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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.
| 51251. |
2H_(2(g))+O_(2(g))to2H_(2)(O)_((l)) This reaction is ____ . |
| Answer» <html><body><p>Redox<br/>Decomposition<br/>Oxidation<br/>Reduction</p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_CHE_XI_P2_C08_E03_101_S01.png" width="80%"/> <br/> `therefore` Oxidation <br/> In the <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a> Oxidation number of .H. <a href="https://interviewquestions.tuteehub.com/tag/atom-887280" style="font-weight:bold;" target="_blank" title="Click to know more about ATOM">ATOM</a> increases and at the same time - <a href="https://interviewquestions.tuteehub.com/tag/decreases-946143" style="font-weight:bold;" target="_blank" title="Click to know more about DECREASES">DECREASES</a> of .O. atom. <br/> `therefore` This is a redox reaction.</body></html> | |
| 51252. |
2H_(2(g))+CO_(2(g))hArrCH_(3)OH_((g)),DeltaH=-92.2kJ. Which of the following condition will shift the equilibrium in the forward direction ? |
| Answer» <html><body><p><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 <a href="https://interviewquestions.tuteehub.com/tag/system-1237255" style="font-weight:bold;" target="_blank" title="Click to know more about SYSTEM">SYSTEM</a> is increased <br/>CO is removed <br/>`CH_(3)OH` is <a href="https://interviewquestions.tuteehub.com/tag/added-367625" style="font-weight:bold;" target="_blank" title="Click to know more about ADDED">ADDED</a> <br/>The <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 the system is increased </p>Answer :D</body></html> | |
| 51253. |
2H_(2) O_(2) to 2 H_(2)O_ + O_(2) . This reaction is |
| Answer» <html><body><p>Decomposition <br/>Combination <br/><a href="https://interviewquestions.tuteehub.com/tag/disproportionation-440717" style="font-weight:bold;" target="_blank" title="Click to know more about DISPROPORTIONATION">DISPROPORTIONATION</a> reactions <br/>Both <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>) and 3) </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> :C</body></html> | |
| 51254. |
2H_(2(g)) + O_(2(g)) rarr 2H_(2)O_((l)) , DeltaH= -ve and DeltaG= -ve. Then the reaction is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/spontaneous-632196" style="font-weight:bold;" target="_blank" title="Click to know more about SPONTANEOUS">SPONTANEOUS</a> and endothermic <br/>Spontaneous and slow <br/><a href="https://interviewquestions.tuteehub.com/tag/non-580180" style="font-weight:bold;" target="_blank" title="Click to know more about NON">NON</a> spontaneous and slow <br/>Non spontaneous ans slow </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/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 51255. |
2g of H_2 and 17g of NH_3 are placed in a 8.21 litre flask at 27^@C. The total pressure of the gas mixture is? |
| 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>><a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a> atm<br/>5 atm<br/>6 atm<br/>3 atm.</p>Solution :`m=n_(H_2)+n_(NH_3)=2/2+17/17=2` <br/> `P=(<a href="https://interviewquestions.tuteehub.com/tag/nrt-1109890" style="font-weight:bold;" target="_blank" title="Click to know more about NRT">NRT</a>)/V=(2xx0.0.0821xx300)/8.21=6`atm</body></html> | |
| 51256. |
Two gram of hydrogen diffuse from a container in 10 minutes. How many grams of oxygen would diffuse through the same container in the same time under similar conditions? |
| Answer» <html><body><p>0.5g <br/><a href="https://interviewquestions.tuteehub.com/tag/4g-318729" style="font-weight:bold;" target="_blank" title="Click to know more about 4G">4G</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/6g-331964" style="font-weight:bold;" target="_blank" title="Click to know more about 6G">6G</a> <br/>8g</p>Answer :D</body></html> | |
| 51257. |
2g of Al is treated separately with excess dilute H_2SO_4 andexcess NaOH.The ratio of volumes of Hydrogen evolved under similar condition is x y. Find x/y |
| 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> :`2Al+3H_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)SO_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>) to Al_(2)(SO_(4))_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)+3H_(2)` <br/> `2Al+2NaOH+2H_(2) 2NaAlO_(2)+3H_(2)` <br/> Ratio is 1:1</body></html> | |
| 51258. |
2g of a non electrolyte solute dissolved in 75g of benzene lowered the freezing point of benzene by 0.20 k. The freezing point depression constant of benzene is 5.12 K Kg mol ^(-1). Find the molar mass of the solute. |
| Answer» <html><body><p></p>Solution :`W_(2) =2g, W_(1) =<a href="https://interviewquestions.tuteehub.com/tag/75g-335510" style="font-weight:bold;" target="_blank" title="Click to know more about 75G">75G</a>, Delta T _(f)=0.2 <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>, k _(f) =5.12 kg mol ^(-1) , M _(2)` = ? <br/> `M _(2) = (K _(f) <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> W _(2) xx 1000)/( Delta T_(f) xx W_(1)) = (5.12 xx 2 xx 1000)/( 0.2 xx <a href="https://interviewquestions.tuteehub.com/tag/75-334971" style="font-weight:bold;" target="_blank" title="Click to know more about 75">75</a>) = 682.66 g mol ^(-1)`</body></html> | |
