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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your Class 11 knowledge and support exam preparation. Choose a topic below to get started.
| 51851. |
One gram - atom of oxygen is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> of <a href="https://interviewquestions.tuteehub.com/tag/oxygen-1144542" style="font-weight:bold;" target="_blank" title="Click to know more about OXYGEN">OXYGEN</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/16g-277851" style="font-weight:bold;" target="_blank" title="Click to know more about 16G">16G</a> of oxygen<br/>22.4 g of oxygen<br/>8g of oxygen</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 51852. |
1.00 gm of a mixture having equal number of moles of carbonates of two alkali metals required 44.4 ml of 0.5 N HCl for complete reaction. Atomic weight of one of the metals is 7.00 The number of moles of each metal carbonate in |
| Answer» <html><body><p>`0.1`<br/>`0.0111`<br/>`0.0055`<br/>`0.00275`</p>Solution :Equal moles of carbonates & <br/> <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a>. `"<a href="https://interviewquestions.tuteehub.com/tag/base-892693" style="font-weight:bold;" target="_blank" title="Click to know more about BASE">BASE</a>"_(1)+` eq `"base"_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)` = eq HCl <br/> <a href="https://interviewquestions.tuteehub.com/tag/implies-1037962" style="font-weight:bold;" target="_blank" title="Click to know more about IMPLIES">IMPLIES</a> eq. of each base = 0.0111 <br/> implies moles = `(0.0111)/(2)=0.0055`</body></html> | |
| 51853. |
(1) Give electronic configuration of the following species and show which of these contain same number of electrons: Cl^(-),N^(3-),P^(3-),K^(+),Na^(+),Mg^(2+),Ar,S^(2-),Ne,O^(2-),Al^(3+),Ca^(2+). (2) In an atom of an inert gas, the difference between the number of p-electrons and s-electrons is equal to the number of d-electrons present in that atom. identify the inert gas and indicate its atomic number. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :(1) <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CHY_CHE_XI_P1_U02_E03_065_S01.png" width="80%"/> <br/> So, `N^(3-),Na^(+),Mg^(2+),O^(2-)Ne,Al^(3+)` have the <a href="https://interviewquestions.tuteehub.com/tag/samme-7626862" style="font-weight:bold;" target="_blank" title="Click to know more about SAMME">SAMME</a> number of electrons (10), so they are isoelectronic. Electronic configuration of each of them `<a href="https://interviewquestions.tuteehub.com/tag/1s-283076" style="font-weight:bold;" target="_blank" title="Click to know more about 1S">1S</a>^(2)<a href="https://interviewquestions.tuteehub.com/tag/2s-301125" style="font-weight:bold;" target="_blank" title="Click to know more about 2S">2S</a>^(2)2p^(6)`. <br/> Again, `Cl^(-),P^(3-),K^(+),Ar,S^(2-)` and `Ca^(2+)` have the same number of electrons (18), so they are isoelectronic. their electronic configuration is: <br/> `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)` <br/> (2) The inerrt gas is krypton (Kr) whose atomic number is 36, has electronic configuration: <br/> `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(10)<a href="https://interviewquestions.tuteehub.com/tag/4s-318811" style="font-weight:bold;" target="_blank" title="Click to know more about 4S">4S</a>^(2)4p^(6)` <br/> Total number of s-electrons=8 and total number of p-electrons=18 <br/> `therefore`Difference in the number of s and p- electrons <br/> `therefore`Difference in the number of s and p-electrons=18-8=10=number of d-electrons.</body></html> | |
| 51854. |
1 g of X has atoms arranged in cubic packing so as to give best packing efficiency. The possible arrangement is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/simple-1208262" style="font-weight:bold;" target="_blank" title="Click to know more about SIMPLE">SIMPLE</a> cubic<br/>face centred cubic<br/>body certred cubic<br/>hexagonal close <a href="https://interviewquestions.tuteehub.com/tag/packing-1145400" style="font-weight:bold;" target="_blank" title="Click to know more about PACKING">PACKING</a></p>Solution :Packing efficiency of ccp is 74% so it <a href="https://interviewquestions.tuteehub.com/tag/best-390038" style="font-weight:bold;" target="_blank" title="Click to know more about BEST">BEST</a> packing is cubic packing</body></html> | |
| 51855. |
1 g of the complex [Cr(H_(2)O)_(5)Cl]Cl_(2). H_(2)O was passed through a cation exchanger to produce HCl. The acid liberated was diluted to 1 litre. What is the normality of this acid solution ? |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :N//A</body></html> | |
| 51856. |
1 g of Mg is burnt in a closed vessel containing 0.5 g of O_(2). Which of the following statements is correct ? |
