Explore topic-wise InterviewSolutions in Class 11.

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.

52001.

0.37 gm of alcohol when treated with CH_3MgI liberates 112 ml of CH_4 at STP. The molecularformula of alcohol is

Answer» <html><body><p>`C_4H_9OH`<br/>`C_5H_(<a href="https://interviewquestions.tuteehub.com/tag/11-267621" style="font-weight:bold;" target="_blank" title="Click to know more about 11">11</a>)<a href="https://interviewquestions.tuteehub.com/tag/oh-585115" style="font-weight:bold;" target="_blank" title="Click to know more about OH">OH</a>`<br/>`C_3H_7OH`<br/>`C_2H_5OH` </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> :Molecilar <a href="https://interviewquestions.tuteehub.com/tag/formula-464310" style="font-weight:bold;" target="_blank" title="Click to know more about FORMULA">FORMULA</a></body></html>
Discussion
52002.

0.376g of aluminium reacted with an acid to displace 0·468 litre of hydrogen at NTP. Find the equivalent volume of hydrogen if the equivalent weight of Al is 9.

Answer» <html><body><p></p>Solution :<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 Al = equivalent 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/> `therefore ("weight of Al")/("<a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a>. weight of Al")=("volume of hydrogen at NTP")/("equivalent volume of hydrogen")` <br/> `(0.376)/(<a href="https://interviewquestions.tuteehub.com/tag/9-340408" style="font-weight:bold;" target="_blank" title="Click to know more about 9">9</a>) = (0.468)/("V(litre)")` <br/> `therefore"" V=11.2` <a href="https://interviewquestions.tuteehub.com/tag/litres-1075876" style="font-weight:bold;" target="_blank" title="Click to know more about LITRES">LITRES</a>.</body></html>
Discussion
52003.

0.36g of an organic compound on combustion gave 1.1g of CO_(2) and 0.54g of H_(2)O. The percentages of carbon and Hydrogen in the compound are

Answer» <html><body><p>75,25<br/>60,40<br/>83.33,16.67<br/>77.8,22.2</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
52004.

0.365 g of HCl gas was passed through 100 cm^(3) of 0.2 MNaOH solution. Th e pH of the resulting solution would be

Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a><br/>5<br/>8<br/><a href="https://interviewquestions.tuteehub.com/tag/13-271882" style="font-weight:bold;" target="_blank" title="Click to know more about 13">13</a></p>Solution :`0.365 g <a href="https://interviewquestions.tuteehub.com/tag/hcl-479502" style="font-weight:bold;" target="_blank" title="Click to know more about HCL">HCL</a> =(0.365)/(36.5) "mole" = 0.01 "mole"` <br/> `100 cm^(3) "of" 0.2 M NaOH=(0.2)/(1000)xx100` <br/> `=0.02 ` mole <br/> NaOH left unneutralized = 0.01 mole <br/> Volume of solution = 100 ml <br/> `:.` Molarity of NaOH in the solution<br/> `=(0.01)/(100)xx1000=0.1M = 10^(-1)M` <br/> `:. [<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>^(+)]=(10^(-14))/(10^(-1)M)=10^(-13)M :. pH=13`</body></html>
Discussion
52005.

0.35 g of an organic substance was Kjeldahlised and the ammonia obtained was passed into 100 ml of N/5 H_(2)SO_(4). The excess acid required 154 ml of N/10 NaOH for neutralisation. Calculate the percentage of nitrogen in the compound.