| 51259. |
2CuIrarrCu+CuI_(2), the reaction is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/disproportionation-440717" style="font-weight:bold;" target="_blank" title="Click to know more about DISPROPORTIONATION">DISPROPORTIONATION</a> reaction<br/>Neutralisation reaction<br/>Oxidation reaction<br/>Reduction reaction</p>Answer :A</body></html> | |
| 51260. |
2CuI rarr Cu + CuI_2the reaction is |
| Answer» <html><body><p>disproportionation <br/>Neutralisation <br/><a href="https://interviewquestions.tuteehub.com/tag/oxisation-2906605" style="font-weight:bold;" target="_blank" title="Click to know more about OXISATION">OXISATION</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/reduction-621019" style="font-weight:bold;" target="_blank" title="Click to know more about REDUCTION">REDUCTION</a></p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_A_C04_E01_026_S01.png" width="80%"/> <br/>Same element undergo <a href="https://interviewquestions.tuteehub.com/tag/oxidation-588780" style="font-weight:bold;" target="_blank" title="Click to know more about OXIDATION">OXIDATION</a> and reduction , so it is disproportionation <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a> .</body></html> | |
| 51261. |
Justify that the reaction : 2Cu_(2) O(s) + Cu_(2)S (s) rarr 6 Cu (s) + SO_(2) (g) is a redox reaction. Identify the species oxidised/reduced, which acts as an oxidant and which acts as a reductant. |
| Answer» <html><body><p></p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/overset-2905731" style="font-weight:bold;" target="_blank" title="Click to know more about OVERSET">OVERSET</a>(+1)(2Cu_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))overset(-2)(O_((s)))+overset(+1)(Cu_(2))overset(-2)(S_((s)))rarroverset(<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>)(6Cu_((s)))+overset(+4-2)(SO_(2))` <br/> In this reaction copper is reduced from `+ 1` state to zero oxidation state and <a href="https://interviewquestions.tuteehub.com/tag/sulphur-1234343" style="font-weight:bold;" target="_blank" title="Click to know more about SULPHUR">SULPHUR</a> is oxidised from `- 2` state to `+ 4` state. The above reaction is thus a redox reaction. <br/> Further, `Cu_(2)S` helps sulphur in `Cu_(2)S` to <a href="https://interviewquestions.tuteehub.com/tag/increase-1040383" style="font-weight:bold;" target="_blank" title="Click to know more about INCREASE">INCREASE</a> its oxidation number,therefore `Cu(I)` is an oxidant, and sulphur of `Cu_(2)S` helps copper both in `Cu_(2)S` itself and `Cu_(2)O` to decrease its oxidation number, therefore, sulphur of `Cu_(2)S` is reductant.</body></html> | |
| 51262. |
2C_((s))+ 1/2O_(2(gg) rarr CO_(2(g)) , DeltaH = -787 KJ H_(2(g)) + 1/2O_(2(g)) rarr H_2O_((l)) , DeltaH = -286 KJ C_2H_(2(g)) + 2 1/2O_(2(g)) rarr 2CO_(2(g)) + H_2O_((l)) , DeltaH = -1301 KJ. Heat of formation of acetylene in KJ is |
| Answer» <html><body><p>`-1802`<br/>`+1802`<br/>`-<a href="https://interviewquestions.tuteehub.com/tag/800-338448" style="font-weight:bold;" target="_blank" title="Click to know more about 800">800</a>`<br/>`+<a href="https://interviewquestions.tuteehub.com/tag/228-1825063" style="font-weight:bold;" target="_blank" title="Click to know more about 228">228</a>`</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> :D</body></html> | |
| 51263. |
2CO(g) + O_(2)(g) hArr 2CO_(2)(g) at 1000 K . What is theK_(C) for this reaction ? Predict the extent of this reaction. |
| 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> :`2CO(g) + O_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)(g) hArr 2CO_(2)(g) at 1000 K ` <br/> `K_(C) = 2.2 xx 10^(22)` <br/> `K_(C) gt 10^(3).` So [products ] >> [Reactants]<br/> Reaction nearly goes to completion and <a href="https://interviewquestions.tuteehub.com/tag/forward-464460" style="font-weight:bold;" target="_blank" title="Click to know more about FORWARD">FORWARD</a> reaction is favoured.</body></html> | |
| 51264. |
2ClCH_(2)CH_(2)OH("excess") underset(415K)overset(Conc. H_(2)SO_(4))rarr X underset("Fuse")overset(KOH)rarr Y The compound Y is |
| Answer» <html><body><p>Ethylene glycol<br/>Ethanol<br/>Divinyl ether<br/>Vinyl chloride</p>Solution :`2ClCH_(2)CH_(2)OH ("<a href="https://interviewquestions.tuteehub.com/tag/excess-978535" style="font-weight:bold;" target="_blank" title="Click to know more about EXCESS">EXCESS</a>") underset(Delta//4.15 K)overset(<a href="https://interviewquestions.tuteehub.com/tag/conc-927968" style="font-weight:bold;" target="_blank" title="Click to know more about CONC">CONC</a>. H_(2)SO_(4))rarr ClCH_(2)underset((X))(CH_(2)-)O-CH_(2)CH_(2)<a href="https://interviewquestions.tuteehub.com/tag/cl-408888" style="font-weight:bold;" target="_blank" title="Click to know more about CL">CL</a> overset(KOH//"<a href="https://interviewquestions.tuteehub.com/tag/fuse-1002649" style="font-weight:bold;" target="_blank" title="Click to know more about FUSE">FUSE</a>")rarr underset("Divinyl ether (Y)")(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>-O-CH=CH_(2))`</body></html> | |
