52671 + Interview Questions in Chemistry in Class 11 Page 1042 InterviewSolution

52051.	0.25 g of an organic compound gave 30 cm^(3) of moist dinitrogen at 288 K and 745 mm pressure. Calculate the percentage of nitrogen . (Aq tension of 288 K =12.7 mm)
Answer» <html><body><p></p>Solution :Mass of substance =0.25 g <br/> Volume of moist dinitrogen =`30cm^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)` <br/> Temperature =288 K <br/> Pressure= `745-12.7 =732.3 mm`<br/> <a href="https://interviewquestions.tuteehub.com/tag/step-25533" style="font-weight:bold;" target="_blank" title="Click to know more about STEP">STEP</a> I . To reduce the volume of `N_(2)` at S.T.P <br/> We <a href="https://interviewquestions.tuteehub.com/tag/know-534065" style="font-weight:bold;" target="_blank" title="Click to know more about KNOW">KNOW</a>, `(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>)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` <br/> `P_(2),T_(2),V_(2)` refer to S.T.P conditions)<br/> `V_(2)=(P_(1)V_(1))/(T_(1))xx (T_(2))/(P_(2))==(732.3xx30xx273)/(288xx760) =27.4 cm^(3)` <br/> Step II: Calculation of percentage of <a href="https://interviewquestions.tuteehub.com/tag/nitrogen-1118291" style="font-weight:bold;" target="_blank" title="Click to know more about NITROGEN">NITROGEN</a> <br/> `22400 cm^(3)` of dinitrogenat S.T.P weight =28 g <br/> `27.4 cm^(3) ` of dinitronge at S.T.P weight `(28xx274)/22400 =0.034 g ` <br/> Percentage of nitrogen in organic compound `=0.034/0.25 xx100=13.6`</body></html>

Discussion

52052.	0.2475 g of an organic substance gave on combustion 0.495 g of CO_(2) and 0.2025g of H_(2)O. Calculate the percentage of carbon and hydrogen in it.
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/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> = 54.54%, <a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a> = 9.09%</body></html>

Discussion

52053.	0.24g of a gas dissolves in 1 L of water at 1.5 atm pressure. Calculate the amount of dissolved gas when the pressure is raised to 6.0 atm at constant temperature.
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> :`P _("solute") = K_(H)x _("solute in solution")` <br/> At pressure `<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.5` atm, `P_(1) = K_(H) x _(1)""…(1)` <br/> At pressure `<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>.0` atm,` P _(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> _(H) x _(2)""…(2)` <br/> Dividing equation (1) by (2) <br/> We get `(p _(1))/( p _(2)) = (x _(1))/( x _(2))` <br/> `(1.5)/(6.0) = (0.24)/(x _(2))` <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 _(2) = (0.24 xx 6. 0)/(1.5) =0.96 g //L`</body></html>

Discussion

52054.	0.2475 g of an organic compound gave on combustion 0.4950 g of carbon dioxide and 0.2025 g of water. Calculate the percentage of carbon and hydrogen in it.
Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/wt-1462289" style="font-weight:bold;" target="_blank" title="Click to know more about WT">WT</a>. of organic <a href="https://interviewquestions.tuteehub.com/tag/compound-926863" style="font-weight:bold;" target="_blank" title="Click to know more about COMPOUND">COMPOUND</a> `= 0.2475` g <br/> Wt. of `CO_2` produced `= 0.4950` g <br/> Wt. Of `H_2 O` produced `= 0.20 25` g <br/> Percentage of carbon `= (<a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a>)/(44) xx ("Wt. Of" CO_2)/("Wt. <a href="https://interviewquestions.tuteehub.com/tag/ofcompound-2885260" style="font-weight:bold;" target="_blank" title="Click to know more about OFCOMPOUND">OFCOMPOUND</a>") xx 100` <br/> `= (12)/(44) xx (0.34950)/(0.2475) xx 100 = 54.54` <br/> Percentage of hydrogen `= (<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)/(18) xx ("Wt. of" H_2 O )/("Wt. of compound") xx 100` <br/> `= (2)/(18) xx (0.2025)/(0.2475) xx 100 = 9.09`</body></html>

Discussion

52055.	0.246 g of the organic compound gave 22.4 cm^(3) of nitrogen gas at STP as determined by Dumas method. The percent of nitrogen in the compound is
Answer» <html><body><p>11.38<br/>17.07<br/>22.76<br/>34.14</p>Answer :A</body></html>

Discussion

52056.	0.246 g of an organic substance when heated with excess of fuming nitric acid and silver nitrate gave 0.2584g of silver bromide residue. Calculate the weight percentage of bromine in the organic compound
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> :0.4467</body></html>

Discussion

52057.	0.24 gm of an organic substance gave 45.6 c.c. of nitrogen at N.T.P. The percentage of nitrogen is
Answer» <html><body><p>`63.6%`<br/>`23.8%`<br/>`53.6% `<br/>`53.6%` </p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/percentage-13406" style="font-weight:bold;" target="_blank" title="Click to know more about PERCENTAGE">PERCENTAGE</a> of `<a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>=(45.6 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 28)/(22400)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>)/(0.24) = 23.8 %`</body></html>