Answer» <html><body><p></p>Solution :Volume of `N//<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a> H_(2)SO_(4)` solution taken = 100 mL <br/> The volume of `N//5 H_(2)SO_(4)` neutralised by N/10 <br/> <a href="https://interviewquestions.tuteehub.com/tag/naoh-572531" style="font-weight:bold;" target="_blank" title="Click to know more about NAOH">NAOH</a> can be obtained as follows : <br/> `N_("acid")xxV_("acid")=N_("<a href="https://interviewquestions.tuteehub.com/tag/alkali-853286" style="font-weight:bold;" target="_blank" title="Click to know more about ALKALI">ALKALI</a>")xxV_("alkali")` <br/> `(N)/(5)xx V_("acid")=(N)/(10)xx154mL` <br/> `V_("acid")=(154)/(10)xxmL` <br/> = 77 <a href="https://interviewquestions.tuteehub.com/tag/nl-571159" style="font-weight:bold;" target="_blank" title="Click to know more about NL">NL</a>. <br/> Therefore, <br/> Volume of `N//% H_(2)SO_(4)` used for <br/> neutralisting ammonia `=(100-77)mL` <br/> `=23mL` <br/> Then, <br/> `"Percentage of nitrogen in the sample "="1.4"xx"Normality volume of acid"` <br/> `("used for neutralising "NH_(3))/("Mass of the compound taken")` <br/> Percentage of nitrogen in the sample <br/> `=(1.4xx1//5xx23)/(0.35)=18.4`</body></html>
Discussion
52006.

0.35 g of an organic acid was kjeldahlised and the ammonia obtained was passed into 100 cm^(3) of N//5H_(2)SO_(4). The excess of the acid required 154 cm^(3) of N//10 NaOH for neutralisation. Calculate the percentage of nitrogen in the compound.

Answer» <html><body><p><br/></p>Solution :Volume of unused <a href="https://interviewquestions.tuteehub.com/tag/acid-847491" style="font-weight:bold;" target="_blank" title="Click to know more about ACID">ACID</a>. <br/> `ubrace(N_(1)V_(1))_("Acid") <a href="https://interviewquestions.tuteehub.com/tag/equiv-2618698" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIV">EQUIV</a> ubrace(N_(2)V_(2))_("Base")` <br/> `1/5xxV=1/10xx154` or `V=77 cm^(3)`. <br/> Step II. <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/nitrogen-1118291" style="font-weight:bold;" target="_blank" title="Click to know more about NITROGEN">NITROGEN</a>. <br/> Volume of acid used `=100-77=23 cm^(3)` <br/> Percentage 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>=(1.4xx"Normality of acid used"xx"Volume of acid used")/("Mass of compound")=(1.4xx0.2xx23)/035=18.4 %`</body></html>
Discussion
52007.

0.35 g of a metal was dissolved in 50 mL N-acid. The whole solution then required 20.85 mL of normal alkaline solution to neutralise the excess of the acid. Calculate the equivalent mass of the metal.

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>
Discussion
52008.

0.33g of an organic compound containing phosphorous gave 0.397 g of Mg_(2)P_(2)0_(7) by the analysis. Calculate the percentage of P 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/weight-1451304" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHT">WEIGHT</a> of organic compound = 0.33g , 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>)`= 0.397g <br/> 222 g of `Mg_(2)P_(2)0_(7)` contains 62 g of phosphorous. <br/>`:. 0.397 "g of" Mg_(2)P_(2)0_(7) "will contain" (62)/(222) xx 0.397 "g of P"`. <br/> `0.33 "g of organic compound contains" (62)/(222) xx 0.397 "g of P"` <br/> `:. <a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a> "g of organic compound will contain" (62)/(222) xx (0.397)/(0.33) xx 100=(2,461.4)/(73.26)=33.59%` <br/> Percentage of phosphorous=33.59%</body></html>
Discussion
52009.

0.32g of metal gave on treatment with an acid 112 mL of hydrogen at STP. Calculate the equivalent weight of the metal

Answer» <html><body><p>58<br/>32<br/>11.2<br/>24</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/equivalent-446407" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENT">EQUIVALENT</a> <a href="https://interviewquestions.tuteehub.com/tag/weight-1451304" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHT">WEIGHT</a> `=("Mass of <a href="https://interviewquestions.tuteehub.com/tag/metal-1094457" style="font-weight:bold;" target="_blank" title="Click to know more about METAL">METAL</a>" xx 11200)/("Volume in mL 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/> Given, mass of metal `= 0.32g` <br/> volume of hydrogen at NTP `= 112` mL<br/> Equivalent weight`=(0.32xx11200)/(112)=32`</body></html>
Discussion
52010.