| 51265. |
2CaSO_4(s)hArr2CaO(s)+2SO_2(g)+O_2(g),DeltaH gt 0 Above equilibrium is established by taking some amount of CaSO_4(s) in a closed container at 1600K. Then which of the following may be correct option? |
| Answer» <html><body><p>Moles of CaSO(s) will increase with increase in temperature. <br/>If the volume of the container is doubled at equilibrium then partical <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 `SO_2(g)` will change at new equilibrium <br/>If the volume of the container is halved pressure of `O_2(g)` at new equilibrium will remain same. <br/>If <a href="https://interviewquestions.tuteehub.com/tag/two-714195" style="font-weight:bold;" target="_blank" title="Click to know more about TWO">TWO</a> moles of the He gas is added at constant pressure then the moles of <a href="https://interviewquestions.tuteehub.com/tag/cao-411869" style="font-weight:bold;" target="_blank" title="Click to know more about CAO">CAO</a>(s) will increase. </p>Answer :A::<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>::D</body></html> | |
| 51266. |
2CaSO_(4)(s)hArr22CaO(s)+2SO_(2)(g)+O_(2)(g), DeltaHgt0 Above equilibrium is established by taking some amout of CaSO_(4)(s) in a closed container at 1600K Then which of the following may be correct option. |
| Answer» <html><body><p>Moles of `<a href="https://interviewquestions.tuteehub.com/tag/cao-411869" style="font-weight:bold;" target="_blank" title="Click to know more about CAO">CAO</a>(s)` will increase with the increase in temperature<br/>If the voulme of the container is doubled at equilibrium then partial pressure of `SO_(2)(g)` will <a href="https://interviewquestions.tuteehub.com/tag/change-913808" style="font-weight:bold;" target="_blank" title="Click to know more about CHANGE">CHANGE</a> at new equilibrium<br/>If the volume of the container is halved partial pressure of `O_(2)(g)` at new equilibrium will remain same <br/>If two moles of the He gas is added at constant pressure then the moles of `CaO(s)` will increase.</p>Solution :(A) As reaction is endothermic therefore it will <a href="https://interviewquestions.tuteehub.com/tag/go-468886" style="font-weight:bold;" target="_blank" title="Click to know more about GO">GO</a> in the forward direction hence moles of `CaO` will increase. <br/> (B) With the increase or decrease of volume partial pressure of the gases will remain same. <br/> (C) Due to the <a href="https://interviewquestions.tuteehub.com/tag/addition-367641" style="font-weight:bold;" target="_blank" title="Click to know more about ADDITION">ADDITION</a> of inert gas at constant pressure reaction will proceed in the direction in which more number of <a href="https://interviewquestions.tuteehub.com/tag/gaseous-467815" style="font-weight:bold;" target="_blank" title="Click to know more about GASEOUS">GASEOUS</a> moles are formed.</body></html> | |
| 51267. |
2B(OH)_(3) +2NaOHhArr NaBO_(2) + Na[B(OH)_(4)]+2H_(2)O How can this reaction be made to proceed in forwards direction ? |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/addition-367641" style="font-weight:bold;" target="_blank" title="Click to know more about ADDITION">ADDITION</a> of cis-1, <a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>-diol<br/>addition of borax<br/>addition of trans-1, 2-diol<br/>addition of `Na_(2)HPO_(4)`</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> | |
| 51268. |
298 K of N_2O_(4(g)) hArr 2NO_(2(g))reaction is in equilibrium K_p = 0.14 atm. Calculate of K_c (R=0.082 L atm K^(-1) "mol"^(-1)) |
| 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> :`5.73xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/9-340408" style="font-weight:bold;" target="_blank" title="Click to know more about 9">9</a>)`M</body></html> | |
| 51269. |
29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl's method and the evolved ammonia was absorbed in 20 mL of 0.1 M HCl solution. The exess of the acid required 15 mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen in the compound is |
| Answer» <html><body><p>29.5<br/>`59.0`<br/>47.4<br/>23.7</p>Solution :Weight of substance <a href="https://interviewquestions.tuteehub.com/tag/taken-659096" style="font-weight:bold;" target="_blank" title="Click to know more about TAKEN">TAKEN</a> `= 29.5 xx 10^(-3) g` <br/> Volume of <a href="https://interviewquestions.tuteehub.com/tag/acid-847491" style="font-weight:bold;" target="_blank" title="Click to know more about ACID">ACID</a> taken <a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a> mL of 0.1 <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> HCl <br/> `:.` Volume of acid used = 20 - <a href="https://interviewquestions.tuteehub.com/tag/15-274069" style="font-weight:bold;" target="_blank" title="Click to know more about 15">15</a> = 5 mL of 0.1 N HCl <br/> `%N= (1.4 xx "Volume" xx "Normality")/("Weight of substance taken") = (1.4 xx 5 xx 0.1)/(29.5 xx 10^(-3))`<br/> `= 23.73`</body></html> | |
| 51270. |