| Answer» <html><body><p>`O_(2)` is the limiting <a href="https://interviewquestions.tuteehub.com/tag/reagent-1178480" style="font-weight:bold;" target="_blank" title="Click to know more about REAGENT">REAGENT</a> and Mg is in excess by 0.25 g<br/>Mg is the limiting reagent and is in excess by 0.5 g<br/>`O_(2)` is the limiting reagent and is in excess by 0.25 g<br/>`O_(2)` is the limiting reagent and Mg is in excess by 0.75 g</p>Solution :`underset(2xx24)(2Mg)+underset(2xx16)(O_(2))rarr2MgO` <br/> <a href="https://interviewquestions.tuteehub.com/tag/48-318016" style="font-weight:bold;" target="_blank" title="Click to know more about 48">48</a> g of Mg require `O_(2)=32` g <br/> 1g of Mg require `O_(2)=(32)/(48) = 0.66 g` <br/> `O_(2)` actually <a href="https://interviewquestions.tuteehub.com/tag/available-888895" style="font-weight:bold;" target="_blank" title="Click to know more about AVAILABLE">AVAILABLE</a> = 0.5 g <br/> This <a href="https://interviewquestions.tuteehub.com/tag/means-1091780" style="font-weight:bold;" target="_blank" title="Click to know more about MEANS">MEANS</a> that `O_(2)` is the limiting reactant <br/> 0.5 g of `O_(2)` will react with `Mg = (48)/(32) <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 0,5 = 0.75 g` <br/> Excess of `Mg = 1 -0.75 =0.25 g`.</body></html> | |
| 51857. |
1 g of hydrogen is found to combine with 80g of bromine and 1g of calcium combines with 4 g of bromine. Equivalent weight of calcium is |
| Answer» <html><body><p>16<br/>20<br/>40<br/>80<br/></p>Solution :1g `H_(2)` combines with `Br_(2)` = 80g <br/> Equivalent <a href="https://interviewquestions.tuteehub.com/tag/weight-1451304" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHT">WEIGHT</a> of `Br_(2)` = 80 <br/> 4g of `Br_(2)` combine with = 1g Ca <br/> `therefore` 80 g of `Br_(2)` will combine with (`=<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> xx 80)/<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a> = 20g Ca</body></html> | |
| 51858. |
1 g of graphiteis burnt in a bombcalorimeter in excess of oxygen at 298 K and 1 atmosphericpressure accordingto the equation C ( graphite ) + O_(2)(g) rarr CO_(2)(g) During the reaction, temperature rises from298K to 299 K. If the heat capacity ofthe bomb calorimeter is 20.7 kJ // K , what is the enthalpy change for the above reaction at 298K and 1 atm ? |
| Answer» <html><body><p></p>Solution :Rise in <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a> of the calorimeter `= 299- 298K = <a href="https://interviewquestions.tuteehub.com/tag/1k-282789" style="font-weight:bold;" target="_blank" title="Click to know more about 1K">1K</a>` <br/> Heat <a href="https://interviewquestions.tuteehub.com/tag/capacity-908575" style="font-weight:bold;" target="_blank" title="Click to know more about CAPACITY">CAPACITY</a> of the calorimeter `= 20.7 k J K^(-1)` <br/> `:. `Heat absorbed by the calorimeter `= C_(v) xxDelta T= (20.7 k J K^(-1)) (1K) = 20.7 kJ` <br/> This s the heat evolved in the combustionof1 g of graphite. <br/> `:. `Heat evolved inthe combustionof 1 <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> of graphite , `i.e., 12g` of graphite `= 20.7 xx 12 kJ = 248.4 kJ` <br/>As this is the heat evolved and the vessel is <a href="https://interviewquestions.tuteehub.com/tag/closed-919469" style="font-weight:bold;" target="_blank" title="Click to know more about CLOSED">CLOSED</a>, therefore, enthalpy change of the reaction `( Delta U )` <br/> `= - 248.4 kJ mol^(-1)`</body></html> | |
| 51859. |
1 g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the equation C_("(graphite)") + O_(2(g)) to CO_(2(g)) During the reaction, temperature rises from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7 kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm ? |
| Answer» <html><body><p></p>Solution :Suppose q is the quantity of heat from the reaction <a href="https://interviewquestions.tuteehub.com/tag/mixture-1098735" style="font-weight:bold;" target="_blank" title="Click to know more about MIXTURE">MIXTURE</a> and `C_(V)` is the heat capacity of the calorimeter, then the quantity of heat absorbed by the calorimeter : <br/> `q=C_(v) xx Delta T` <br/> Quantity of heat from the reaction will have the same magnitude but opposite <a href="https://interviewquestions.tuteehub.com/tag/sign-1207134" style="font-weight:bold;" target="_blank" title="Click to know more about SIGN">SIGN</a> because the heat lost by the system is equal to the heat gained by the calorimeter. <br/> `q=-C_(v) xx Delta T` <br/> `=-20.7 "kJ/K" K xx (299- 298) K = -20.7 "kJ"` <br/> Here, negative sign indicates the exothermic nature of the reaction. <br/> <a href="https://interviewquestions.tuteehub.com/tag/thus-2307358" style="font-weight:bold;" target="_blank" title="Click to know more about THUS">THUS</a>, `Delta U` for the combustion of the 1g of graphite `= -20.7 "kJ K"^(-1)` <br/> For combustion of 1 <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> of graphite, <br/> `= (12.0 g "mol"^(-1) xx (-20.7 "kJ") )/(1g)` <br/> `= -2.48 xx 10^(2) "kJ mol"^(-1)` since `Deltan_(g) =0` <br/> `therefore Delta H = Delta U = - 2.48 xx 10^(2) "kJ mol"^(-1)`</body></html> | |
| 51860. |
1 g of graphite is burnt in a bomb calorimeter in excess of O_(2) at 298 K and 1 atm. Pressure according to the equations. C_("graphite")+O_(2(g)) to CO_(2(g)) During the reaction the temperature rises from 298 K to 200K. Heat capacity of the bomb calorimeter is 20.7KJK^(-1). What is the enthalpy change for the above reaction at 298 K 1 atm? |