Discussion

52058.	0.24 g of organic compound containing phosphorous gave 0.66 g of Mg_(2)P_(2)0_(7) by the usual analysis. Calculate the percentage of phosphorous in the compound
Answer» <html><body><p></p>Solution :Weight of an <a href="https://interviewquestions.tuteehub.com/tag/organic-1138713" style="font-weight:bold;" target="_blank" title="Click to know more about ORGANIC">ORGANIC</a> compound = <a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>.24 <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/> Weight of `Mg_(2)P_(2)0_(<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>)`= 0.66 g <br/> 222 g of `Mg_(2)P_(2)0_(7)` contains= 62 g of P <br/> `"0.66 g contains"=(62)/(222) xx 0.66 "g of P"` <br/> `"<a href="https://interviewquestions.tuteehub.com/tag/percentage-13406" style="font-weight:bold;" target="_blank" title="Click to know more about PERCENTAGE">PERCENTAGE</a> of P"=(62)/(222) xx (0.66)/(0.24) xx 100=76.80%`</body></html>

Discussion

52059.	0.24 g of an organic compound on complete combustion gave 0.198 g of carbondioxide and 0.1014 g of water, then the percentage composition of carbon and hydrogen in the compound respectively
Answer» <html><body><p>4.58, 21.95<br/>21.95,4.58<br/>45.8,2.195<br/>2.195,45.8</p>Solution :% of `C = ("wtof" CO_(2))/(<a href="https://interviewquestions.tuteehub.com/tag/44-316683" style="font-weight:bold;" target="_blank" title="Click to know more about 44">44</a>) xx (100)/(w) xx <a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a>` <br/> % of `<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a> = ("wtof" H_(2)O)//(18) xx (100)/(w) xx2`</body></html>

Discussion

52060.	0.24 g of an organic compound gave 0.287 g of AgCl in the carius method. Calculate the percentage of chlorine in the compound.
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/cl-408888" style="font-weight:bold;" target="_blank" title="Click to know more about CL">CL</a> = <a href="https://interviewquestions.tuteehub.com/tag/25-296426" style="font-weight:bold;" target="_blank" title="Click to know more about 25">25</a>%`</body></html>

Discussion

52061.	0.234g of an organic compound on heating with conc nitric acid and silver nitrate in carius furnace gace 0.84g or siver chloride. Calculate the weight percentageof chlorine in the compound.
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> :80.8</body></html>

Discussion

52062.	0.2346 g of an organic compound containing C, H & 0, on combustion gives 0.2754 g of H_(2)0 and 0.4488 g CO_(2) . Calculate the % composition of C, H & 0 in the organic compound.
Answer» <html><body><p></p>Solution :<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 organic substance (<a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a>) = 0.2346 g <br/> Weight of water (x) = 0.2754 g<br/> Weight of `CO_(2)` (y) = 0.4488 g <br/> `"Percentage of <a href="https://interviewquestions.tuteehub.com/tag/carbon-16249" style="font-weight:bold;" target="_blank" title="Click to know more about CARBON">CARBON</a>" = (12)/(44) xx (y)/(w) 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> = (12)/(44) xx (0.4488)/(0.2346) xx 100 = 52.17%` <br/> `"Percentage of hydrogen" =(2)/(18) xx (x)/(w) xx 100=(2)/(18) xx (0.2754)/(0.2346) xx 100=13.04%` <br/> Percentage of oxygen =[100-(52.17+13.04)]=100-65.21=34.79%</body></html>

Discussion

52063.	0.2325 g oforganiccompound wasanalysedfornitrogenby Dumas method. 31.7mLofmoist nitrogenwascolletedat 25^(@) Cand 755.8 mm of Hgpressure, Calculatethepercentageof nitrogenin thesample(Aqueoustensionof waterat 25^@Cis 23.8mm of Hg)
Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/18-278913" style="font-weight:bold;" target="_blank" title="Click to know more about 18">18</a>%`<br/>`19.7%`<br/>`15.1%`<br/>`20.2%`</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>

Discussion

52064.	0.2325 g of an organic compound was analysed for nitrogen by Duma's method. 31.7 mL of moist nitrogen gas was collected at 25^(@)C and 755.8 mm Hg pressure. Determine the percentage of nitrogen in the compound. The aqueous tension of water is 23.8 mm Hg at 25^(@)C.
Answer» <html><body><p><br/></p>Solution :Step I. Volume of `N_(2)` at N.T.P. <br/> `{:("Experimental Conditions","N.T.P. Conditions"),(V_(1)=31.7 mL,V_(2)=?),(P_(1)=755.8-23.8=732 mm,P_(2)=760 mm),(T_(1)=25+273=<a href="https://interviewquestions.tuteehub.com/tag/298-1834947" style="font-weight:bold;" target="_blank" title="Click to know more about 298">298</a> K,T_(2)=273 K):}` <br/> `(P_(1)V_(1))/T_(1)=(P_(2)V_(2))/T_(2)` or `V_(2)=(P_(1)V_(1)T_(2))/(T_(1)P_(2))=((732mm)<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>(31.7 mL)xx(273 K))/((298 K)xx(760 mm))=27.97 mL` <br/> Step II. Percentage of <a href="https://interviewquestions.tuteehub.com/tag/nitrogen-1118291" style="font-weight:bold;" target="_blank" title="Click to know more about NITROGEN">NITROGEN</a> <br/> `=28/22400xx("Volume of "N_(2)" at N.T.P.")/("<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 <a href="https://interviewquestions.tuteehub.com/tag/compound-926863" style="font-weight:bold;" target="_blank" title="Click to know more about COMPOUND">COMPOUND</a>")xx100=28/22400xx27.97/0.2325xx100=15.1 %`</body></html>

