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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.
| 93101. |
10 ml of H_(2)O_(2) solution on treatment with Kl and titration of liberated lz required 20 mL of 1 N hypo solution. What is the normality of H_(2)O_(2) solution? |
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Answer» Solution :2 `H_(2)O_(2)+2ItoI_(2)` `I_(2)+2S_(2)O_(3)^(2-)toS_(4)O_(4)^(3-)+2I` `H_(2)O_(2)=I_(2)=2S_(2)O_(3)^(2-)` `N_(1)V_(1)=N_(3)V_(2)` `H_(2)O_(2)` `N_((N_(1)O_(2))=(20xx1)/(10)=2N` |
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| 93102. |
10 ml of H_(2)O_(2) solution on treatment with Kl and titration of liberated l_(2) required 20 mL of 1 N hypo solution. What is the normality of H_(2)O_(2) solution ? |
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Answer» |
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| 93103. |
10 ML. of gaseous hydrocarbonon combustion gives 40 ml . ofCO_(2) (g) and 50 Ml . of H_(2) O (vap) . Thehydrocarbon is |
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Answer» `C_(6) H_(5)` |
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| 93104. |
10 ml of gaseous hydrocarbon on combustion gives 40 ml of CO_2_(g) and 50 ml of H_2O (vap) . The hydrocarbon is : |
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Answer» `C_4H_5` |
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| 93105. |
10 mL of gas having density 1.6 g mL^(-1) liberated by passing 10 amp for 100 minute . The equivalent mass of gas is |
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Answer» 2 `W = (E xxi xx t)/(96500)` and , `W = d xx V = 1.6 xx 10 = 16 g` `therefore 16 = (E xx 10 xx 100 xx 60)/(96500)` `E = 25.8 -= 26` |
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| 93106. |
1.0 mL of ethyl was added to 25 mL of N/2 HCl, 2.0mL of the mixture was withdrawn from time to time during the progress of easter hydrolysis and titrated against standard NaOH solutions. The amount of NaOH required for titration at various intervals is given below: The value of at infinite time was obtained by completing the hydrolysis on boiling. Show that the reaction is of first order. Also find the average value of rate constant. |
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Answer» Solution :INITIAL concentration of ethyl acetate (a) = `V_(infty)-V_(0) = (43.95 -20.24)=23.71mL` `k=(2.303)/tlog a/(a-x) = 2.303/tlog (V_(infty)-V_(0))/(V_(infty-V_(t))` i) At t=20 min,`k=(2.303)/(20min) log (23.71)/(43.95 - 21.73) = (2.303)/(20 min)log(23.71)/(20 min) xx 0.28` `0.00322 min^(-1)` (ii) At t=75 min, k`=(2.303)/(75min) log(23.71)/(43.95-25.20)=(2.303)/(75 min)log(23.71)/(18.75)` `=(2.303)/(75min)(log 23.71 - log 18.75) = (2.303)/(75min)(1.375-1.273)=2.303/(75min) xx 0.102` `=0.00313min^(-1)` iii) At t=110 min, `k=(2.303)/(110min) log23.71/(43.95-27.60) = 2.303/(110min) log(23.71)/(43.95-27.60)= 2.303/(110min) log (23.71)/(16.35)` `=(2.303)/(110min)(log23.71-log16.35) = 2.303/(110min) (1.375-1.244)` `=2.303/(110 min) xx 0.151 = 0.00316 min^(-1)` IV) At t=183min, `k=(2.303)/(183 min)log(23.71)/(43.95-30.22) = 2.303/(183min) log(23.71)/(13.73)` `=2.303/(183 min) (log 23.71 - log 13.73) = 2.303/(183 min)(1.375-1.138)` `=(2.303).(183 min) xx 0.237 = 0.00299 min^(-1)` Since the value of k COMES out to be nearly constant, the hydrolysis of ethyl is a FIRST order reaction: Average value of RATE constant(k)`=(0.00322 + 0.0313 + 0.00316 + 0.00299)/(4) = 0.00312 min^(-1)` |
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| 93107. |