| Answer» <html><body><p></p>Solution :`q = C_v xx Delta` <br/> Quantity of heat from the reaction will have the same magnitude but opposite sign because the heat lost by the system (reaction mixture) is equal to the heat gained by the calorimeter <br/> `q= - C_v xx DeltaT = -20.7kJ//<a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a> xx (299-298) K = -20.7kJ ` <br/> For combustion of 1 <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> of <a href="https://interviewquestions.tuteehub.com/tag/graphite-15721" style="font-weight:bold;" target="_blank" title="Click to know more about GRAPHITE">GRAPHITE</a> <br/> `DeltaE = -2.48 xx 10^2 kJ mol^(-1)` <br/> `DeltaH = DeltaE = -2.48 xx 10^2 kJmol^(-1)`, <a href="https://interviewquestions.tuteehub.com/tag/since-644476" style="font-weight:bold;" target="_blank" title="Click to know more about SINCE">SINCE</a> `Deltan = 0`.</body></html> | |
| 51861. |
1 g of an impure sample of magnesium carbonate (containing no thermally decomposable impurities) on complete thermal decomposition gave 0.44 g of carbon dioxide gas.To percentage of impurity in the sample is…........... |
| Answer» <html><body><p>0<br/>`4.4%`<br/>0.16<br/>`8.4%`</p>Solution :`MgCO_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)toMgO+CO_(2)<a href="https://interviewquestions.tuteehub.com/tag/uarr-3241817" style="font-weight:bold;" target="_blank" title="Click to know more about UARR">UARR</a>`<br/> `MgCO_(3):(1xx24)+(1xx12)+(3xx16)=84 g`<br/> `CO_(2):(1xx12)+(2xx16)=44 g`<br/> 100% pure 84 g `MgCO_(3)` on heating <a href="https://interviewquestions.tuteehub.com/tag/gives-1007647" style="font-weight:bold;" target="_blank" title="Click to know more about GIVES">GIVES</a> 44 g `CO_(2)`<br/> Given that <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> g of `MgCO_(3)` on heating gives `0.44 g CO_(2)`<br/> Therefore, 84 g `MgCO_(3)` sample on heating gives `36.96 g CO_(2)=84%`<br/> Percentage of purity of the sample =`(100%)/(44 g CO_(2))xx36.96 g CO_(2)=84%`<br/> Percentage of impurity=16%</body></html> | |
| 51862. |
1 g of an impure sample of magnesium carbonate (containing no thermally decomposable impurities) on complete thermal decomposition gave 0.44 g of carbon dioxide gas. The percentage of impurity in the sample is |
| Answer» <html><body><p>0<br/>0.044<br/>0.16<br/>0.084</p>Solution :`MgCO_(3) to MgO +CO_(2) <a href="https://interviewquestions.tuteehub.com/tag/uarr-3241817" style="font-weight:bold;" target="_blank" title="Click to know more about UARR">UARR</a>` <br/>`MgCO_(3): (1xx24) + (1xx12) + (3xx16)=84g` <br/><a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a>% <a href="https://interviewquestions.tuteehub.com/tag/pure-1172806" style="font-weight:bold;" target="_blank" title="Click to know more about PURE">PURE</a> 84 g of `MgCO_(3)` on heating gives 44g `CO_(2)` <br/>Given that 1g of `MgCO_(3)` on heating gives 0.44g `CO_(2)` <br/>Therefore, 854 g `MgCO_(3)` <a href="https://interviewquestions.tuteehub.com/tag/sample-1194587" style="font-weight:bold;" target="_blank" title="Click to know more about SAMPLE">SAMPLE</a> on heating gives 36.96 g `CO_(2)` <br/>Percantaqge of <a href="https://interviewquestions.tuteehub.com/tag/purity-609119" style="font-weight:bold;" target="_blank" title="Click to know more about PURITY">PURITY</a> of the sample= `(100%)/(44 g CO_(2)) xx36.96g CO_(2)` <br/>=84% <br/>Percentage of impurity=16%</body></html> | |
| 51863. |
1 g of a metal required 50 mL of 0.5 N HCl to dissovle it. The equivalent mass of the metal is : |
| Answer» <html><body><p>25<br/>50<br/>20<br/>40</p>Solution :N//A</body></html> | |
| 51864. |
1 g of a mixture containing equal no.of moles of carbonates of two alkali metals, required 44.4 mL of 0.5 N HCl for complete reaction. The atomic weight of one metal is 7, find the atomic weight of other metal. Also calculate amount of sulphate formed on quantitative conversion of 1.0g of the mixture in two sulphates. |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :`<a href="https://interviewquestions.tuteehub.com/tag/23-294845" style="font-weight:bold;" target="_blank" title="Click to know more about 23">23</a>, 1.3998 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>`;</body></html> | |
| 51865. |
1 g of a carbonate (M_2CO_3) on treatment with excess HCl produces 0.01186 mole of CO_2. The molar mass of M_2CO_3 is g "mol"^(-1) : |
| Answer» <html><body><p>1186<br/>84.3<br/>118.6<br/>11.86</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 51866. |
(1) Consider the following cases- The nature of flow of energy in case (I) is same as that in- (A) II , (B) III , (C ) II and III, (D) None |