Discussion

52065.	0.22g of a metal chloride required 0.51g of AgNO_(3) to precipitate chloride completely. If the specific heat of metal is 0.057 ca g^(-1), find the molecular formula of metal chloride?
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> :`MCl_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)`</body></html>

Discussion

52066.	0.222 g of iron ore was brought into solution, Fe^(3+) is reduced to Fe^(2+) with SnCl_(2). The reduced solution required 20 mL of 0.1 N KMnO_(4) solution. The percentage of iron present in the ore is (equivalent weight of iron is 55.5 ) :
Answer» <html><body><p>`55.5 %`<br/>`45.0%`<br/>`50.0 %`<br/>`40.0 %`</p>Solution :Number of <a href="https://interviewquestions.tuteehub.com/tag/equivalent-446407" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENT">EQUIVALENT</a> of <a href="https://interviewquestions.tuteehub.com/tag/iron-1051344" style="font-weight:bold;" target="_blank" title="Click to know more about IRON">IRON</a><br/> =Number of equivalents of `KMnO_(4)` <br/> `(<a href="https://interviewquestions.tuteehub.com/tag/nv-582326" style="font-weight:bold;" target="_blank" title="Click to know more about NV">NV</a>)/(1000)=(0.1xx20)/(1000)=0.002` <br/> Mass of iron (pure)`=0.002xx55.5=0.111 <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/> % purity `=("Mass of pure iron")/("Mass of iron ore")xx100=(0.111)/(0.222)xx100=50%`</body></html>

Discussion

52067.	0.22g of an organic compound on combustion in an atomosphere of carbon -dioxide gave 34 cm^(3) of moist nitrogen at 17^(0)C and 733.4mm pressure. It the aqueous tension at 17^(0)C is 13.4 mm, calculate the percentage of nitrogen in the compound
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> :0.1727</body></html>

Discussion

52068.	0.2174 g of the substance gave 0.5825 g of BaSO_(4) by Carius method. Calculate the percentage of sulphur.
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> :`36.78%`</body></html>

Discussion

52069.	0.2056 g of an organic compound gave on combustion 0.114 g of H_(2)O and 0.880 g of CO_(2_. Find the percentage of hydrogen and carbon in the organic compound.
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/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> = 93.76%, <a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a> = 6.128`</body></html>

Discussion

52070.	0.2046 g of an organic compound gave 30.4 mL of moist nitrogen measured at 288 K and 732.7 mm pressure. Calculate the percentage of nitrogen in the substance (Aqueous tension at 288 K is 12.7 mm).
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> :`16.68%`</body></html>

Discussion

52071.	0.2046 g of an organic compound gave 30.4 cm^(3) of moist nitrogen measured at 288 K and 732.7 mm pressure. Calculate the percentage of nitrogen in the compound (Aqueous tension at 288 K is 12.7 mm)
Answer» <html><body><p><br/></p>Solution :Step I. Volume of `N_(2)` at N.T.P. <br/> `{:("Experimental Conditions","N.T.P. Conditions"),(V_(1)=30.4 cm^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>),V_(2)=?),(P_(1)=732.7-12.7=720 mm,P_(2)=<a href="https://interviewquestions.tuteehub.com/tag/760-335611" style="font-weight:bold;" target="_blank" title="Click to know more about 760">760</a> mm),(T_(1)=<a href="https://interviewquestions.tuteehub.com/tag/288-1834218" style="font-weight:bold;" target="_blank" title="Click to know more about 288">288</a> K,T_(2)=273 K):}` <br/> `(P_(1)V_(1))/T_(1)=(P_(2)V_(2))/T_(2)` or `V_(2)=(P_(1)V_(1)T_(2))/(T_(1)P_(2))=((720mm)<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>(30.4 cm^(3))xx(273 K))/((288 K)xx(760 mm))=27.3 cm^(3)` <br/> Step II. Percentage of nitrogen <br/> `=28/22400xx("Volume of "N_(2)" at N.T.P.")/("Mass of <a href="https://interviewquestions.tuteehub.com/tag/compound-926863" style="font-weight:bold;" target="_blank" title="Click to know more about COMPOUND">COMPOUND</a>")xx100=28/22400xx27.3/0.2046xx100=16.68 %`</body></html>

