBR: Weight of AgBr = 0.94g % of Bromne `= (09.4 xx 80 xx 100)/(1 xx 188) = 40` |
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| 27. |
One gram molecule of oxygen is |
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Answer» 16 GMS of OXYGEN |
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| 28. |
One gram mole, one gram molecule, and one gram molecular mass have the........meaning. |
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| 29. |
One gram mole of a monoatomic gas occupies 22.4 L at S.T.P |
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| 30. |
One gram mole of a gas at NTP occupies 22.4 litre as volume. This fact was derived from : |
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Answer» DALTON's theory |
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| 31. |
One gram limestone is heated and quickline so formed is dissolved in one litre of water. The normality of solution is 0.01xx x. What is value of x. |
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Answer» 100 g ________ 56 g 1 g ________ 0.56g Normality = `(0.56)/(28)xx(1)/(1)=0.02N` Eq. wt of Quicklime `(CAO)=28` |
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| 32. |
One gram formula weight of copper sulphate (CuSO_4) contains |
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Answer» one ATOM of copper |
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| 33. |
One gas having ratio of density is 1 : 2 and temperature ratio is 2 : 1 then what is ratio of pressure ? |
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Answer» Solution :1 : 1 `d_(1)=(Mp_(1))/(RT_(1)) "" d_(2)=(Mp_(2))/(RT_(2))` `THEREFORE p_(1)=(d_(1)RT_(1))/(M) p_(2)=(d_(2)RT_(2))/(M)` `therefore (p_(1))/(p_(2))=(1xx R xx2)/(M)XX(M)/(2xx R xx1)=(1)/(1)` (MOLAR mass M = constant, gas can.t change but state can change). |
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| 34. |
One following reaction which product is formed HC -= CH+NaNH_(2) overset(NaNH_(3))rarr ? |
| Answer» Answer :D | |
| 35. |
One element has atomic weight 39. Its electronic configuration is 1s^(2), 2s^(2) 2p^(6),3s^(2) 3p^(6) 4s^(1). The true statement for that element is: |
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Answer» Hight VALUE of IE |
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| 36. |
One Debye (D) equal to |
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Answer» `1xx10^(-4)` ESU. Cm |
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| 37. |
One container having 0.4 g O_(2) and 0.06 g H_(2) at 100^(@)C then (a) What will be total pressure of container ?(b) This mixture of gas produce water at 100^(@)C temperature then which gas will left in container ? Calculate their partial pressure. (R = 6.0821 L atm mol^(-1)K^(-1)) |
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Answer» <P> |
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| 38. |
One compound reaction with Cl_(2), gives product B and product B on (Zn + dil. HCl) treated then we get our compound back A. Give the structure of A and B. |
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Answer» Solution :If COMPOUND (A) = Ethane, then it is possible. `UNDERSET("Ethane")underset((A))(CH_(3)CH_(3))underset(-HCl)overset(Cl_(2), hv)rarr underset("Chloroethane")underset((B))(CH_(3)CH_(2)CL) underset("REDUCTION -HCl")overset(Zn, H^(+), +H_(2))rarr underset("Ethane")underset((A))(CH_(3)CH_(3))` |
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| 39. |
One compoud contain 4.07%,Hydrogen24.47% carbon and71.65% chlorine. Its molar mass is 98.96 g. Find empirical formula and molecular formula. |
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Answer» Molecular formula`C_(2)H_(4)Cl_(2)` |
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| 40. |
One commercial system removes SO_(2) emission from smoke at 95^(@)C by the following set of reactions: SO_(2)(g) + Cl_(2)(g) rightarrow SO_(2)Cl_(2)(g) ……………..(i) SO_(2)Cl_(2) + 2H_(2)O rightarrow H_(2)SO_(4) + 2HCl ……………………….(ii) H_(2)SO_(4) + Ca(HO)_(2) rightarrow CaSO_(4) + 2H_(2)O...............(iii) Assuming the process to be 95% efficient, how many moles of CaSO_(4) may be produced from 128g SO_(2)? [Ca = 40, S=32, O=16] |
| Answer» Answer :A | |
| 41. |
One carbonyl hydrocarbon (X) on addition of H_(2) gas addition product on ozonolysis of 'X' gives (i) Ethanal (ii) Propanone and (iii) Butane-1, 4-diol. Give the structure of 'X' and IUPAC name. |
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Answer» Solution :(i) No of `pi`-bond in= addition of `H_(2)` `therefore` Thee are `2pi`-bond in .X. HYDROCARBON. (II) On ozonolysis, 3 products are formed. (a) Ethanal `CH_(3)CHO` or `CH_(3)CH=O` (b) Propanone `CH_(3)COCH_(3)` or `CH_(3)-underset(CH_(3))underset(|)(C)=O` (c) Butane-1, 4-diol `CHOCH_(2)CH_(2)CHO` or `O=CHCH_(2)CH_(2)CH=O` (iii) On removing .O. and PUTTING double bond is (a), (b) and (c) above structure. The structure .X. can be written as. [(a) and (b) at end and (c) at middle] `(CH_(3)CH)/((a))=(CH-CH_(2)CH_(2)CH)/((b))=underset((CH_(3))/((c)))underset(|)(C)-CH_(3)` IUPAC NAME : 2-Methyl-xota-2, 6-diene. |
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| 42. |
One can distinguish between HCOOH and CH_(3)COOH with : |
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Answer» `NaHCO_(3)` |
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| 44. |
Explain water gas shift reaction. |
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| 45. |
One atom of mercury has a mass of 333 xx 10^(-24) g . Its atomic weight is |
| Answer» Answer :A | |
| 46. |
One atom of an element X weighs 6.644 xx 10^(-23) g. Calculate the number of gram-atoms in 40 kg of it. |
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Answer» SOLUTION :Wt. of 1 mole of atoms of X = wt. of 1 atom x Av. const. `=6.44 xx 10^(-23) xx 6.022 xx 10^(23)` = 40 G Thus, the at. Wt. Of X=40 No. Of moles (or gram-atoms) of X `=("WEIGHT in g")/("at. Wt")` `=(40 xx 1000)/40 = 1000` |
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| 47. |
One atom of an element weigs 6.644xx10^(-26)kg.How many gram atoms are present in 40kg of the element ? |
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| 48. |
One atom of an element weighs 1.8 xx 10^(-22)g, its atomic mass is |
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Answer» 18 `=6.022 xx 10^(23) xx 1.8 xx 10^(-23)` =108.39 |
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| 49. |
Oneatom has(z=25) states electron |
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Answer» Solution :ZS= 25 ,`1S^(2)2s^(2) 2p^(6)3s^(2)3p^(6)3d^(5)4s^(2)` Selectronare `=4XX 2 =8` |
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