| Answer» <html><body><p></p>Solution :(D) None, Because in `II` and `III`, the flow of <a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a> or <a href="https://interviewquestions.tuteehub.com/tag/matter-22749" style="font-weight:bold;" target="_blank" title="Click to know more about MATTER">MATTER</a> is taking place only in one direction.While in <a href="https://interviewquestions.tuteehub.com/tag/equilibrium-974342" style="font-weight:bold;" target="_blank" title="Click to know more about EQUILIBRIUM">EQUILIBRIUM</a> <a href="https://interviewquestions.tuteehub.com/tag/state-21805" style="font-weight:bold;" target="_blank" title="Click to know more about STATE">STATE</a>, the flow of energy takes place in both <a href="https://interviewquestions.tuteehub.com/tag/directions-955033" style="font-weight:bold;" target="_blank" title="Click to know more about DIRECTIONS">DIRECTIONS</a> equally. Thus `(I)` is a dynamic equilibrium while states in `II` and `III` are called steady state (static equilibrium).</body></html> | |
| 51867. |
1 cm^(3) of 0.01 N HCl is added to one litre of sodium chloride solution . What will be the pH of the resulting solution ? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :As NaCl solution is neutral, it simply dilutes the <a href="https://interviewquestions.tuteehub.com/tag/hcl-479502" style="font-weight:bold;" target="_blank" title="Click to know more about HCL">HCL</a> solution from <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> <a href="https://interviewquestions.tuteehub.com/tag/cc-911065" style="font-weight:bold;" target="_blank" title="Click to know more about CC">CC</a> to 1000 cc.<br/> Now `[H^(+)]=0.01//1000=10^(-5)M :. pH = 5`.</body></html> | |
| 51868. |
1-Chlorobutane on reaction with alcoholic potash gives |
| Answer» <html><body><p>1-Butene<br/>1-Butanol<br/>2-Butene<br/>2-Butanol.</p>Answer :A</body></html> | |
| 51869. |
1-Butyne reacts with KMnO_4//OH//Delta to give |
| Answer» <html><body><p>`CH_3-CH_2 - CH_2 - COOH`<br/>`CH_3-CH_2 - COOH`<br/>`CH_3 - CH_2 - COOH + CO_2`<br/>`CH_3 - CH_2 - COOH + HCOOH` </p>Solution :`CH_3 - CH_ 2- <a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> -= C- H <a href="https://interviewquestions.tuteehub.com/tag/underset-3243992" style="font-weight:bold;" target="_blank" title="Click to know more about UNDERSET">UNDERSET</a>(Delta)<a href="https://interviewquestions.tuteehub.com/tag/overset-2905731" style="font-weight:bold;" target="_blank" title="Click to know more about OVERSET">OVERSET</a>(KMnO_4//overset(Ɵ)OH) to CH_3 * CH_2 * COOH + Hunderset(CO_2)underset(darr(O))COOH`</body></html> | |
| 51870. |
1-Butyne reacts with hot alkaline KMnO_4 to produce |
| Answer» <html><body><p>`CH_3CH_2CH_2COOH`<br/>`CH_3CH_2COOH`<br/>`CH_3CH_2COOH+CO_2`<br/>`CH_3CH_2COOH + <a href="https://interviewquestions.tuteehub.com/tag/hcooh-479470" style="font-weight:bold;" target="_blank" title="Click to know more about HCOOH">HCOOH</a>`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Terminal alkynes on oxidation <a href="https://interviewquestions.tuteehub.com/tag/give-468520" style="font-weight:bold;" target="_blank" title="Click to know more about GIVE">GIVE</a> a mixture of `CO_2` and a carboxylic acid with one C-atom less than that of the starting alkyne.</body></html> | |
| 51871. |
1-Butyne can be converted into 1- bromo-1 -butene by reacting it with which of the following reagent ? |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/hbr-479388" style="font-weight:bold;" target="_blank" title="Click to know more about HBR">HBR</a> <br/>`HBr and (C_6H_5COO)_2`<br/>`Br_2 and H_2O_2`<br/>`Br_2 and C Cl_4` </p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`CH_3CH2 C -= CH + HBr <a href="https://interviewquestions.tuteehub.com/tag/underset-3243992" style="font-weight:bold;" target="_blank" title="Click to know more about UNDERSET">UNDERSET</a>("Anti-Mark <a href="https://interviewquestions.tuteehub.com/tag/addn-3531500" style="font-weight:bold;" target="_blank" title="Click to know more about ADDN">ADDN</a>")<a href="https://interviewquestions.tuteehub.com/tag/overset-2905731" style="font-weight:bold;" target="_blank" title="Click to know more about OVERSET">OVERSET</a>((C_6H_5COO)_2) to CH_3CH_2 CH = CHBr`</body></html> | |
| 51872. |
1-butylene reacts with cold alkaline KMnO_(4) to produce |
| Answer» <html><body><p>`CH_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)CH_(2)COOH`<br/>`CH_(3)CH_(2)CH_(2)COOH`<br/>`CH_(3)CH_(2)COOH+CO_(2)`<br/>`CH_(3)CH_(2)COOH + HCOOH`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`CH_(3)-CH_(2)-C-=CH underset("alk." KMnO_(4))overset("Cold")rarr CH_(3)CH_(2)COOH + CO_(2)`</body></html> | |
| 51873. |
1-Butene may be converted to butane by reaction with : |
| Answer» <html><body><p>`Zn-Hg` <br/>`<a href="https://interviewquestions.tuteehub.com/tag/pd-590584" style="font-weight:bold;" target="_blank" title="Click to know more about PD">PD</a>//H_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)` <br/>`Zn-HCl` <br/>`Sn//HCl` </p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`CH_(3)CH_(2)<a href="https://interviewquestions.tuteehub.com/tag/ch-913588" style="font-weight:bold;" target="_blank" title="Click to know more about CH">CH</a>=CHJ_(2) <a href="https://interviewquestions.tuteehub.com/tag/overset-2905731" style="font-weight:bold;" target="_blank" title="Click to know more about OVERSET">OVERSET</a>(H_(2)//Pd)toCH_(3)CH_(2)CH_(2)CH_(3)`</body></html> | |