Discussion

52072.	0.2033 g of an organic compound in Dumas method gave 31.7 mL of moist N_(2) at 14^(@)C and 758 mm pressure. Percentage of N_(2) in the compound is (Aq. Tension at 14^(@)C = 14mm)
Answer» <html><body><p>0.1844<br/>0.1689<br/>0.156<br/>0.16</p>Solution :`p_(N_(2)) = <a href="https://interviewquestions.tuteehub.com/tag/758-1920112" style="font-weight:bold;" target="_blank" title="Click to know more about 758">758</a> - 14` <br/> `("at" 14^(@)C) (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>)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` (at <a href="https://interviewquestions.tuteehub.com/tag/stp-633132" style="font-weight:bold;" target="_blank" title="Click to know more about STP">STP</a>) <br/> % N ` = ("Volume of" N_(2) "at STP")/(22,400) xx (100)/(<a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a>) xx 28`</body></html>

Discussion

52073.	0.200g of iodine is stirred in 100mL of water. After equilibrium is reached, we add 150mL of water to the system. How much iodine will be left undissolved?
Answer» <html><body><p>`1.3g`<br/> `0.130g`<br/> `0.013g`<br/> `13g`</p>Solution :After an equilibrium between `0.200g` of iodine and `100mL` of water is reached, 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 `<a href="https://interviewquestions.tuteehub.com/tag/150ml-1791979" style="font-weight:bold;" target="_blank" title="Click to know more about 150ML">150ML</a>` of water will result in the further dissolution of `I_2`. <br/> <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 solution after the addition of `150mL` of water <br/> `=100mL+150mL` <br/> `=<a href="https://interviewquestions.tuteehub.com/tag/250ml-297852" style="font-weight:bold;" target="_blank" title="Click to know more about 250ML">250ML</a>` <br/> According, `100mL` of solution contains `0.28g I_2`. <br/> `1mL` solution contains `(0.28g)/(<a href="https://interviewquestions.tuteehub.com/tag/1000ml-1771304" style="font-weight:bold;" target="_blank" title="Click to know more about 1000ML">1000ML</a>)` <br/> Therefore, `250mL` of solution will contain <br/> `(0.28g)/(1000ml)xx250mL=0.070g` of `I_2` <br/> Undissolved iodine `=(0.200g)-(0.070g)` <br/> `=0.130g`</body></html>

Discussion

52074.	0.20 g of an organic compound forms 0.5764 g of CO_(2) and 0.1512 g of water on combustion. Calculate the percentage of carbon and hydrogen in the compound.
Answer» <html><body><p></p>Solution :Mass of the organic <a href="https://interviewquestions.tuteehub.com/tag/compound-926863" style="font-weight:bold;" target="_blank" title="Click to know more about COMPOUND">COMPOUND</a> =0.20 g <br/> Mass of `CO_(2)` <a href="https://interviewquestions.tuteehub.com/tag/formed-464209" style="font-weight:bold;" target="_blank" title="Click to know more about FORMED">FORMED</a> = 0.5764 g <br/> Mass of `H_(2)O` formed = 0.1512 g <br/> Percentage of `C=12/44 xx("Mass of <a href="https://interviewquestions.tuteehub.com/tag/carbon-16249" style="font-weight:bold;" target="_blank" title="Click to know more about CARBON">CARBON</a> <a href="https://interviewquestions.tuteehub.com/tag/dioxide-954355" style="font-weight:bold;" target="_blank" title="Click to know more about DIOXIDE">DIOXIDE</a> formed")/("Mass of compound")xx100` <br/> `=12/44xx((0.5764 g))/((0.20 g))xx100=78.6 %` <br/> Percentage of `H=2/18xx("Mass of water formed")/("Mass of compound")xx100` <br/> `=2/18xx((0.1512 g))/((0.20 g))xx100=8.4 %`</body></html>

Discussion

52075.	0.2004 g of glucose gave on combustion 0.2940 g of CO_(2) and 0.1202 g of H_(2)O. Find the percentage composition.
Answer» <html><body><p></p>Solution :Weight of <a href="https://interviewquestions.tuteehub.com/tag/organic-1138713" style="font-weight:bold;" target="_blank" title="Click to know more about ORGANIC">ORGANIC</a> compound = 0.2004 g <br/> Weight of carbon dioxide = 0.2940 g <br/> Weight of water = 0.1202 g <br/> Percentage of carbon <br/> 44 g of <a href="https://interviewquestions.tuteehub.com/tag/co2-409421" style="font-weight:bold;" target="_blank" title="Click to know more about CO2">CO2</a> contains, carbon = 12 g <br/> `"0.2940 g of "CO_(2)" contains, carbon"=(12xx0.2940)/(44)xx(100)/(0.2004)=40.01` <br/> Percentage of <a href="https://interviewquestions.tuteehub.com/tag/hydrogen-22331" style="font-weight:bold;" target="_blank" title="Click to know more about HYDROGEN">HYDROGEN</a> <br/> <a href="https://interviewquestions.tuteehub.com/tag/18-278913" style="font-weight:bold;" target="_blank" title="Click to know more about 18">18</a> g of `H_(2)O` contains, hydrogen = 2 g <br/> 0.1202 of `H_(2)O` contains, hydrogen `=(2xx0.1202)/(18)` <br/> Percentage of hydrogen `=(2xx0.1202)/(18)xx(100)/(0.2004)=6.66` <br/> Percentage 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> is therefore by difference, <br/> `=[100-(40.01+6.66)]=53.33`</body></html>