1.0 ml of ethyl acetate was added to 25 ml of N/2 HCl. 2 ml of the mixture were withdrawn from time to time during the progress of the hydrolysis of the ester and titrated against standard NaOH solution. The amount to NaOH required for titration at various intervals is given below : {:("Time (min)",,,:,,,0,,,20,,,75,,,119,,,183,,,oo),("NaOH used (ml)",,,:,,,20.24,,,21.73,,,25.20,,,27.60,,,30.22,,,43.95):} The value at oo time was obtained by completing the hydrolysis on boiling. Show that it is a reaction of the firstorder and find the average value of the velocity constant. |
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Answer» Solution :As explained above, if the REACTION is of the first order, it must obey the EQUATION `k=(2.303)/(t)log""(a)/(a-x)=(2.303)/(t)log""(V_(oo)-V_(0))/(V_(oo)-V_(t))` In the present case, `V_(0)=20.24" ml", V_(oo)=43.95" ml":.V_(oo)-V_(0)=43.95-20.24=23.71" ml"` The value of k at different instants can be calculated as FOLLOWS : `{:("t (min)",,,V_(t),,,V_(oo)-V_(t),,,k=(2.303)/(t)log""(V_(oo)-V_(0))/(V_(oo)-V_(t))),(20,,,21.73,,,43.95-21.73=22.22,,,k=(2.303)/(20min)log""(23.71)/(22.22)=0.00324min^(-1)),(75,,,25.20,,,43.95-25.20=18.75,,,k=(2.303)/(75min)log""(23.71)/(18.75)=0.00313min^(-1)),(119,,,27.60,,,43.95-27.60=16.35,,,k=(2.303)/(119min)log""(23.71)/(16.35)=0.00312min^(-1)),(183,,,30.22,,,43.95-30.22=13.73,,,k=(2.303)/(183min)log""(23.71)/(13.73)=0.00299min^(-1)):}` Since the value of k COMES out to be nearly constant, it is a reaction of the first order. The mean value of `k=0.00312min^(-1).` |
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| 93108. |
10 ml of concentrated HCl were diluted to 1 litre. 20 ml of this diluted solution required 25 ml of 0.1 N sodium hydroxide solution for complete neutralization, the normality of the concentrated acid will be |
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Answer» 8 `N_(1)xx20ml=0.1xx25impliesN_(1)=(0.1xx25)/(20)=0.125` If one LITRE HCl present in 0.125 Therefore in 10 ml `=(0.125)/(1000)xx100=12.5` The normality of CONC. `HCl` is 12.5 N. |
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| 93109. |
1.0 mL of ethyl acetate was added to 25 mL of N//2 HCl, 2 mL of the mixture were withdraw form time to time during the progress of the hydrolyiss of the ester and titrated against standard NaOH solution. The amount of NaOH required for titration at various intervals is given below: The value at oo time was obtained by completing the hydrolyiss on boiling. Show that it is a reaction of the first order and find the average value of the velocity constant. |
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Answer» Solution :Amount of `NaOH` USED at `t = 0` `(V_(0))propHCl` present Amount of `NaOH` used at any time `t=t`, `(V_(t)) PROP HCl "present" +CH_(3)COOH` FORMED. `CH_(3)COOH` formed at any time, `t = prop` Ethyl acetate reacted `= (V_(t)-V_(0)) prop x`. Amount of `NaOH` used at time `t = oo` `(V_(oo)) prop HCl "present + MAXIMUM"CH_(3)COOH` formed maximum `CH_(3)COOH` formed `prop` Initial concentration of ethyl acetate `(V_(oo)-V_(0)) prop a` Hence, if the given reaction is of the first order, it must obey the equation `k = (2.303)/(t)log.(a)/(a-x) = (2.303)/(t)log.(V_(oo)-V_(0))/(V_(oo)-V_(t))` In the present case, `V_(0) = 20.24 mL, V_(oo) = 43.95 mL` `:. V_(oo) - V_(0) = 43.95 - 20.24 = 23.71 mL` The value of `k` at difference instant can be calculated as follows:  Since the value of `k` comes out to be nearly constant, it isa reaction of first order. The value of `k = 0.00312 min^(-1)`. |
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| 93110. |