0.33 g of an organic compound was analysed for nitrogen by Kjeldhal's method. The ammonia gas evolved was absorbed in 50 ml of 0.05 M H_(2)SO_(4).The excess acid required 25 ml of 0.1 M NaOH for neutralization. The Percentage of nitrogen in the given compound is

Answer» <html><body><p>11.5<br/>23<br/>12. 5 <br/>14.5</p>Solution :<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 `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>)SO_(4)` used = 25 ml <br/> <a href="https://interviewquestions.tuteehub.com/tag/normality-1124423" style="font-weight:bold;" target="_blank" title="Click to know more about NORMALITY">NORMALITY</a> of `H_(2)SO_(4) = 0.1` N</body></html>
Discussion
52011.

0.33 mol cholesterol gives 9 mole CO_(2) on combustion. It was observed that cholesterol contains 83.85%C, 12%H and 4.15%O. Find its molecular formula and molecular mass.

Answer» <html><body><p><br/></p>Answer :`C_(27)H_(<a href="https://interviewquestions.tuteehub.com/tag/46-317673" style="font-weight:bold;" target="_blank" title="Click to know more about 46">46</a>)O; <a href="https://interviewquestions.tuteehub.com/tag/386-1860926" style="font-weight:bold;" target="_blank" title="Click to know more about 386">386</a>`</body></html>
Discussion
52012.

0.32g of an organic compound containing sulphur produces 0.233g of BaSO_(4). Percentage of sulphur in the compound is

Answer» <html><body><p>20<br/>50<br/>10<br/>80</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
52013.

0.324 g of copper was dissolved in nitric acid and the copper nitrate so produced was burnt till all copper nitrate converted to 0.406 g of copper oxide. Calculate the equivalent weight of copper.

Answer» <html><body><p></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>(0.324 "g")(<a href="https://interviewquestions.tuteehub.com/tag/cu-428205" style="font-weight:bold;" target="_blank" title="Click to know more about CU">CU</a>+ HNO_3) to "copper nitrate" tounderset(0.406 "g") "copper oxide"` <br/> Suppose the <a href="https://interviewquestions.tuteehub.com/tag/equivalent-446407" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENT">EQUIVALENT</a> weight of copper `(Cu)` is `E_(Cu)`. <br/> Equivalent of Cu `=` equivalent of copper oxide <br/> `(0.324)/(E_(Cu) )= (0.406)/("eq. wt. of copper oxide")`<br/> `(0.324)/( E_(Cu) ) = ( 0.406)/("eq. wt. of Cu + eq. wt. of O")= ( 0.406)/( E_(Cu ) +8)` <br/> Hence the equivalent weight of Cu is `31.60`.</body></html>
Discussion
52014.

0.32 g of metal on treatment with acid gave 112mL of hydrogen at NTP. The equivalent weight of metal is

Answer» <html><body><p>58<br/><a href="https://interviewquestions.tuteehub.com/tag/32-307188" style="font-weight:bold;" target="_blank" title="Click to know more about 32">32</a><br/>11.2<br/>24</p>Solution :0.32 gm <a href="https://interviewquestions.tuteehub.com/tag/metal-1094457" style="font-weight:bold;" target="_blank" title="Click to know more about METAL">METAL</a> `<a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a>` <a href="https://interviewquestions.tuteehub.com/tag/112-268456" style="font-weight:bold;" target="_blank" title="Click to know more about 112">112</a> ml of hydrogen <br/> ? `"" rarr` 11,200 ml of hydrogen <br/> = 32 gm</body></html>
Discussion
52015.