Discussion

52076.	0.2 of an organic compound containing phosphorusgave 1.877 g of ammonium phosphomolybdate by usual analysis. Calculate the percentage of phosphorus in the organic compound.
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>></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>(1877g)((NH_(3))_(3)PO_(4). 12MoO_(3)) -= underset(31g)(P)` <br/> `:. % P = (<a href="https://interviewquestions.tuteehub.com/tag/31-306830" style="font-weight:bold;" target="_blank" title="Click to know more about 31">31</a>)/(1877) xx (1.877)/(0.2) = 15.5`</body></html>

Discussion

52077.	0.2 mole of an alkane on complete combustion gave 26.4g of CO_(2). The molecular weight of alkane is
Answer» <html><body><p>16<br/>30<br/>44<br/>58</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>

Discussion

52078.	0.2 gram -mole of an unsaturated hydrocarbon on complete combustion produces 26.4 g of CO_(2) . The molecular weight of the hydro-carbon is
Answer» <html><body><p>42<br/>88<br/>46<br/>30</p>Answer :A</body></html>

Discussion

52079.	0.2 g off an organic compound contains C, H and O. On combustion, it yields 0.15 g CO_(2) and 0.12" g "H_(2)O. The percentage of C, H and O respectively is
Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>=15%,<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>=20%,O=65%`<br/>`C=10%,H=8.2%,O=81.8%`<br/>`C=12.2%,H=8.8%,O=79%`<br/>`C=20%,H=6.66%,O=73.34%`</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> :% of `C=(12)/(44)xx(0.15)/(0.2)xx100=20%` <br/> % of `H=(2)/(18)xx(0.12)/(0.2)xx100=6.66%` <br/> % of `O=100-(20+6.66)=73.34%`</body></html>

Discussion

52080.	0.2g of an organic compound on analysis give 0.147g of carbondioxide, 0.12 g of water and 74.6 c,c of nitrogen at S.T.P. Calculate the weight percentages of constituents.
Answer» <html><body><p></p>Solution :Weight of <a href="https://interviewquestions.tuteehub.com/tag/compound-926863" style="font-weight:bold;" target="_blank" title="Click to know more about COMPOUND">COMPOUND</a> = w = 0.<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> h , volume of `N_(2)` at STP = `v_(2)` = 74.6 cc <br/> `%C = (w_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)xx12xx100)/(wxx44) = (0.147xx12xx100)/(0.2xx44) = 20.04%` <br/> `%H = (w_(2)xx2xx100)/(wxx18) = (0.12xx2xx100)/(0.2xx18) = 6.66%` <br/> `%N = (v_(2)xx25 xx100)/(wxx22400) = (0.74.6)/(8xx0.2) = 46.63%` <br/> Weight percentage of oxygen is obtained <a href="https://interviewquestions.tuteehub.com/tag/indirectly-7311994" style="font-weight:bold;" target="_blank" title="Click to know more about INDIRECTLY">INDIRECTLY</a> <br/> `%O = <a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a> = (%C+%H+%N) = 100-73.33 = 26.67%`</body></html>

Discussion

52081.	0.2 g of an organic compound on complete combustion produces 0.18 g of water. The precentage of hydrogen in it is:
Answer» <html><body><p>5<br/>10<br/>15<br/>20</p>Solution :Percentage of `<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>` <br/> `=2/18xx("<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 "H_(2)O)/("Mass of <a href="https://interviewquestions.tuteehub.com/tag/compound-926863" style="font-weight:bold;" target="_blank" title="Click to know more about COMPOUND">COMPOUND</a>")xx100` <br/> `=2/18xx0.18/0.20xx100=10`</body></html>

Discussion

52082.	0.2 g ofa gas 'x' occupy a volume of 440 ml . If 0.1 g of CO_2 gas occupy a volume of 320 ml at the same temperature and pressure , then the gas 'x' could be
Answer» <html><body><p>`C_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>) H_(10)`<br/>`NO_(2)`<br/>`O_(2)`<br/>`SO_(2)`</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>

Discussion

52083.	0.1M CH_(3)COOH solution is titrated against 0.05M NaOH solution. Calculate pH at 1//4th and 3//4th stages of neutralization of acid. The pH for 0.1M CH_(3)COOH is 3.
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> :`4.5228, 5.4771;`</body></html>

Discussion

52084.	In an estimation of sulphur by Carius method, 0.2175 g of the substance gave 0.5825 g of BaSO_(4) , calculate the percentage composition of S in the compound.
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/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> = 30.6%, <a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a> = 3.8%, <a href="https://interviewquestions.tuteehub.com/tag/cl-408888" style="font-weight:bold;" target="_blank" title="Click to know more about CL">CL</a> = 45.2%, O = 20.4%`</body></html>