10 mL of an HCl solution gave 0.1435 g of AgCl when treated with excess of AgNO_(3) .The normality of the resulting solution is |
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Answer» `0.1` |
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| 93111. |
10 mL of a solution containing Na_(2)CO_(3)andNaHCO_(3) is titrated by HCl using phenolphthalein and then methyl orange (added after first end point). The first and second end points were found after adding 10 mL and 15 mL of N/10 HCl respectively. The ratio of m.e. of Na_(2)CO_(3)andNaHCO_(3) in the solution is |
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Answer» `2//1` |
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| 93112. |
10 ml of a solution of H_(2)O_(2) labelled '10 volume'just decolourizes 100 ml of potassium permanganate solution acidified with dilute H_(2)SO_(4). Calculate the amount of potassium permanganate in the given Solution: |
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Answer» `0.1563` GM |
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| 93113. |
10 ml of a mixture of carbon monoxide, march gas and hydrogen exploded with excess of oxygen gave a contraction of 6.5 CC/. There was further contraction of 7 CC when the residual gas was treated with canstic potash.The volume of march gas present in original mixture as |
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Answer» 5CC |
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| 93114. |
10 mL of a gaseous hydrocarbon require 30 mL of oxygen for complete combustion. The hydrocarbon is |
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Answer» `C_2 H_4` (II) `underset(10 mL) (C_2 H_2) + underset(25 mL) (5//2 O_2) to underset(-) (2 CO_2) + underset(-) (H_(2)O)` (iii) `underset(10mL) (C_2 H_6)+ underset(35 mL) (7//2 O_2) to 2 CO_2 + 3H_2O` (iv) `underset(10 mL) (C_3 H_6) + underset(45 mL) (9//2 O_2) to 3 CO_2 + 3 H_2O` It is evident from EQ. (i) that 10 mL of `C_2 H_4` will require 30 mL of `O_2` ,therefore , hydrocarbon is `C_2 H_4`. |
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| 93115. |
10 mL of 1.0 M aqueous solution of Br_(2)is added to excess of NaOH in order to disproportionate quantitatively to Br^(-) " and " BrO_(3)^(-) . The resulting soluting is made free from Br^(-)ion by extraction and excess of OH^(-) neutralized by acidifying the solution. This solution requires 1.5g of an impure CaC_(2)O_(4) sample for complete redox change . Calculate% purity of CaC_(2)O_(4) sample. |
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Answer» Solution :`2Br_(2) + 6OH^(-) to 5Br^(-) + BrO_(3)^(-) + 3H_(2)O` reacts to give redox CHANGE as `Br^(5+) + 6eto Br^(-) ` (VALENCE factor = 6) `C_(2)^(3+) to 2C^(4+) + 2e^(-)` MILLIMOLE of `BrO_(3)^(-)` FORMED ` = 10/3` Meq. of `CaC_(2)O_(4) = ` Meq. of `BrO_(3)^(-) = 10/3 xx 6 = 20` `W_(CaC_(2)O_(4))/((128)/2) xx 1000 = 20 "" W_(CaC_(2)O_(4)) = 1.28 ` g ` :. " of " CaC_(2)O_(4) = (1.28)/(1.5) xx 100 = 85.33` |
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| 93116. |
10 Ml of 1M of H_(2)SO_(4) will completely neutralise |
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Answer» 10 ML of 1M NAOH solution `NaOH hArr Na^(+) + OH^(-)` 1 MOLE of `H_(2)SO_(4)` acid gives 2 MOLES of `H_(3)O^(+)` ions. So 2 moles of `OH^(-)` are required for complete neutralization. |
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| 93117. |
10 mL of 0.1 M-HCI solution is added in 90 mL of a buffer solution having 0.1 M-NH_(4)OH and 0.1 M-NH_(4)CI. The percentage change in pH of solution is (K_(a) " of " NH_(4)^(+) = 5xx10^(-10))[log 2 = 0.3] : |