| 51874. |
1-"Butene"+HBrunderset(hupsilon)overset(H_(2)O_(2))rarr 1-Bromobutane. This reaction is based as |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/markownikoff-2170489" style="font-weight:bold;" target="_blank" title="Click to know more about MARKOWNIKOFF">MARKOWNIKOFF</a>'s <a href="https://interviewquestions.tuteehub.com/tag/rule-1192038" style="font-weight:bold;" target="_blank" title="Click to know more about RULE">RULE</a> <br/>Saytzeff's rule <br/><a href="https://interviewquestions.tuteehub.com/tag/anti-379578" style="font-weight:bold;" target="_blank" title="Click to know more about ANTI">ANTI</a> Markownikoff's rule <br/><a href="https://interviewquestions.tuteehub.com/tag/hoffmann-2113339" style="font-weight:bold;" target="_blank" title="Click to know more about HOFFMANN">HOFFMANN</a>'s rule </p>Answer :C</body></html> | |
| 51875. |
1-Butene and cyclobutane show |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/position-1159826" style="font-weight:bold;" target="_blank" title="Click to know more about POSITION">POSITION</a> isomerism<br/>ring-chain isomerism<br/>functional isomerism<br/>metamerism</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 51876. |
1-butene cannot show geometrical isomerism. |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :T</body></html> | |
| 51877. |
1-Butanol is heated withconcentrated sulphuricacidat 443 k. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :But-2-eneis <a href="https://interviewquestions.tuteehub.com/tag/obtained-7273275" style="font-weight:bold;" target="_blank" title="Click to know more about OBTAINED">OBTAINED</a> asthemajorproduct. Firstn-butyl `(1^(@))` carbocationis formed. Thisbeinglesssstablerearranges to a morestable `2^(@)` <a href="https://interviewquestions.tuteehub.com/tag/carbocation-909158" style="font-weight:bold;" target="_blank" title="Click to know more about CARBOCATION">CARBOCATION</a> whichsubsequently losesa protonto formbut-2-ene inaccordance withSaytzeff rule.<br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PR_CHE_02_XI_APP_04_E02_003_Q01.png" width="80%"/> <br/> `underset(("More stable "))underset(2^(@) "Butyl carbocation ")(CH_(3)-CH_(2) -<a href="https://interviewquestions.tuteehub.com/tag/overset-2905731" style="font-weight:bold;" target="_blank" title="Click to know more about OVERSET">OVERSET</a>(+)(CH) -CH_(3)) overset(-H^(+))underset(("Saytzeff elimiN/Ation"))(to) underset("But -2-en (80 %)")(CH_(3)CH =CHCH_(3))+ underset("But -1-ene (20%)")(CH_(3) CH_(2) CH=CH_(2)`</body></html> | |
| 51878. |
1-bromo-3-chlorocyclobutane when treated with two equivalents of Na, in the presence of ether which of the following will be formed ? |
| Answer» <html><body><p><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/VMC_JEE_XI_QZ_PC_04A_E03_004_O01.png" width="30%"/> <br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/VMC_JEE_XI_QZ_PC_04A_E03_004_O02.png" width="30%"/> <br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/VMC_JEE_XI_QZ_PC_04A_E03_004_O03.png" width="30%"/> <br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/VMC_JEE_XI_QZ_PC_04A_E03_004_O04.png" width="30%"/> </p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/VMC_JEE_XI_QZ_PC_04A_E03_004_S01.png" width="80%"/></body></html> | |
| 51879. |
1-Bromo-3-chlorocyclobutane is treated with two equivalents of Na, in the presence of ether. Which of the following compounds will be formed. |
| Answer» <html><body><p><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_OBJ_FING_CHE_XI_C13_E02_001_O01.png" width="30%"/><<a href="https://interviewquestions.tuteehub.com/tag/br-390993" style="font-weight:bold;" target="_blank" title="Click to know more about BR">BR</a>><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_OBJ_FING_CHE_XI_C13_E02_001_O02.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_OBJ_FING_CHE_XI_C13_E02_001_O03.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_OBJ_FING_CHE_XI_C13_E02_001_O04.png" width="30%"/></p>Solution :Since alkyl bromides are more reactive than alkyl <a href="https://interviewquestions.tuteehub.com/tag/chlorides-915862" style="font-weight:bold;" target="_blank" title="Click to know more about CHLORIDES">CHLORIDES</a>, therefore, intramolecular wurtz <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a> <a href="https://interviewquestions.tuteehub.com/tag/occurs-1127848" style="font-weight:bold;" target="_blank" title="Click to know more about OCCURS">OCCURS</a> on the <a href="https://interviewquestions.tuteehub.com/tag/side-1206832" style="font-weight:bold;" target="_blank" title="Click to know more about SIDE">SIDE</a> of Br atom. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_OBJ_FING_CHE_XI_C13_E02_001_S01.png" width="80%"/></body></html> | |
| 51880. |
1-Bromo-3-chlorocyclobutane is treated with two equivalents of Na, in the presence of either. Which of the following will be formed? |