0.3160 g of an organic substance was heated with fuming nitric acid in a carius tube to convert phosphorus present into phosphoric acid. Addition of magnesia mixture formed MgNH_(4)PO_(4) which upon heating gave 0.1697 g Mg_(2)P_(2)O_(7). Calculate the percentage of phosphorus in the compound.

Answer» <html><body><p>&lt;<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>&gt;<br/></p>Solution :Percentage of <a href="https://interviewquestions.tuteehub.com/tag/phosphorus-1153299" style="font-weight:bold;" target="_blank" title="Click to know more about PHOSPHORUS">PHOSPHORUS</a> `=62/222xx("Mass of "Mg_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)P_(2)O_(<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>))/("Mass of compound")xx100=62/222xx0.1697/0.3160xx100=15 %`</body></html>
Discussion
52016.

0.316 g of an organic compound gives 0.466 g of barium sulphate by carius method. Calculate the percentage of sulphur?

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> :`S = 20.25%`</body></html>
Discussion
52017.

0.316 g of an organic compound, after heating with fuming nitric acid and barium nitrate crystals in a sealed tube gave 0.466 g of the precipitate of barium sulphate. Determine the percentage of sulphur in the compound. (Atomic masses : Ba = 137, S = 32, O = 16, C = 12, H = 1).

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 the substance taken = 0.316 g <br/> Mass of `BaSO_(4)` <a href="https://interviewquestions.tuteehub.com/tag/formed-464209" style="font-weight:bold;" target="_blank" title="Click to know more about FORMED">FORMED</a> = 0.466 g <br/> Molecular mass of `BaSO_(4) = <a href="https://interviewquestions.tuteehub.com/tag/137-272694" style="font-weight:bold;" target="_blank" title="Click to know more about 137">137</a> + <a href="https://interviewquestions.tuteehub.com/tag/32-307188" style="font-weight:bold;" target="_blank" title="Click to know more about 32">32</a> + <a href="https://interviewquestions.tuteehub.com/tag/64-330663" style="font-weight:bold;" target="_blank" title="Click to know more about 64">64</a> = 233` <br/> Then, mass of S in 0.466 g of `BaSO_(4)` <br/> `=(0.466xx32)/(233)g` <br/> Percentage of S in compound <br/> `=(0.466xx32xx100)/(233xx0.316)` <br/> = 20.25</body></html>
Discussion
52018.

0.303 g of an organic compound was analysed for nitrogen by Kjeldahl's method. The ammonia evovled was absorbed in 50 ml. of 0.1N H_(2)SO_(4). The excess acid required 25ml of 0.1N NaOH for neutralisation. Calculate the percentage of nitrogen present 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> :0.1155</body></html>
Discussion
52019.

0.303 g of an organic compound was analysed for nitrogen by Kjedahl's method. The ammonia gave 0.2584 g of silver bromide 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> :0.1155</body></html>
Discussion
52020.

0.303 g fo sample was analysed for nitrogen by Kjeldahl's method. The ammonia gas evolved was absorbed in 50 ml of 0.05 M H_(2)SO_(4). The excess acid required 25 ml of 0.1 M NaOH for neutralisation.

Answer» <html><body><p>`11.5`<br/>23<br/>`12.5`<br/>`14.5`</p>Solution :No. of milli eq of `H_(2)SO_(4)` <a href="https://interviewquestions.tuteehub.com/tag/left-1070879" style="font-weight:bold;" target="_blank" title="Click to know more about LEFT">LEFT</a> after <a href="https://interviewquestions.tuteehub.com/tag/neutralisation-577399" style="font-weight:bold;" target="_blank" title="Click to know more about NEUTRALISATION">NEUTRALISATION</a>.<br/>= No. of m. eq of NaOH<br/>`= 25xx0.1xx1=2.5` m eq.<br/>No. of m eq of `H_(2)SO_(4)` taken<br/>`= 50xx0.05xx2=5m. eq.`<br/>No. of eq of `H_(2)SO_(4)` ultilised for neutralisations<br/>`= 5-2.5=2.5` m.eq.<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>` No. of m. eq. of `NH_(3)` <a href="https://interviewquestions.tuteehub.com/tag/liberated-2789268" style="font-weight:bold;" target="_blank" title="Click to know more about LIBERATED">LIBERATED</a> = 2.5 m.q.<br/>% of `N=("No. of meq. of "NH_(3)xx1.4)/("wt.of org. <a href="https://interviewquestions.tuteehub.com/tag/compound-926863" style="font-weight:bold;" target="_blank" title="Click to know more about COMPOUND">COMPOUND</a>")`<br/>`= (2.5xx1.4)/(0.303)=11.55%`.</body></html>
Discussion
52021.