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Answer» `(10)/(4.7) %` INCREASE |
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| 93118. |
10 ml mixture of H_(2), CH_(4) and CO_(2) was exploded with 15 ml of oxygen. After treatment with KOH the vloume reduced by 6 ml and again on treatment with alkaline pyrogallol, the volume further reduced by 3 ml. Then volume of H_(2) in mixture. |
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Answer» Solution :`{:(H_(2) (g),+,(1)/(2)O_(2) (g),rarr,H_(2)O(l),),(x,,(x)/(2),,,),(-,-,-,,,):}` `{:(CH_(2) (g),+,2O_(2) (g),rarr,CH_(4) (g),+,2H_(2)O(g),),(y,,2y,,-,,-,),(-,,-,,y,,-,):}` `CO_(2) (g) + O_(2) (g) rarr x` `10 - (x + y)` `V_(CO_(2))` (total) = 6 mL (absorbed by KOH) `y + 10 - (x - y) = 6` `10 - x = 6` `x = 4 mL` `V_(H_(2)) = 4 mL` `V_(O_(2)) = 15 - ((x)/(2) + 2y) = 3` (absorbed by alkaline PYROGALLOL) `V_(O_(2)) = 12 = (x)/(2) + 2y` `x + 4y = 24` `4y = 20` `y = 5` `V_(O_(2)) = 5 mL` |
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| 93119. |
10 ml mixture of H_(2), CH_(4) and CO_(2) was exploded with 15 ml of oxygen. After treatment with KOH the vloume reduced by 6 ml and again on treatment with alkaline pyrogallol, the volume further reduced by 3 ml. Percentage (%) composition of CH_(4) in the mixture. |
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Answer» 0.4 `{:(CH_(2) (g),+,2O_(2) (g),rarr,CH_(4) (g),+,2H_(2)O(g),),(y,,2y,,-,,-,),(-,,-,,y,,-,):}` `CO_(2) (g) + O_(2) (g) rarr x` `10 - (x + y)` `V_(CO_(2))` (total) = 6 mL (absorbed by KOH) `y + 10 - (x - y) = 6` `10 - x = 6` `x = 4 mL` `V_(H_(2)) = 4 mL` `V_(O_(2)) = 15 - ((x)/(2) + 2y) = 3` (absorbed by alkaline pyrogallol) `V_(O_(2)) = 12 = (x)/(2) + 2y` `x + 4y = 24` `4y = 20` `y = 5` `V_(O_(2)) = 5 mL` |
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| 93120. |
10 millimoles of a diacidic base exactly neutralises 100ml of an acid. Then the Normality of that acid is |
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Answer» `0.2 N` |
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| 93121. |
10 mg of an adsorbate gets adsorbed on a surface. This cause the release of 3J of heat constant pressure and at 27^(@)C. ["Molar mass of adsorbate" =100 g//mol]. (i) Find DeltaH_(AD) (ii) Argue whether the adsorption is physical or chemical? (iii) If 20 mg of adsorbate is adsorbed a temperature T_(0). Then compare T_(0) and 27^(@)C. |
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Answer» Adsorption is physical. Hence, the AMOUNT of substance adsorb will temperature. Therefore, `T_(0) lt 27^(@)C`. |
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| 93122. |
10 milimoles of a solid A when dissolved in 10 noles of a liquid solvent B starts trimerisation process obeying first order kinetics (with the trimer also soluble in the solvent B). On addition of 5 mili-moles of another soluble substance C after 10 minutes the trimarisation compeletely stops. The reresulting solution is cooled to some temperature lower than 49.96^(@) C (melting point of solution obtained after mixing C)to cause solidification of some B. The remaining solution after removal of solid B is heated to 70^(@) Cwhere vapour pressure was found to be 200 mm of Hg. Given: Vapour pressure of liquid B at 70^(@) C =201 mm Of Hg, Normal melting point of B = 50^(@) C, Cryoscopic constant = 2 K kg mol^(-1), molar mass of B = 50 g mol^(-1). Calcualate a four digit number abcd where ab=half life of trimarisation of A (in min) cd=moles of B solidified. |
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Answer» |
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| 93123. |