| Answer» <html><body><p><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_D_C04_E02_008_O01.png" width="30%"/> <br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_D_C04_E02_008_O02.png" width="30%"/> <br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_D_C04_E02_008_O03.png" width="30%"/> <br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_D_C04_E02_008_O04.png" width="30%"/> </p>Solution :Since alkyl bromides are more reactive than alkyl chlorides, therefore, <a href="https://interviewquestions.tuteehub.com/tag/intramolecular-2746576" style="font-weight:bold;" target="_blank" title="Click to know more about INTRAMOLECULAR">INTRAMOLECULAR</a> Wurtz reaction <a href="https://interviewquestions.tuteehub.com/tag/occurs-1127848" style="font-weight:bold;" target="_blank" title="Click to know more about OCCURS">OCCURS</a> on the <a href="https://interviewquestions.tuteehub.com/tag/side-1206832" style="font-weight:bold;" target="_blank" title="Click to know more about SIDE">SIDE</a> of Br atom. <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_D_C04_E02_008_S01.png" width="80%"/></body></html> | |
| 51881. |
1 bar means …….. Pa ? |
| Answer» <html><body><p></p>Solution :1 <a href="https://interviewquestions.tuteehub.com/tag/bar-892478" style="font-weight:bold;" target="_blank" title="Click to know more about BAR">BAR</a> `= <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(5)` Pa<br/>It is <a href="https://interviewquestions.tuteehub.com/tag/atmospheric-887165" style="font-weight:bold;" target="_blank" title="Click to know more about ATMOSPHERIC">ATMOSPHERIC</a> <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a> of sea level.</body></html> | |
| 51882. |
1 BM is equal to: |
| Answer» <html><body><p>`(hc)/(<a href="https://interviewquestions.tuteehub.com/tag/mpie-3730002" style="font-weight:bold;" target="_blank" title="Click to know more about MPIE">MPIE</a>^(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>))`<br/>`(hc)/(4pim)`<br/>`(e^(2)hc)/(<a href="https://interviewquestions.tuteehub.com/tag/4m-318788" style="font-weight:bold;" target="_blank" title="Click to know more about 4M">4M</a>)`<br/>`(ehc)/(<a href="https://interviewquestions.tuteehub.com/tag/pim-591493" style="font-weight:bold;" target="_blank" title="Click to know more about PIM">PIM</a>)`</p>Answer :A</body></html> | |
| 51883. |
1 amu is ........... part of atomic mass of one C atom. |
| Answer» <html><body><p>`1/2`<br/>`1/12`<br/>12<br/>2</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 51884. |
(1 ) Among the alkaline earth metals BeO is insoluble in water but other oxides are soluble. Why? (ii) State Diffusion Law. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Lattice energy of <a href="https://interviewquestions.tuteehub.com/tag/beo-390019" style="font-weight:bold;" target="_blank" title="Click to know more about BEO">BEO</a> is comparatively higher than the hydrogen energy . Therefore it is almost insoluble in water. Whereas other alkaline earth metals oxides have high ? Ydration energy. Hence they are soluble in water. <br/> (ii)Diffusion is the spreading of molecules of a substance throughout ·a space or a second substance. Diffusion <a href="https://interviewquestions.tuteehub.com/tag/refers-1181652" style="font-weight:bold;" target="_blank" title="Click to know more about REFERS">REFERS</a> to the ~bility of the gases to mix with each other. E.g., Spreading of something such as brown tea <a href="https://interviewquestions.tuteehub.com/tag/liquid-1075124" style="font-weight:bold;" target="_blank" title="Click to know more about LIQUID">LIQUID</a> spreaqing through the water in a tea cup</body></html> | |
| 51885. |
1-4-dichlorohexane (1 mole ) +Nal (1"mole" ) overset("Acetone ")toproduct of the reaction is : |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/cl-408888" style="font-weight:bold;" target="_blank" title="Click to know more about CL">CL</a>-CH_(2)-CH_(2)-<a href="https://interviewquestions.tuteehub.com/tag/underset-3243992" style="font-weight:bold;" target="_blank" title="Click to know more about UNDERSET">UNDERSET</a>(I)underset(|)<a href="https://interviewquestions.tuteehub.com/tag/ch-913588" style="font-weight:bold;" target="_blank" title="Click to know more about CH">CH</a>-CH_(2)-CH_(3)`<br/>`Cl-CH_(2)-CH_(2)-underset(<a href="https://interviewquestions.tuteehub.com/tag/ci-408488" style="font-weight:bold;" target="_blank" title="Click to know more about CI">CI</a>)underset(|)CH-CH_(2)-CH_(3)`<br/>`H_(2)C=CH-underset(CI)underset(|)CH-CH_(2)-CH_(3)`<br/>`I-CH_(2)-CH_(2)-CH_(2)-underset(Cl)underset(|)CH-CH_(2)-CH_(3)`</p>Solution :<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/MSC_ORG_CHM_C05C_E01_245_S01.png" width="80%"/></body></html> | |
| 51886. |
1, 3-butadiene + maleic acidto X |