Discussion

52085.	0.196 g of an organic compound gave 0.22 g of CO_(2) and 0.0675 g of H_(2)O. In carius determination, 0.3925 g of the substance gave 0.717 g of dry AgCl. Find the percentage composition of the substance.
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/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> = 30.6%, <a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a> = 3.8%, <a href="https://interviewquestions.tuteehub.com/tag/cl-408888" style="font-weight:bold;" target="_blank" title="Click to know more about CL">CL</a> = 45.2%, O = 20.4%`</body></html>

Discussion

52086.	0.189g of an organic substance gave in a Carius determination 0.287 g of silver chloride. What is the percentage of chlorine in the given compound ?
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> :`37.57%`</body></html>

Discussion

52087.	0.1890 g of organic compound containing chlorine gave in carius method 0.2870 g of silver chloride. Find the percentage of chlorine in the compound.
Answer» <html><body><p><br/></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/percentage-13406" style="font-weight:bold;" target="_blank" title="Click to know more about PERCENTAGE">PERCENTAGE</a> of bromine `=35.5/143.5xx ("<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 <a href="https://interviewquestions.tuteehub.com/tag/agcl-368919" style="font-weight:bold;" target="_blank" title="Click to know more about AGCL">AGCL</a>")/("Mass of compound")xx100=35.5/143.5xx0.2870/0.1890xx100=37.56%`</body></html>

Discussion

52088.	0.185 g of an organic compound when treated with Conc. HNO_(3) and silver nitrate gave 0.320g of silver bromide. Calculate the % of bromine in the compound. (Ag=108, Br=80).
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> :Weight of organic substance (w) = 0.185 g , Weight of silver bromide (<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>) = 0.320 g <br/> `"<a href="https://interviewquestions.tuteehub.com/tag/percentage-13406" style="font-weight:bold;" target="_blank" title="Click to know more about PERCENTAGE">PERCENTAGE</a> of bromine"=(80)/(<a href="https://interviewquestions.tuteehub.com/tag/188-280322" style="font-weight:bold;" target="_blank" title="Click to know more about 188">188</a>) xx (x)/(w) 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>=(80)/(188) xx (0.32)/(0.185) xx 100=73.6%`</body></html>

Discussion

52089.	0.16g of methane was subjected to combustion at 27^(@)C in a bomb calorimeter system. The temperature of the calorimeter system ( including water ) was found to rise by 0.5^(@)C. Calculate the heat of combustion of methane at (i) constant volume, and (ii) constant pressure . The thermal capacity of the calorimeter system is17.7 k J K^(-1) ( R = 8.314 kJ K^(-1) mol^(-1))
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>><br/></p>Answer :`q_(v) = - 885 kJ mol^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>) , q_(p) = -<a href="https://interviewquestions.tuteehub.com/tag/890-1929934" style="font-weight:bold;" target="_blank" title="Click to know more about 890">890</a> kJ mol^(_1)`</body></html>

Discussion

52090.	0.16g of an organic substance was heated in Carius tube and the sulphuric acid formed was precipitated as BaSO_(4) with BaCl_(2). The weight of the dry BaSO_(4) was 0.35 g. Find the percentage of sulphur.
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> :`30.04%`</body></html>

Discussion

52091.	0.1688 g when analysed by the Dumas method yield 31.7 mL of moist nitrogen measured at 14^(@)C and 758 mm mercury pressure. Determine the % of N in the substance (Aqueous tension at 14^(@)C =12 mm of Hg).
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> :Weight of Organic compound =0.168 g <br/> `"Volume of moist nitrogen" (V_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>))=31.7 ml=31.7 xx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>) L` <br/> `"Temperature" (T_(1)) = 14^(@)C = 14 + 273 = 287 K` <br/> Pressure of Moist nitrogen (P) =758 mm Hg <br/> Aqueous tension at `14^(@)C` = 12 mm of Hg <br/> `:.` Pressure of <a href="https://interviewquestions.tuteehub.com/tag/dry-960094" style="font-weight:bold;" target="_blank" title="Click to know more about DRY">DRY</a> nitrogen=(`P-P^(1)`)=758-12=746 mm of Hg <br/>`(P_(1)V_(1))/(T_(1))=(P_(0)V_(0))/(T_(0))` <br/> `V_(0)=(746 xx 31.7 xx 10^(-3))/(287) xx (273)/(<a href="https://interviewquestions.tuteehub.com/tag/760-335611" style="font-weight:bold;" target="_blank" title="Click to know more about 760">760</a>)` <br/> `V_(0)=29.8 xx 10^(-3) L` <br/> `"Percentage of nitrogen"=((28)/(22.4) xx (V_(0))/(W)) xx 100` <br/> `=(28)/(22.4) xx (29.58 xx 10^(-3))/(0.1688) xx 100=21.90%`</body></html>