1.0 M solution of which of the following salts is most basic ? |
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Answer» NaClO Basic nature order `= NaClO gt NaClO_(2) gt NaClO_(3) gt NaClO_(4)` |
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| 93124. |
1.0 M aqueous solutions of AgNO_(3), Cu(NO_(3))_(2) and Au(NO_(3))_(3) are electrolyzed in the apparatus shown, so that same amount of electricity passes through each solutions. If 0.10 mols of solid Cu are formed how many "mole"s of Ag and Au are formed? |
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Answer» 0.10 "MOLE"s AG, 0.10 "mole"s Au |
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| 93125. |
10 litre solution of urea contains 240g urea. The active mass of urea will be |
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Answer» `0.04` `therefore` ACTIVE MASS `=(240)/(60XX10)=0.4`. |
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| 93126. |
10 litres of 0.1 MAgNO_3 is electrolysed with 2 faradays . '5.6 x' litres of O_2 is released as STP what is x ? |
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Answer» |
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| 93127. |
10 grams of a solute is dissolved in 90 grams of a solvent. Its mass percent in solution is |
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Answer» `0.01` `=(10)/(90+10)xx100=10` |
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| 93128. |
1.0 gram of a monobasic acid HA in 100 gram H_2O lower the freezing point by 0.155 K.0.45 gram of same acid require 15 ml of 1/5M KOH solution for complete neutralisation.If the degree of dissociation of acid is alpha, then value of '20alpha' is : (K_f for H_2O =1.86 K.Kg/mole) |
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Answer» `(M_("acid")_(exp)=120` Also `N_1V_1=N_2V_2` `0.45/((M_("acid")_"thor"))=1/5xx15xx1/1000 IMPLIES (M_("acid")_("thor"))=(0.45xx5xx1000)/15=150` `i=(M_("thor")/M_("exp"))=150/120=1.25 implies thereforealpha=(i-1)=1.25-1=0.25` |
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| 93129. |
1.0 gm of moist sample of mixture of potassium chlorate (KClO_(3)) and potassium chloride (KCl) was dissolved in water and solution was made upto 250 ml. This solution was treated with SO_(2) to reduce all ClO_(3)^(-) " to " Cl^(-) and excess of SO_(2) was removed by boiling . The total chloride was precipitated as silver chloride . The weight of precipitate was found to be 0.1435 gm. In another experiment , 25 ml of original solutions was heatod with 30 ml 0.2 N FeSO_(4) and unused FeSO_(4)required 37.5ml of 0.08 N KMnO_(4) solutions . Calculate the molar ratio of the ClO_(3)^(-) to the Cl^(-) in the given mixture Given that , ClO_(3)^(-) + 6l^(-) e^(2+) + 6H^(+) to Cl^(-) + 6Fe^(3+) + 3H_(2)O 3SO_(2) + ClO_(3)^(-) + 3H_(2)O to Cl^(-) + 3SO_(4)^(2-) + 6H^(+) |
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Answer» Solution :`ClO_(3)^(-)` is reduced to `Cl^(-) " by " SO_(2) " and " ClO_(3)^(-)` is also reduced to `Cl^(-) " by " FE^(2+) ` , hence AgCl is formed due to TOTAL `Cl^(-)` Meq. of `Fe^(2+) `initially taken ` = 30 xx 0.2 = 6` Meq. of `Fe^(2+)` unused` = 37 . 5 xx 0.08 = 3` ` :. ` Meq of `Fe^(2+) " used " = 6.0 - 3.0 = 3.0` Thus, Meq. of `ClO_(3)^(-) " in " 25 ml= 3.0` Moles of `ClO_(3)^(-) " in " 25 ml = (3.0)/(1000 xx 6) = 0.0005` `.^(+5)ClO_(3)^(-) to .^(-1)Cl^(-) `(n-factor 6) 0.N `5 .... -1` Thus , moles of `ClO_(3)^(-)` in 25 ml solution ` = 0.0005` `ClO_(3)^(-)` is also reduced to `Cl^(-)` by `SO_(2)` in first experiment andprecipitated as AgCl. Thus , `Cl^(-)` formed from `ClO_(3)^(-) =AgCl` from `ClO_(3)^(-) = 0.0005` Total AgCl formed both from actual and `Cl^(-) " from " ClO_(3)^(-) = 0.1435` GM ` = (0.1435)/(143.5)=0.0010` mol Thus, AgCl formed due to `Cl^(-) ` only = `0.0010 - 0.0005 = 0.0005` mol Thus , `ClO_(3)^(-) " and " Cl^(-)` are in molar ratio ` = 1 : 1 `. |