| Answer» <html><body><p><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_TRG_AO_CHE_XI_V01_D_C04_E02_190_O01.png" width="30%"/> <br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_TRG_AO_CHE_XI_V01_D_C04_E02_190_O02.png" width="30%"/> <br/> It is a <a href="https://interviewquestions.tuteehub.com/tag/syn-1236892" style="font-weight:bold;" target="_blank" title="Click to know more about SYN">SYN</a>. <a href="https://interviewquestions.tuteehub.com/tag/addition-367641" style="font-weight:bold;" target="_blank" title="Click to know more about ADDITION">ADDITION</a> <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a> <br/>The above reaction is diene-dienophile addition reaction </p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Diels - <a href="https://interviewquestions.tuteehub.com/tag/alder-7380241" style="font-weight:bold;" target="_blank" title="Click to know more about ALDER">ALDER</a> reaction</body></html> | |
| 51887. |
(1)/(2)N_(2)(g)+O_(2)rarrNO_(2)(g),Delta_(r)H^(@)=-40KJ//mol Given: C_(P.m)(NO_(2),g)=40J//mol//K,C_(p,m) (O_(2),g)=30JK^(-1)mol^(-1) C_(P,m)N_(2)(g)=30JK^(-1)mol^(-1) What is the enthalpy of formation of NO_(2)(g) at 1298K ? |
| Answer» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>>`-40KJ//mol`<br/>`-50KJ//mol`<br/>`-45KJ//mol` <br/>`-6KJ//mol` </p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`Delta_(r)H_(T)=Delta_(r)H^(@)+int_(298)^(1298)Delta_(r)C_(P)dT` <br/> At `1298KDelta_(r)H=-40KJ-5` <br/> `DeltaT=-40KJ-5xx1000xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)KJ=-45KJ//mol`</body></html> | |
| 51888. |
Given that (1)/(2)S_(8(s)) + 6O_(2(g)) rarr 4SO_(3(g)), Delta H^(0) = - 1590kJ. The standard enthalpy of formation of SO_(3) is |
| Answer» <html><body><p>`-1590 <a href="https://interviewquestions.tuteehub.com/tag/kj-1063034" style="font-weight:bold;" target="_blank" title="Click to know more about KJ">KJ</a> <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a>^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)`<br/>`-397.5 KJ mol^(-1)`<br/>`-3.975 KJ mol^(-1)`<br/>`+397.5 KJ mol^(-1)`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 51889. |
1/2H_(2(g)) + 1/2I_(2(g)) hArr HI_((g)) in this equilibriumwhat is the relation between K_c and K'_c |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`K_c=1/(K._c)`</body></html> | |
| 51890. |
1/2 litre of 2xx10^(-3) M AlCl_3 and 1/2 litre of 4 xx 10^(-2) M solution of NaOH are mixed and the solution is diluted to 10^2 litres with water at room temperature. Will a precipitate form ? Given K_(sp)Al(OH)_3=5xx10^(-33) |
| Answer» <html><body><p></p>Solution :The solubility equilibrium of `Al(OH)_3` is <br/> `Al(OH)_3 hArr Al^(3+) + 3OH^(-)` <br/>On mixing `1/2` litre of `AlCl_3`and `1/2` litre of NaOH, the total solution is made `10^2` litres by adding <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a>.<br/> Concentration of `AlCl_3=(2xx10^(-3))/10^2 xx1/2=1xx10^(-5)` M <br/> Concentration of NaOH =`(4xx10^(-2))/10^2 xx1/2=2xx10^(-4)` M <br/> Since `AlCl_3` <a href="https://interviewquestions.tuteehub.com/tag/ionises-2748319" style="font-weight:bold;" target="_blank" title="Click to know more about IONISES">IONISES</a> as : <br/>`AlCl_3 hArr Al^(3+) +3Cl^-` <br/>`[Al^(3+)]=[AlCl_3]=1xx10^(-5)` M <br/> Since NaOH ionises as : <br/> `NaOH hArr <a href="https://interviewquestions.tuteehub.com/tag/na-572417" style="font-weight:bold;" target="_blank" title="Click to know more about NA">NA</a>^(+) + OH^(-)` <br/> `[OH^-]=[NaOH]=2xx10^(-4)` <br/> Now, ionic product =`[Al^(3+)][OH^-]^3` <br/> `=1xx10^(-5)xx(2xx10^(-4))^3 =8xx10^(-17)` <br/> As the product the concentration of <a href="https://interviewquestions.tuteehub.com/tag/ions-1051295" style="font-weight:bold;" target="_blank" title="Click to know more about IONS">IONS</a> of the salt is more than `K_(sp)`, the salt will be precipitated .</body></html> | |
| 51891. |
(1)/(2) H_(2(g)) + (1)/(2) Cl_(2(g)) rarr HCl_((g)) , Delta H^(0) = - 92.4 kJ/mole, HCl_((g)) + nH_(2)O rarr H_((aq))^(+) + Cl_((aq))^(-) , Delta H^(0) = -74.8 kJ/mole Delta H^(0)f for Cl_((aq))^(-) is |
| Answer» <html><body><p>`-17.6` kJ/mole<br/>`-167.<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>` kJ/mole<br/>`+17.6` kJ/mole<br/>`-35.2` kJ/mole</p>Solution :`HCl_((<a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>)) + H_(2)O <a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> H_((aq))^(+) + Cl_((aq))^(-)` <br/> `Delta H = Delta H_(f_(Cl^(-))) - Delta H_(f_(<a href="https://interviewquestions.tuteehub.com/tag/hcl-479502" style="font-weight:bold;" target="_blank" title="Click to know more about HCL">HCL</a>))` <br/> `-74.8 = Delta H_(f_(HCl)) - (-92.4) rArr Delta H_(f_(HCl)) = -167.2`</body></html> | |
| 51892. |
(1)/(12)the mass of .C-12. atom is equal to |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/amu-25369" style="font-weight:bold;" target="_blank" title="Click to know more about AMU">AMU</a><br/>`(1)/(16)th` mass of .O-16. <a href="https://interviewquestions.tuteehub.com/tag/atom-887280" style="font-weight:bold;" target="_blank" title="Click to know more about ATOM">ATOM</a><br/>`(1)/(16)th`mass of `CH_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)` molecule <br/>1 gm</p>Solution :`(1)/(12)xxC-12=1" amu "=(1)/(16)xx16` amu</body></html> | |
| 51893. |
0^(@)C is known as absolute zero temperature. |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a></body></html> | |
| 51894. |