293 K = .......... ""^(@)F. |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/273-1832932" style="font-weight:bold;" target="_blank" title="Click to know more about 273">273</a><br/>68<br/><a href="https://interviewquestions.tuteehub.com/tag/293-1834823" style="font-weight:bold;" target="_blank" title="Click to know more about 293">293</a><br/>77</p>Solution :`293K= 293 - 273 = 20 ""^(@)<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>` <br/> `F = 9/5xx(C^(@))+32= 9/5xx(20)+32=68""^(@)F`</body></html> | |
| 51271. |
2.9 g of a gas at 95^(@)C occupied the same volume as 0.184 g of hydrogen at 17^(@)C at the same pressure. What is the molar mass of the gas ? |
| Answer» <html><body><p></p>Solution :As `"" P_(1)=P_(<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/> and `"" V_(1)=V_(2)`, But PV=nRT<br/> `:. "" P_(1)V_(1)=P_(2)V_(2)`, i.e., `n_(1)RT_(1)=n_(2)RT_(2)` <br/> `:. "" n_(1)T_(1)= n_(2)T_(2) "or" (w_(1))/(M_(1))T_(1)=(w_(2))/(M_(2))T_(2)` <br/> i.e, `"" (2.9)/(M_(X))<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>(95+273)=(0.184)/(2)xx(17+273)` <br/> or `"" M_(X)=(2.9xx368xx2)/(0.184xx290)=40" <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> mol"^(-1)`</body></html> | |
| 51272. |
2.9 g of a gas at 95^@C occupied the same volume as 0.184 g of dihydrogen at 17^@C, at the same pressure. What is the molar mass of the gas? |
| Answer» <html><body><p></p>Solution :Suppose the molar mass of the <a href="https://interviewquestions.tuteehub.com/tag/gas-1003521" style="font-weight:bold;" target="_blank" title="Click to know more about GAS">GAS</a> is M. According to the gas equation , `PV = <a href="https://interviewquestions.tuteehub.com/tag/nrt-1109890" style="font-weight:bold;" target="_blank" title="Click to know more about NRT">NRT</a>`. Since P and V are the same, we have <br/> For the gas : <br/> `PxxV = (2.9)/M xxRxx(273 +95)` ...(i) <br/> For `H_2 : PxxV= (0.184)/2 xx R xx (273+17)` ...(ii)<br/> From <a href="https://interviewquestions.tuteehub.com/tag/equations-452536" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATIONS">EQUATIONS</a> (i) and (ii), we have <br/> `(2.9)/M xx368 = (0.184)/2 xx <a href="https://interviewquestions.tuteehub.com/tag/290-1834725" style="font-weight:bold;" target="_blank" title="Click to know more about 290">290</a>` <br/> or `M = (2.9 xx 368 xx 2)/(0.184xx290)= 40 g <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a>(-1)`</body></html> | |
| 51273. |
2.9 g of a gas at 95^(@)C occupied the same volume as 0.184 g of dihydrogen at 17^(@)C, at the same pressure. What is the molar of the gas ? |
| 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> :According to ideal gas equations,<br/>`pV=nRT=(wRT)/(M) "" therefore M=(wRT)/(pV)`<br/>`therefore (pV)/(R )=(<a href="https://interviewquestions.tuteehub.com/tag/wt-1462289" style="font-weight:bold;" target="_blank" title="Click to know more about WT">WT</a>)/(M)` where `(pV)/(R )=` constant of number<br/>for <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> gas<br/>`w=w_(1)=2.9` gram<br/>R = Gas constant<br/>`T=T_(1)=95^(@)<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>`<br/>`= (95+273)=<a href="https://interviewquestions.tuteehub.com/tag/368-1859650" style="font-weight:bold;" target="_blank" title="Click to know more about 368">368</a> K`<br/>p = p bar<br/>V = V L<br/>`M = M_(1)=M_(1)g mol^(-1)`<br/>For `H_(2)` gas :<br/>`w_(2)=0.184` gram<br/>R = Gas constant<br/>`T_(2)=(17+273)K=290 K`<br/>p = p bar<br/>V = V L<br/>`M=M_(2)=2.0 g mol^(-1)`<br/>`therefore (pV)/(R )=(w_(1)T_(1))/(M_(1))=(w_(2)T_(2))/(M_(2))=` constant<br/>`therefore M_(1)=(w_(1)T_(1))((M_(2))/(w_(2)T_(2)))=((2.9g)(368 K)(2g mol^(-1)))/((0.184 g)(290 K))`<br/>`= 40 g mol^(-1)`</body></html> | |
| 51274. |
2.84 g of methyl iodide was completely converted into methyl magnesium iodide and was decomposed by excess of ethanol. The volume of the gaseous hydrocarbon proudced at NTP will be |
| Answer» <html><body><p>22.4 litre<br/>224 mL<br/>0.448 litre<br/>0.224 litre</p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/underset-3243992" style="font-weight:bold;" target="_blank" title="Click to know more about UNDERSET">UNDERSET</a>("one <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a>")(CH_(3)I)overset(Mg)tounderset("onemole"(CH_(3)Mgl)overset(<a href="https://interviewquestions.tuteehub.com/tag/etoh-2073590" style="font-weight:bold;" target="_blank" title="Click to know more about ETOH">ETOH</a>)tounderset("one mole")(CH_(4))`</body></html> | |
| 51275. |
2.82g of glucose is dissolved in 30g of water. The mole fraction of glucose in the solution is |
| Answer» <html><body><p>0.01<br/>`0.99`<br/>`0.52`<br/>1.66</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> :No. of moles of glucose `=(2.82)/(<a href="https://interviewquestions.tuteehub.com/tag/180-279527" style="font-weight:bold;" target="_blank" title="Click to know more about 180">180</a>)=0.01567` <br/> No. of moles of <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a> `=30/18=1.667` <br/> Total of moles of solution =0.01567+1.667=1.683 <br/> Moles fraction glucose `=(0.01567)/(1.683)=0.0093=0.01`</body></html> | |
| 51276. |