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| 93130. |
1.0 gm of mixture of CaCO_3 and MgCo_3 on one complete of decomposition gave 240 ml of CO_2 at NTP. Calculate the percentage composition of mixture. |
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Answer» |
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| 93131. |
10 gm of Helium at 127^(@) C is expanded isothermally from 100 atm to 1 atm. Calculate the work done when the expansion is carried out (i) in single step (ii) in three steps the intermediate pressure being 60 and 30 atm respectively and (iii) reversibly. |
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Answer» <P> SOLUTION :(i) Work donue = `V.DeltaP``V=10/4 xx (8.314 xx 400)/(100 xx 10^(5)) = 83.14 xx 10^(-5) m^(3)` So `W = (83.14)/10^(5) (100-1) xx 10^(5) = 8230.86` J (ii) In three steps. `V_(1) = 83.14 xx 10^(-5) m^(3)` `W_(1) =(83.14 xx 10^(-5)) xx (100-60) xx 10^(5)` `=3325.6` Jules. `V_(u) = (2.5 xx 8.314 xx 400)/(60 xx 10^(3)) = 138.56 xx 10^(-5) m^(3)` `W_(R) = 138.56 xx 10^(-5) (60-30) xx 10^(5)` `=4156.99 = 4157` J. `V_(m) =(2.5 xx 8.314 xx 400)/(30 xx 10^(5)) = 277.13 xx 10^(-5) m^(3)` `W_("total") = W_(I) + W_(II) + W_(III)` `=3325.6 + 4156.909 + 8036.86 = 15519.45` J (iii) For reversible PROCESS `W = 2.303 nRT log P_(1)/P_(2)` `=2.303 xx 10/4 xx 8.314 xx 400 xx log (100/1)` `W = 38294.28` Joules. |
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| 93132. |
1.0 gm of an alloy of aluminium and magnesium when treated with excess of dilute HCl forms magnesium chloride aluminium chloride and hydrogen. The evolved hydrogen collected over mercury at 0^@C has a volume of 1.20 litres at 0.92 atm pressure Calculate the composition of alloy.[Al = 27 , Mg=24.3] |
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Answer» |
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| 93133. |
10 gm of helium at 127^(@) C is expanded isothermally from 100 atm to 1 atm.Calculate the work done when the expansion is carried out (i) in single step, (ii) In three steps, the intermediate pressure being 60 and 30 atm respectively and (iii) reversibly. |
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Answer» Solution :(i) Work DONE = `V.DeltaP` `V = 10/4 xx (8.314 xx 400)/(100 xx 10^(5)) = 83.14 xx 10^(-5) m^(3)` So, `W = (83.14)/10^(5) (100-1) xx 10^(5) = 8230.86` J (ii) Ini three steps, `V_(i) = 83.14 xx 10^(5) m^(3)` `w_(i)= (83.14 xx 10^(-5)) xx (100-60) xx 10^(5)` `=3325.6` J `V_(3) = (2.5 xx 8.314 xx 400)/(60 xx 10^(5)) = 138.56 xx 10^(-5) m^(3)` `w_(a) =V.DeltaP` `w_(1) = 138.56 xx 10^(-5)(60-30) xx 10^(5)` `=4156.99 = 4157` J `V_(|||) = (2.5 xx 8.314 xx 400)/(30 xx 10^(5)) = 277. 1.3 xx 10^(-5) m^(3)` `w_(|||) = 8036.88` J `W_("total") =w_(1) + w_(2) + w_(3)` `=3325.6 + 4156.909 + 8036.86 = 15519.45` J (iii) For reversible process. `w = 2.303 nRT LOG P_(1)/P_(2)` `=2.303 xx (10/4) xx 8.314 xx 400 xx log(100/1)` `w = 38294.28` JULES. |
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| 93134. |
1.0 g radioactive sodium on decay becomes 0.25 g in 16 hours. How much tie 48 g of same radioactive sodium will need to become 3.0 g |
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Answer» 48hours |
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| 93135. |
10 g of S reacts with excess of O_2 to form 15 g of SO_2. The % yield of the reaction is |