Discussion

52092.	0.16 g of N_(2)H_(4) are dissolved in water and the total volume made upto 500 ml. Calculate the percentageof N_(2)H_(4) that has reacted with water in this solution . The K_(b) for N_(2)H_(4) is 4.0xx10^(-6)M.
Answer» <html><body><p></p>Solution :`N_(2)H_(4)+H_(2)O hArr N_(2)H_(5)^(+) + OH^(-)`<br/> Conc. Of `N_(2)H_(4) = 0.16 g ` in500 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> = 0.<a href="https://interviewquestions.tuteehub.com/tag/32-307188" style="font-weight:bold;" target="_blank" title="Click to know more about 32">32</a> g `L^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>) = (0.32)/(32) ` mol `L^(-1) = 0.01 M = 10^(-2) M` <br/> Suppose x mol `L^(-1)` is the amount of hydrazine reacted. Then <br/> `{:(,N_(2)H_(4) ,+ ,H_(2)O,hArr,N_(2)H_(5)^(+),+,OH^(-),,),("Initial conc.",10^(-2)M,,,,,,,,),("Conc. after reaction",10^(-2) - x ,,x,,x,,,,):}` <br/> `K_(b) = ([N_(2)H_(5)^(+)][OH^(-)])/([N_(2)H_(4)])` <br/> `4.0xx10^(-6) = (x^(2))/(10^(2)-x) ~= (x^(2))/(10^(2)) or x^(2)=4.0xx10^(-8) or x=2.0xx10^(-4)` mol `L^(-1)` <br/> `:.` % of hydrazine reacted with water `=(2.0xx10^(-4))/(10^(-2))xx100=2%`</body></html>

Discussion

52093.	0.16 g of an organic compound was heated in a carius tube and H_(2)SO_(4) acid formed was precipitated with BaCl_(2). The mass of BaSO_(4) was 0.35 g. Find the percentage of sulphur.
Answer» <html><body><p></p>Solution :Weight of organic <a href="https://interviewquestions.tuteehub.com/tag/substance-1231528" style="font-weight:bold;" target="_blank" title="Click to know more about SUBSTANCE">SUBSTANCE</a> (w) = 0.16 <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/> Weight of <a href="https://interviewquestions.tuteehub.com/tag/barium-393995" style="font-weight:bold;" target="_blank" title="Click to know more about BARIUM">BARIUM</a> <a href="https://interviewquestions.tuteehub.com/tag/sulphate-1234306" style="font-weight:bold;" target="_blank" title="Click to know more about SULPHATE">SULPHATE</a> (x) = 0.35 g <br/> `"Percentage of Sulphur"=(<a href="https://interviewquestions.tuteehub.com/tag/32-307188" style="font-weight:bold;" target="_blank" title="Click to know more about 32">32</a>)/(233) xx (x)/(w) xx 100=(32)/(233) xx (0.35)/(0.16) xx 100=30.04%`</body></html>

Discussion

52094.	0.16 g a dibasic acid required 25 mL of decinormal NaOH solution for complete neutralisation. The molecular mass of the acid is :
Answer» <html><body><p>32<br/>64<br/>128<br/>256</p>Solution :N//A</body></html>

Discussion

52095.	0.156g of an organic compound in carius tube method gave 0.235g of silver iodide. Calculate the percent weight of halogen in the compound .
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> :81.4</body></html>

Discussion

52096.	0.157g of a certain gas collected over water occupies a volume of 135ml at 27°C and 750mm of Hg. Assuming ideal behaviour, the molecular weight of the gas is (aqueous tension at 27°C is 26.7 mm of Hg)
Answer» <html><body><p>30<br/>32<br/>28<br/>16</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>

Discussion

52097.	0.1570 g of an organic compound was heated with fuming nitric acid in carius tube. Addition of excess of barium chloride yielded 0.4813 g of barium sulphate as a white precipitate. Determine the percentage of sulphur in the compound.
Answer» <html><body><p><br/></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/percentage-13406" style="font-weight:bold;" target="_blank" title="Click to know more about PERCENTAGE">PERCENTAGE</a> of <a href="https://interviewquestions.tuteehub.com/tag/sulphur-1234343" style="font-weight:bold;" target="_blank" title="Click to know more about SULPHUR">SULPHUR</a> `=32/233xx("<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 "BaSO_(4))/("Mass of <a href="https://interviewquestions.tuteehub.com/tag/compound-926863" style="font-weight:bold;" target="_blank" title="Click to know more about COMPOUND">COMPOUND</a>")xx100=32/233xx0.4813/0.1570xx100=42.1 %`</body></html>

Discussion

52098.	0.15 mole of pyridinium chloride has been added into 500 cm^(3) of 0.2 M pyridine solution. Calculate pH of the resulting solution, assuming no change in volume. (K, for pyridine = 1.5 xx 10^(-9)M)
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> :Pyridiniumchloride = 0.15 mole <br/>Pyridine <a href="https://interviewquestions.tuteehub.com/tag/500-323288" style="font-weight:bold;" target="_blank" title="Click to know more about 500">500</a> ` cm ^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)"of "0.2 M = 0.1 ` mole<br/>` pOH = pK_b +log "" ([s]) /(["Base" ]) ` <br/>` =- log (1.5 xx 10 ^(-<a href="https://interviewquestions.tuteehub.com/tag/9-340408" style="font-weight:bold;" target="_blank" title="Click to know more about 9">9</a>) ) +log"" (0.15)/(0.1) ` <br/> ` =9- log 1. 5+ log 1.5 =9` <br/> ` <a href="https://interviewquestions.tuteehub.com/tag/ph-1145128" style="font-weight:bold;" target="_blank" title="Click to know more about PH">PH</a> =14- 9 = 5`</body></html>