2.82 g of glucose is dissolved in 30 g of water. Calculate the mole fraction of glucose and water. |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of glucose `= 2. 82 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>` <br/> No. of moles of glucose `= (2. 82)/( 180) = (<a href="https://interviewquestions.tuteehub.com/tag/30-304807" style="font-weight:bold;" target="_blank" title="Click to know more about 30">30</a>)/(18) = 1. 67` <br/> ` <a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a> _(H _(2)O)= (1. 67)/( 1. 67 + 0.016) = (1. 67)/( 1. 686) =0.99` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> x _(H _(2)O) + x _("glucose") =1` <br/> ` 0.99 + x _("glucose")=1` <br/> `x _("glucose")=1-0.99 =0.01`</body></html> | |
| 51277. |
28 NO_(3)^(-) + 3As_(2)S_(3) + 4H_(2)O to 6AsO_(4)^(3-)+28 NO+ 9SO_(4)^(2-) + 8H^(+). What will be the equivalent mass of As_(2)S_(3) in above reaction ? |
| Answer» <html><body><p>`(M. <a href="https://interviewquestions.tuteehub.com/tag/wt-1462289" style="font-weight:bold;" target="_blank" title="Click to know more about WT">WT</a>.)/(<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/>`(M. wt.)/(<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/>`(M. wt.)/(24)`<br/>`(M. wt.)/(28)`</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> | |
| 51278. |
28 gm KOH is dissolved in 90 g water. Find mole Fraction of KOH and water respectively. |
| Answer» <html><body><p> <br/> <br/> <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/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> <a href="https://interviewquestions.tuteehub.com/tag/fraction-458259" style="font-weight:bold;" target="_blank" title="Click to know more about FRACTION">FRACTION</a> of `KOH=0.0909`, <br/> Mole fraction of `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>)O=0.909`</body></html> | |
| 51279. |
28 g of nitrogen and 6 g of hydrogen were mixed in a 1 litre closed container. At equilibrium 17 g NH_(3) was produced. Calculate the weight of nitrogen, hydrogen at equilibrium. |
| 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> :<a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a>`m_(N_(2))=<a href="https://interviewquestions.tuteehub.com/tag/28g-1834386" style="font-weight:bold;" target="_blank" title="Click to know more about 28G">28G</a>,m_(H_(2))=6g,` <br/> `V=1L` <br/> `(n_(N_(2)))_("<a href="https://interviewquestions.tuteehub.com/tag/initial-499269" style="font-weight:bold;" target="_blank" title="Click to know more about INITIAL">INITIAL</a>")=(28)/(28)="1 mol"` <br/> `(n_(H_(2)))_("Initial")=6/2=3" mol"` <br/> `N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/SUR_CHE_XI_V02_C08_E01_048_S01.png" width="80%"/> <br/> `[NH_(3)]=((17)/(17))="1 mol"` <br/> Weight of `N_(2)` = (no. of moles of `N_(2)` ) `xx` molar mass of `N_(2)` <br/> `=0.5xx28=14g` <br/> Weight of `H_(2)` = (no. of moles of `H_(2)` ) `xx` molar mass of `H_(2)` <br/> `=1.5xx2=3g`</body></html> | |
| 51280. |
28 g of nitrogen and 6 g hydrogen were mixed in a1 litre closed container. At equilibrim 17 g NH_3 was produced. Calculate the weight of weight of nitrogen, hydrogen at equilibrium. |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> `m_(N_2) = <a href="https://interviewquestions.tuteehub.com/tag/28g-1834386" style="font-weight:bold;" target="_blank" title="Click to know more about 28G">28G</a>, m_(H_2) = 6 g , V = 1L`<br/> `(n_(N_2))_("initial")= <a href="https://interviewquestions.tuteehub.com/tag/28-299271" style="font-weight:bold;" target="_blank" title="Click to know more about 28">28</a>/28 = 1mol` <br/> `(n_(H_2))_("initial")= 6/2 = 3mol`<br/> `N_2(g)+ 3H_2(g) hArr 2NH_3(g)`<br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_CHE_XI_V02_C08_E01_046_S01.png" width="80%"/> <br/> `[NH_3] = (17/17) = 1mol = 2x`<br/> `rArrx = 0.5` mol <br/> At equilibrium, `[N_2] = 1 - x = 0.5 mol`<br/> `[H_2] = 3 -3x =3-3 (0.5) = 1.5 mol` <br/> <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`N_2` = (no. of moles of `N_2`)` xx`molar <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of `N_2 = 0.5 xx 28 = 14g` <br/> Weight of `H_2` = (no. of molesof `H_2`) `xx` molar mass of `H_2= 1.5 xx 2 = 3g`</body></html> | |
| 51281. |
28 mg of an organic compound containing nitrogen was digested according to Kjeldahl's method and the evolved ammonia was absored in 20 mL of 0.1 M HCl solution. The excess of the acid required 15 mL of 0.1 M NaOH solution for complete neutralization. Then the percentage of 'N' is the compound is |
| Answer» <html><body><p>50<br/>30<br/>25<br/>40</p>Solution :Excess <a href="https://interviewquestions.tuteehub.com/tag/acid-847491" style="font-weight:bold;" target="_blank" title="Click to know more about ACID">ACID</a> `= 15xx0.1xx1` m. eq.<br/>Acid taken initially `= 20xx0.1xx1` m.eq.<br/>Acid used for neutralisation<br/>`=(20xx0.1xx1)-(15xx0.1xx1)=0.5` m. eq.<br/>`% <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> =(NH_(3)" liberated"xx1.4)/("wt.of <a href="https://interviewquestions.tuteehub.com/tag/org-1138670" style="font-weight:bold;" target="_blank" title="Click to know more about ORG">ORG</a>. <a href="https://interviewquestions.tuteehub.com/tag/compound-926863" style="font-weight:bold;" target="_blank" title="Click to know more about COMPOUND">COMPOUND</a>")=(0.5xx1.4)/(0.028)=25%`.</body></html> | |