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Answer» 0.25 |
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| 93136. |
10 g of neon initinally at a pressure of 506.625 kPa and temperature of 473 K expandadiabaticallyto a pressure of 202.65 kPa .Calculateentropyofthe system and oftotalentropy changefor the following waysof carryingout is this expamsion .(i) Expansion is carriedout expansion . (ii) Expansionoccursaganista constant external pressureof 202.65 kPa . (iii) Expansion is a free expansion. |
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Answer» `DeltaS_(surr)= 0""DeltaS_(sure)= - DeltaS_(SYSTEM) = 0` `DeltaS_("total") = DeltaS _(SYS)+DeltaS = 0 ` Hence `DeltaS_(sys)= DeltaS_(surr)= DeltaS_("total") =0` (ii)`DeltaS_(surr)= 0` `DeltaS_(sys)= nC_(V)In ((T_(2))/(T_(1)))+ nR In ((V_(2))/(V_(1)))` `= nC_(V) In ((T_(2))/(T_(1))) + nR In ((P_(1))/(P_(2))(T_(1))/(T_(2)))` `= N[C_(V) In ((T_(2))/(T_(1))) + nR In ((T_(2))/(T_(1)))+ In((P_(1))/(P_(2)))]` `= n[C_(V) In ((T_(2))/(T_(1))) + nR In((P_(1))/(P_(2)))]""[T_(2)" of irreverrsubile adiabatic will be calculated "]` `= 0.957 J//K""byw = DeltaU rArr - P_(ext)(V_(2)-V_(1))=nC_(V,m)(T_(1)-T_(2))` |
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| 93137. |
1.0 g of Mg is burnt with 0.28 g of O_(2)in a closed vessel. Which reactant is left in excess and how much ? |
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Answer» Mg, 5.8 g 48G of Mg requires 32 g of `O_(2)` ? requires 0.28 g of `O_(2)` Mass of magnesium REQUIRED `=(48xx0.28)/(32)=0.42g` But 1 g of Mg is available. Thus, Mg is the excess REAGENT. Excess of Mg LEFT BEHIND `=1-0.42=0.58g` |
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| 93138. |
1.0 g of magnesium is burnt with 0.56 g O_(2) in a closed vessel. Which reactant is left in excess and how much ? (At. Wt. of Mg = 24, O = 16) |
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Answer» Mg, 0.16 g 1 g of Mg needs `O_(2)=(32)/(48)g=(2)/(3)g=0.67g` which is not present. `0.56` g of `O_(2)` will react with Mg `=(48)/(32)xx0.56=0.84` Thus, Mg is left in excess and amount left `=1-0.84g=0.16g` |
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| 93139. |
10 g of KClO_3 ON heating gave enough oxygen to react completely with hydrogen produced by the action of dil H_2SO_4 on zinc. Required for this purpose. [K=39 , Cl=35.5 , Zn=65] |
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Answer» |
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| 93140. |
10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Maximum amount of water produced in this reaction will be :- |
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Answer» 2 mole |
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| 93141. |
10 g of hydrogen and 64g of oxygen were filledin a steel vessel and exploded. Amount of water produced in this reaction will be |
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Answer» 3 mol LIMITING reagent is `O_2` 1mol of `O_2` gives2 MOLES `H_2O` ` therefore ` 2 moles of `O_2` will give 4 moles `H_2O` |
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| 93142. |
10 g of hydrochloric acid gas occupies 5.6 litre of volume at STP . The emperical formulae of the gas is HF , then its molecular formula in the gaseous state will be |
| Answer» Answer :B | |
| 93143. |
10 g of argon gas is compressed isothermally and reversibly at a temperature of 27^(@)C from 10 L of 5L. Calculate q, W, DeltaU and Delta H for this process. R = 2.0 cal K^(-1) "mol"^(-1), "log"_(10) 2 = 0.30, Atomic wt. of Ar = 40. |