0.9g of an organic compound gave on combustion: (i) 1.584g CO_(2) (ii) 0.648g H_(2)O When 0.24g of the substance was Kjeldahlised and the ammonia formed was absorbed in 50 cm^(3) of (N)/(2)H_(2)SO_(4). The excess acid required 77 cm^(3) of N//10NaOh for complete neutralization. Molecular mass of the compound is estimated to be 100. Molecular formula of the compound will be: |
| Answer» <html><body><p>`C_(<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>)H_(4)NO_(2)`<br/>`C_(4)H_(8)N_(2)O`<br/>`C_(4)H_(8)NO`<br/>`C_(8)H_(4)NO`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :B</body></html> | |
| 51895. |
0.9g of an organic compound gave on combustion: (i) 1.584g CO_(2) (ii) 0.648g H_(2)O When 0.24g of the substance was Kjeldahlised and the ammonia formed was absorbed in 50 cm^(3) of (N)/(2)H_(2)SO_(4). The excess acid required 77 cm^(3) of N//10NaOh for complete neutralization. Molecular mass of the compound is estimated to be 100. The compound has maximum percentage composition of which of the following elements? |
| Answer» <html><body><p>C<br/>H<br/>N <br/>O </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 51896. |
0.9g of an organic compound gave on combustion: (i) 1.584g CO_(2) (ii) 0.648g H_(2)O When 0.24g of the substance was Kjeldahlised and the ammonia formed was absorbed in 50 cm^(3) of (N)/(2)H_(2)SO_(4). The excess acid required 77 cm^(3) of N//10NaOh for complete neutralization. Molecular mass of the compound is estimated to be 100. Which among the following is the empirical formula of the compound? |
| Answer» <html><body><p>`C_(<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>)H_(4)NO_(2)`<br/>`C_(4)H_(8)N_(2)O`<br/>`C_(4)H_(8)NO`<br/>`C_(8)H_(4)NO`</p>Answer :B</body></html> | |
| 51897. |
0.9g of an organic compound gave on combustion: (i) 1.584g CO_(2) (ii) 0.648g H_(2)O When 0.24g of the substance was Kjeldahlised and the ammonia formed was absorbed in 50 cm^(3) of (N)/(2)H_(2)SO_(4). The excess acid required 77 cm^(3) of N//10NaOh for complete neutralization. Molecular mass of the compound is estimated to be 100. Percentage composition of nitrogen in the compound will be: |
| Answer» <html><body><p>16<br/>61<br/>6<br/>28</p>Answer :D</body></html> | |
| 51898. |
An impure sample of sodium chloride which weighed 1.2 g gave on treatment with excess of silver nitrate solution 2.4g of silver chloride as precipitate.Calculate the percentage purity of the sample |
| Answer» <html><body><p></p>Solution :The double <a href="https://interviewquestions.tuteehub.com/tag/decomposition-946042" style="font-weight:bold;" target="_blank" title="Click to know more about DECOMPOSITION">DECOMPOSITION</a> reaction involved is: <br/> `NaCl(aq) + AgNO_(3)(aq) to NaNO_(3)(aq) + AgCl(s)` <br/> Mass of NaCl taken = 0.989 g Mass of AgCl formed = 2.42 g <br/> Using the relation <br/> `("Mass of NaCl")/("Mass of AgCl") = ("<a href="https://interviewquestions.tuteehub.com/tag/equivalent-446407" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENT">EQUIVALENT</a> <a href="https://interviewquestions.tuteehub.com/tag/weight-1451304" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHT">WEIGHT</a> of NaCl")/("Equivalent weight of AgCl") =("Eq. weight of Na + Eq. weight of Cl")/("Eq. weight of Ag + Eq. weight of Cl")` <br/> and substituting the values, we <a href="https://interviewquestions.tuteehub.com/tag/get-11812" style="font-weight:bold;" target="_blank" title="Click to know more about GET">GET</a> <br/> `0.989/2.42 = ("Eq. mass of Na" + 35.5)/(108 + 35.5)` <br/> `=(0.989/2.43 xx 143.5) - 35.5 = 58.4- 35.5 = 22.9`</body></html> | |
| 51899. |
0.9g of an organic compound gave on combustion: (i) 1.584g CO_(2) (ii) 0.648g H_(2)O When 0.24g of the substance was Kjeldahlised and the ammonia formed was absorbed in 50 cm^(3) of (N)/(2)H_(2)SO_(4). The excess acid required 77 cm^(3) of N//10NaOh for complete neutralization. Molecular mass of the compound is estimated to be 100. What is the percentage composition of carbon in the compound? |
| Answer» <html><body><p>0.16<br/>0.08<br/>0.28<br/>0.48</p>Answer :D</body></html> | |
| 51900. |
0.96g chloroplatinate of a diacid base when ingnited gave 0.32 g platinum. The molecular mass of the base is [AW of Pt = 195] |
| Answer» <html><body><p>175<br/>350<br/>`87.5`<br/>210</p>Solution :`(W)/(ZE+410)=(<a href="https://interviewquestions.tuteehub.com/tag/omega-585625" style="font-weight:bold;" target="_blank" title="Click to know more about OMEGA">OMEGA</a>)/(<a href="https://interviewquestions.tuteehub.com/tag/195-281683" style="font-weight:bold;" target="_blank" title="Click to know more about 195">195</a>)`<br/>`(0.96)/(ZE+410)=(0.32)/(195)<a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> E = (175)/(2)`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/mw-550075" style="font-weight:bold;" target="_blank" title="Click to know more about MW">MW</a>=<a href="https://interviewquestions.tuteehub.com/tag/z-750254" style="font-weight:bold;" target="_blank" title="Click to know more about Z">Z</a> xx E` since it is diacidic base<br/>`therefore MW = Z xx (175)/(2)=175`</body></html> | |