| 51282. |
28 g of each of the following gases are taken at 27^(@)Cand 600 mm pressures. Which of these will have the least volume? |
| 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>>HBr<br/>HCl<br/>HF<br/>HI</p>Solution :PV=<a href="https://interviewquestions.tuteehub.com/tag/nrt-1109890" style="font-weight:bold;" target="_blank" title="Click to know more about NRT">NRT</a> or `<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>=(nRT)/(P)`. As P and T are same for all given gases, `V prop n`. As `n=(w)/(M)`, n will be <a href="https://interviewquestions.tuteehub.com/tag/least-7256596" style="font-weight:bold;" target="_blank" title="Click to know more about LEAST">LEAST</a> for which M is maximum viz. HI.</body></html> | |
| 51283. |
27g of Al was treated with NaOHsolution when a whitegelatinous precipitate was obtainedwhichuponstrongheatinggave an oxide. The amount ofoxide (in g ) is |
| 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> :`<a href="https://interviewquestions.tuteehub.com/tag/underset-3243992" style="font-weight:bold;" target="_blank" title="Click to know more about UNDERSET">UNDERSET</a>(54 g)(2Al)overset(NaOH)<a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> 2A(<a href="https://interviewquestions.tuteehub.com/tag/oh-585115" style="font-weight:bold;" target="_blank" title="Click to know more about OH">OH</a>)_(3)overset(Delta)rarrunderset(102 g)(Al_(2)O_(3))` <br/> `54 g` of Al produce `Al_(2)O_(3) = 102//2 = 51.0 g`</body></html> | |
| 51284. |
2.79 g of an organic compound when heated in Carius tube with cone. HNO_3 " and "H_3PO_4 formed converted into MgNH_4.PO_4 ppt. The ppt. on heating gave 1.332 g of Mg_2P_2O_7.The percentage of P in the compound is |
| 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>>23.33<br/>13.33<br/>33.33<br/>26.66</p>Solution :`% of P = (<a href="https://interviewquestions.tuteehub.com/tag/62-330265" style="font-weight:bold;" target="_blank" title="Click to know more about 62">62</a> <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 1.332 xx <a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a>)/(222xx2.79)=13.33`</body></html> | |
| 51285. |
2.76 g of silver carbonate on strong ignition leaves a residue weighing |
| Answer» <html><body><p>2.48g<br/>2.16g<br/>2.32g<br/>2.84g</p>Solution :`2Ag_(2)CO_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)overset(<a href="https://interviewquestions.tuteehub.com/tag/delta-947703" style="font-weight:bold;" target="_blank" title="Click to know more about DELTA">DELTA</a>)rarrunderset("Residue")(4Ag)+2CO_(2)uarr+O_(2)uarr`</body></html> | |
| 51286. |
2.76 g of Ag_2CO_3 on being heated yields a residue weighing |
| Answer» <html><body><p>A)2.16 g<br/>B)2.32 g<br/>C)2.48 g<br/>D)2.64 g </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> :`underset(2 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 276 g)(2Ag_(2)CO_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)) to underset(4 xx 108 g)(4 Ag) + 2CO_(2) uarr + O_(2) uarr` <br/> `therefore` <a href="https://interviewquestions.tuteehub.com/tag/residue-622603" style="font-weight:bold;" target="_blank" title="Click to know more about RESIDUE">RESIDUE</a> given by 2.76 g of <br/> `Ag_(2)CO_(3) =(4 xx 108)/(2 xx 276)xx 2.76 = 2.16 g`</body></html> | |
| 51287. |
2.746 g of a compound on analysis gave 1.94 g of silver, 0.268 g of sulphur and 0.538 g of oxygen. Find the empirical formula of the compound. |
| Answer» <html><body><p></p>Solution :The <a href="https://interviewquestions.tuteehub.com/tag/percent-1150333" style="font-weight:bold;" target="_blank" title="Click to know more about PERCENT">PERCENT</a> composition of elements can be calculated as <a href="https://interviewquestions.tuteehub.com/tag/follows-994526" style="font-weight:bold;" target="_blank" title="Click to know more about FOLLOWS">FOLLOWS</a> : <br/> % Ag `=1.94/2.746 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 100 = 70.6 %` <br/> `5 S = 0.268/2.746 xx 100 = 9.76 %` <br/> `% O = 0.538/2.746 xx 100 = 19.6 %` <br/> Calculation of empirical <a href="https://interviewquestions.tuteehub.com/tag/formula-464310" style="font-weight:bold;" target="_blank" title="Click to know more about FORMULA">FORMULA</a>: <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/NTN_HCS_ISC_CHE_XI_P1_C01_E09_024_S01.png" width="80%"/> <br/> `therefore` The empirical formula of the given compound is `Ag_(2)SO_(4)`.</body></html> | |
| 51288. |
273 mL of a gas at STP was taken to 27^(@)C and 600 mm pressure. The final volume of the gas would be : |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/273-1832932" style="font-weight:bold;" target="_blank" title="Click to know more about 273">273</a> mL<br/>300 mL<br/>380 mL<br/>586 mL</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 51289. |
273 K is equal to _______ degree centigrade. |
| Answer» <html><body><p>0<br/>100<br/>373<br/>1</p>Answer :A</body></html> | |