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Answer» |
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| 93144. |
10 g of an orgainc compound is dissolved per litre of the solution and gave an osmotic pressure of 1.18 atmosphere at 273 K. Calculate the molecular mass of compound. |
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Answer» `M_(B)=(W_(B)xxRxxT)/(pixxV)=((10g)xx(0.0821"L atm K"^(-1)mol^(-1))xx(273 K))/((1.18"atm")xx(1L))=189.94" g mol"^(-1)`. |
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| 93145. |
1.0 g of an oxide of metal M contained 0.5 g of M and 4.0 g of another oxide of M contained 1.6 g of M. These data illustrate the |
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Answer» Law of RECIPROCAL proportion Mass of oxygen = 1.0-0.5 = 0.5 g Mass of oxygen which COMBINES with 0.5 g of metal = 0.5 g Mass of oxygen in second oxide = 4.0 - 1.6 = 2.4 g Mass of oxygen which combines with 1.6 g of metal = 2.4 g Mass of oxygen which combines with 0.5 GOF metal ` = (2.4)/(1.6) xx 0.5 = 0.75 g ` Ratio of different masses of oxygen combining with 0.5 g of metal 0.5 : 0.75 2 : 3 sicne it is simple ratio ,it illustrates law of multiple proportions . |
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| 93146. |
10 g of a substance were dissolved in water and the solution was made up to 250 cm^(3). The osmotic pressure of the solution was found to be 8xx10^(5)Nm^(-2) (pascals) at 288 K. Find the molar mass of the solute. |
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Answer» Solution :Here, `w_(2)=10g, V=250cm^(3)=250xx10^(-6)m^(3)=2.5xx10^(-4)m^(3)` `pi=8xx10^(5)Mn^(-2),R="8.314 J K"^(-1)"mol"^(-1), T = 288K` `THEREFORE""M_(2)=(w_(2)RT)/(piV)=((10g)("8.314 Nm K"^(-1)"mol"^(-1))(288K))/((8xx10^(5)Nm^(-2))XX(2.5xx10^(-4)m^(3)))(1J = 1Nm)="119.7 G mol"^(-1)` |
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| 93147. |
1.0 g of a radioactive isotope was found to reduce to 125 mg after 24 hours. The half-life of the isotope is |
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Answer» 8 HOURS `((1)/(2))^(n) = (125)/(1000) = (1)/(8)` `((1)/(2))^(n) = ((1)/(2))^(3),n = 3`, so number, of `t_(1//2) = 3` TOTAL time = 24 hours, Half-life time = `(24)/(3)` = 8 hours. |
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| 93148. |
10 g of a substance was dissolved in water and the solution was made up to 250 cm^3 . The osmotic pressure of the solution was found to be 8xx10^5 Nm^(-2) at 288 K. Find the molecular weight of the solute. |
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Answer» SOLUTION :1CC = `10^(-6) m^3` 119.72 |
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| 93149. |
10 g of a radioactive isotope is reduced to 1.25 g in 12 years. Therefore, half-life period of the isotope is |
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Answer» Solution :(b) : `[A]=([A]_(0))/(2^(n)):.1.25=(10)/(2^(n))" or "2^(n)=(10)/(1.25)=8=2^(3):.n=3` i.e., 3 half LIVES = 12 yrs. `" or "t_(1//2)=4" yr"` ALTERNATIVELY, `10goverset(t_(1//2))to5goverset(t_(1//2))to2.5goverset(t_(1//2))to1.25g` `3xxt_(1//2)=12" yror"t_(1//2)=4" yr".` |
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| 93150. |
10 g of a piece of marble was put into excess of dilute HCl acid. When the reaction was complete, 1120 cm^(3)of CO_(2)was obtained at STP. The percentage of CaCO_(3)in the marble is |
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Answer» `25%` 1120 cc of `CO_(2)` are obtained from pure `CaCO_(3)``=100/22400 xx 1120 = 50 g` % of `CaCO_(3) = 5/10 xx 100 = 50%` |
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