Discussion

52099.	0.15 mole of pyridinium chloride has been added into 500 cm^(3) of 0.2 M pyridine solution . Calculate pH and hydroxyl ion concentration in the resulting solution assuming no change in the volume (k_(b) forpyridine = 1.5xx10^(-9)).
Answer» <html><body><p></p>Solution :Pyridine is a weak base. Thus, pyridine + pyridine chloride solution is a basic <a href="https://interviewquestions.tuteehub.com/tag/buffer-905159" style="font-weight:bold;" target="_blank" title="Click to know more about BUFFER">BUFFER</a>. Hence, <br/> `pOH = pK_(b) + log .(["Salt"])/(["Base"])` <br/> `pK_(b) = - log K_(b) = - log (1.5xx10^(-9))=9-0.1761=8.8239` <br/> [Pyridine ] = 0.23 M (Given ), [ Pyridinium chloride ] = `(0.15)/(500) xx <a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a> = 0.30 M` <br/> `:. pOH = 8.82+ log. (0.30)/(0.20) = 8.82 + 0.1761 = 8.896` <br/> i.e., `- log [OH^(-)]= 8.896or log [OH^(-)] = - 8.896 = <a href="https://interviewquestions.tuteehub.com/tag/bar-892478" style="font-weight:bold;" target="_blank" title="Click to know more about BAR">BAR</a>(9) . 104 or [OH^(-) ] = 1.271xx10^(-9)` <br/> `[OH^(-)] ` from `H_(2)O = 10^(-<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>)` M cannot be neglected. <br/> Hence, total `[OH^(-)]=1.27xx10^(-9) + 10^(-7)= 10^(-9) (1.27+100) = 111.27 xx 10^(-9) M = 1.1127 xx 10^(-7) M` <br/> `[H^(+)] = (K_(w))/([OH^(-)])=(10^(-14))/(1.1127xx10^(-7))=8.987 xx 10^(-8)M` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/ph-1145128" style="font-weight:bold;" target="_blank" title="Click to know more about PH">PH</a> = - log [H^(+)] = - log (8.987 xx 10^(-8))=8-0.9536= 7.0464`.</body></html>

Discussion

52100.	015mole of CO taken in 25 l flask is maintained at 750 K along with catalyst so that the following reaction can take place : CO (g) + 2 H_(2) (g) hArr CH_(3)OH (g) Hydrogen is introduced until the total pressure of the system is 85 atmosphere at equilibrium and 008 mole of methanol is formed . Calculate (i) K_(p) and K_(c) and (ii) the final pressure if the same amount of CO and H_(2) as before are used but with no catalyst so that the reaction does not take place.
Answer» <html><body><p></p>Solution :`{:((i),CO,+,2H_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>),hArr,CH_(3)<a href="https://interviewquestions.tuteehub.com/tag/oh-585115" style="font-weight:bold;" target="_blank" title="Click to know more about OH">OH</a>) ,("Intialy :",015 "mole",,,,) ,("At eqm:",015 - 008 "mole",,,,), (,=0017 "mole",,,,):}` <br/> Total volume, `V= 25 L`<br/> Total <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a>` P+ 85 "atm", T= 750 "K" `<br/>`" <a href="https://interviewquestions.tuteehub.com/tag/applying-1982651" style="font-weight:bold;" target="_blank" title="Click to know more about APPLYING">APPLYING</a> "PV= nRT,`<br/>we get` 85 xx25 = n xx0* 0821 * 750or n = 0* 345 "mole" `<br/> ` :."No. of moles of " H_(2) " at equilibrium "= 0* 345 - ( 0* 017 + 008 ) = 0 248 " mol "`<br/>` P_(CO)= (0* 017 )/(0345 ) xx 85 "atm"= 0* 42 "atm" ` <br/>` p_(H_(2)) = ( 0248 )/(0 345) xx85 "atm " = 6 11 "atm "`<br/> ` P _(CH_(3)OH) = (08)/(0345) xx 85 "atm"= 1 97 "atm" `<br/> ` K_(p) = (P_(CH_(3)OH))/(P_(CO) xx P_(H_(2)))= (197 )/(042 xx (6* 11)^(2) )= 0* 1256 `<br/> ` K_(c)= ([CH_(3) OH])/([CO][H_(2)]^(2))=(0* 08 //25)/((0 017 //25 )(0248 //25)^(2)) = 478 2` n(ii) No. of moles of `H_(2) " taken intially " = 0* 248 + 2 xx 0* 08 = 0* 308 `<br/> No. of moles of CO taken intially `= 015`<br/> ApplyingPV = nRt , ` P xx 25 = 0* 458 xx 0* 0821 xx 750or P = 11* 28 "atm " `.</body></html>

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Explore topic-wise InterviewSolutions in Class 11.

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