| 51290. |
27 g of Al (at mass=27) will react completely with oxygen equal to |
| Answer» <html><body><p>24 g<br/>8g<br/>40 g<br/>10 g<br/></p>Solution :a) The equation of the reaction of aluminium and oxygen, is as <br/> `4Al + 3O_(2) to 2Al_(2)O_(3)` <br/> therefore 108 g of AI reacts with 96g of `O_(2)`. <br/> therefore <a href="https://interviewquestions.tuteehub.com/tag/27g-1833325" style="font-weight:bold;" target="_blank" title="Click to know more about 27G">27G</a> of AI with react with `=(<a href="https://interviewquestions.tuteehub.com/tag/96-342541" style="font-weight:bold;" target="_blank" title="Click to know more about 96">96</a> <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> <a href="https://interviewquestions.tuteehub.com/tag/27-298706" style="font-weight:bold;" target="_blank" title="Click to know more about 27">27</a>)/(108)` = 24 g of `CO_(2)`</body></html> | |
| 51291. |
26c.c of CO_(2) is passed over red hot coke. The volume of CO evolved is (under the same condition) |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/15-274069" style="font-weight:bold;" target="_blank" title="Click to know more about 15">15</a> c.c<br/>10 c.c<br/>32 c.c<br/>52 c.c</p>Answer :D</body></html> | |
| 51292. |
26.8g of Na_(2)SO_(4). nH_(2)O contains 12.6 g of water. The value of n is: |
| Answer» <html><body><p>1<br/>10<br/>6<br/>7</p>Answer :D</body></html> | |
| 51293. |
2.68 xx10^(-3) moles of a solutioncontaing an ion A^(n+) require 1.61 xx10^(-3) moles MnO_(4)^(-) for the oxidation of A^(n+) to AO_(3)^(-) in acid medium what is the value of n ? |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/step-25533" style="font-weight:bold;" target="_blank" title="Click to know more about STEP">STEP</a> 1 To write the reduction and oxidation half reaction <br/> Reducton : `Mno_(4)^(-) + 8H^(+)+5e^(-)rarrMn^(2+)+4H_(2)O` <br/> oxidation `A^(n+)+3H_(2)OrarrAO_(3)^(-)+6H^(+)+(5-n)E^(-)` <br/> Step 2 To <a href="https://interviewquestions.tuteehub.com/tag/find-11616" style="font-weight:bold;" target="_blank" title="Click to know more about FIND">FIND</a> out the <a href="https://interviewquestions.tuteehub.com/tag/value-1442530" style="font-weight:bold;" target="_blank" title="Click to know more about VALUE">VALUE</a> of n <br/> <a href="https://interviewquestions.tuteehub.com/tag/since-644476" style="font-weight:bold;" target="_blank" title="Click to know more about SINCE">SINCE</a> in a <a href="https://interviewquestions.tuteehub.com/tag/redox-2246707" style="font-weight:bold;" target="_blank" title="Click to know more about REDOX">REDOX</a> reaction number of electrons lost = number of electrons gained therefore multiple oxidant of Eq (i) i.e `MnO_(4)^(-)` by (5-n) and reductant of Eq (ii) i.e `A^(n+)` by 5 adn equate we have (5-n) `MnO_(4)^(-)=5 A^(n+) i.e (5-n)` moles of `MnO_(4)^(-)` will oxidise `An^(n+)` = moles<br/> Equationg the values of moles of `A^(n+)` actually oxidised =`2.68xx10^(-3)` moles <br/> Equating the value of Eq (iii) and (iv) we have `(5)/(5-n) xx1.61xx10^(-3)=2.68xx10^(-3) or5xxx1.61=(5-n)xx2.68 or 2.68 n = 5(2.68-1.61)=5xx1.07=5.35 or n =(5.35)/(2.68)=2`</body></html> | |
| 51294. |
26.8 g of Na_2SO_4.xH_2O gave 12.6 g of water on heating. Determine the value of x in the compound using mole concept. |
| Answer» <html><body><p></p>Solution :Weight of hydrated salt `Na_2SO_4.xH_2O = 26.8 g` <br/>Weight of water given by it = 12.6 g<br/> `therefore` Weight of anhydrous salt `Na_2SO_4 = 26.8 - 12.6 = 14.2g` <br/> Gram molecular mass of `Na_2SO_4 = (2 xx 22.99) + 32.06 + (4 xx 16.0) = 142.04 g`<br/> `therefore` Number of moles of `Na_2SO_4 = 142 04/(142.04) = 0.0999` <br/>and number of moles of `H_(2)O`associated with it `12.6/18.016`<br/> `therefore 0.0999` moles of `Na_2SO_4` <a href="https://interviewquestions.tuteehub.com/tag/attach-2449108" style="font-weight:bold;" target="_blank" title="Click to know more about ATTACH">ATTACH</a> moles of water = `0.699 , therefore` <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/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> of `Na_2SO_4` will attach moles of water <br/> `=0.699/0.0999 xx 1` <br/> = 6.99 <br/> =7, Hence, x=7</body></html> | |
| 51295. |
2.650 g of anhydrous sodium carbonate are dissolved in water and the solution made up to 500 mL. On titration 50 mL of this solution neutralises 50 mL of a solution of sulphuric acid. How much water should be added to 450 mL of this acid as to make it exactly N//12 ? |
| 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> | |
| 51296. |
2.65 g of diacidic base was dissolved in water and made up to 500 mL. 20 of this solutioncompletely neutralised 12 mL of N//6 HCl. Find out the equivalent mass and molecular mass of the base. |
| 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> | |
| 51297. |
2.5mLof 2//5M weak mono-acidic base (K_(b) = 1 xx 10^(-12) at 25^(@)C) is titrated with 2//15M HCI in water at 25^(@)C. Find the concentration of H^(o+) ions at equivalence point. (K_(w) = 1 xx 10^(-14)at 25^(@)C) a. 3.7 xx 10^(-13)M b. 3.2 xx 10^(-7)M c. 3.2 xx 10^(-2)M d. 2.7 xx 10^(-2)M |
| Answer» <html><body><p></p>Solution :a. First find the <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> of `HCI` re2uired to reach the eqivalent point. <br/> mEq of base = mEq of HCI <br/> `rArr (<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>.5 xx (2)/(5) xx 1) = ((2)/(15)xx1)V_(HCI)` <br/> `rArr V_(HCI) = 7.5mL` <br/> The net volume of the solution at the equivalent point <br/> `= V_("base") +V_(HCI) = 2.5 +7.5 = 10mL` <br/> `rArr ["salt"] = c (2//5 xx 2.5)/(10) = 0.1M` <br/> `ph` of an aqueous solution of such a salt is given by : <br/> `pH = 7 - (1)/(2) (pK_(b) +logC)` <br/> `= 7-(1)/(2) (12 + log 0.1) = 1.5` <br/> `rArr [H^(o+)] = "Antilog" (-1.5)` <br/> `= "Antilog" [(-1-0.5 +1-1) = bar(2).5]` <br/> `= 3.2 xx 10^(-2)M` <br/> or <br/> `[H^(o+)] = 10^(-15) = (1)/(10sqrt(10)) = 3.2 xx 10^(-2)M` (Answer 'c') <br/> Please note that answer (c) of the above solution is incorrect. (Why ?) <br/> Actually, `pH = 7 - (1)/(2) (pK_(b) + logC)` is valid only when <br/> `1-h = 1`. <br/> To <a href="https://interviewquestions.tuteehub.com/tag/check-25817" style="font-weight:bold;" target="_blank" title="Click to know more about CHECK">CHECK</a>, claculate `h` using :<br/> `h = sqrt((K_(h))/(c)) = sqrt((K_(w))/(K_(b)C)) = sqrt((10^(-14))/(10^(-12)xx0.1)) = sqrt(0.1)` <br/> `rArr 1 - sqrt(0.1) != 1` <br/> So, we have to solve from basics as follows: <br/> `{:(B^(oplus),+,H_(2)O,hArr,BOH + ,H^(oplus)),(C-Ch,,,,Ch,Ch):}` <br/> `rArr K_(h) = ([BOH][H^(o+)])/([B^(o+)]) = ((Ch)(Ch))/(C-Ch)` <br/> `K_(h) = (K_(w))/(K_(a)) = (10^(-14))/(10^(-12)) = 10^(-2) = 0.01` <br/> Also, <br/> `K_(h) = (Ch^(2))/((1-h))` (Not neglecting `h` and solving, we get) <br/> `0.01 = (0.1 xxh^(2))/(1-h)` <br/> or `h~~ 0.271` <br/> `[H^(o+)] = Ch = 0.27 xx 0.1 = 0.0271 = 2.71 xx 10^(-2) M` <br/> Hence answer is (d). <br/> Note: In such problem, the answer cna be obtained <a href="https://interviewquestions.tuteehub.com/tag/directly-955045" style="font-weight:bold;" target="_blank" title="Click to know more about DIRECTLY">DIRECTLY</a>. The answer (c) `(3.2 xx 10^(-2)M)` is calculated by using the formula of salt of `W_(B)//S_(A)`. The concentration of salt is `0.1M` (which is very high). So `(1-h)` cannot be taken equal to `1`, since the two answer (c) and (d) are very close to each other. <a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>, `[H^(o+)]` would slightly less than the calculated value. <br/> `:. [H^(o+)]` calculated by using formula `= 3.2 xx 10^(-2)M` <br/> `[H^(o+)]` observed (by not using approximation) `= 2.7 xx 10^(-2)M` <br/> In such problem, if the `pH` is asked, calculated it by using formula will be slightly less than the observed `pH` (by not using approcimation) e.g., `pH` (calculated by using formula) `=- log (3.2 xx 10^(-2))` <br/> `=- log [(2)^(5) x 10^(-3)]` <br/> `=-5 log 2+3 = 1.5` <br/> `pH` (observed) (by not usingapproximation <br/> `=- log (27 xx 10^(-3))` <br/> `=- 3log 3+3 = 1.56`</body></html> | |
| 51298. |
25mL of a solution of ferric alum Fe_(2)(SO_(4))_(3) (NH_(4))_(2) (SO_(4))//24H_(2)O containing1.25g of the salt was boiled with iron when the reaction Fe + Fe_(2) (SO_(4))_(3) rarr 3FeSO_(4)occurredtreated with 0.107N KMnO_(4) in acid medium. What is titre value? If Cu has been used in place of Fe what would have been titre value? |
| 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> :`36.36mL, 24.24 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a>`</body></html> | |
| 51299. |
25mLof decimolar solution of a stable divalent transition metal cation is oxidised by 50mL of 0.02M acidified permanganate solution. What is the final oxidation state of the metal? |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a> moles of `M^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>+)` ion =x moles of permanganate , where x is the increase in oxidation number of the metal ion, `M^(2+)` <br/> Using the <a href="https://interviewquestions.tuteehub.com/tag/molarity-1100268" style="font-weight:bold;" target="_blank" title="Click to know more about MOLARITY">MOLARITY</a> <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a> for the <a href="https://interviewquestions.tuteehub.com/tag/titration-710389" style="font-weight:bold;" target="_blank" title="Click to know more about TITRATION">TITRATION</a> <br/> `(V_(1)M_(1))/(5)=(V_(2)M_(2))/(x)` <br/> The final oxidation state of metal `=+2-(-2)=+4`</body></html> | |
| 51300. |
25mL of a solutioncontainingFe^(2+) and Fe^(3+) sulphate acidifed with H_(2)SO_(4) is reducedby 3g of metallliczinc. The dsolution required 34.25mL of N//10 solution of K_(2)Cr_(2)O_(7) for oxidation. Beforereduction with zinc, 25mL of the same solution. Calculate the strengthof FeSO_(4) and Fe_(2) (SO_(4))_(3) in solution. |
| 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> :`FeSO_(4) = 13.64g//"<a href="https://interviewquestions.tuteehub.com/tag/litre-1075864" style="font-weight:bold;" target="_blank" title="Click to know more about LITRE">LITRE</a>'`, <br/> `Fe_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)(SO_(4))_(3) = 9.45g//"litre"`</body></html> | |