0.302 g of organic compound gave 0.268g of silver bromide. The % of bromine in the sample is

Answer» <html><body><p>20<br/>50<br/>37.75<br/>75</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
52022.

0.301 g of an organic compound gave 0.282 g of silver bromide by a halogen estimation method. Find the percentage of bromine 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> :`39.87%`</body></html>
Discussion
52023.

0.301 g of an organic compound gave 0.282 g of silver bromide by carius method. Find the percentage of bromine.

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/br-390993" style="font-weight:bold;" target="_blank" title="Click to know more about BR">BR</a> = 39.83%`</body></html>
Discussion
52024.

0.30 g of an organic compound containing C, H and O on combustion yields 0.44 g CO_(2) and 0.18 g H_(2)O. If one mole of compound weighs 60, then molecular formulaof the compound is

Answer» <html><body><p>`C_(3)H_(8)O`<br/>`C_(2)H_(4)O_(2)`<br/>`CH_(2)O`<br/>`C_(4)H_(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)O`</p>Solution :`%<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> = (12)/(44) <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> (0.44)/(0.30) xx 100 = 40` <br/> `%H = (2)/(18) xx (0.8)/(0.30) xx 100 = 6.66` <br/> `%O = 100 - (40.0 + 6.66) = 53.34` <br/> Now `C:H:O = (40)/(12) : (6.66)/(1.0) : (53.34)/(16)` <br/> = 3.33: 6.66 : 3.33 <br/> `:. E.F. = CH_(2)O` <br/> E.F. wt. = 12+2+16 = 30 <br/> But Mol. wt. = <a href="https://interviewquestions.tuteehub.com/tag/60-328817" style="font-weight:bold;" target="_blank" title="Click to know more about 60">60</a> <br/> `:. M.F. = CH_(2)O xx (60)/(30) = C_(2)H_(4)O_(2)`</body></html>
Discussion
52025.

0.30 g of a substance gives 0.88 g of carbon dioxide and 0.54 g of water. Calculate the percentage of carbon and hydrogen in it

Answer» <html><body><p></p>Solution :(i) Weight of organic compound = 0.30 <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 carbon-dioxide= 0.<a href="https://interviewquestions.tuteehub.com/tag/88-339824" style="font-weight:bold;" target="_blank" title="Click to know more about 88">88</a> g <br/> Weight of water = 0.54 g <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/> 18 g water contains 2 g of hydrogen <br/> `:."0.54" g"` of water contain `=2/(18)xx0.54g` of hydrogen <br/> `:.""%` of hydrogen `=2/(18)xx(0.54)/(0.30)xx100=(2)/(18)xx(54)/(0.3)` <br/> `""%` of H = `0.111xx180=19.888~~20%` <br/> Percentage of carbon : <br/> <a href="https://interviewquestions.tuteehub.com/tag/44-316683" style="font-weight:bold;" target="_blank" title="Click to know more about 44">44</a> g of `CO_(2)` contains 12 g of carbon <br/> 0.88 g of `CO_(2)` contains = `(12)/(44)xx0.88g` of carbon <br/> `:.""%` of carbon `=(12)/(44)xx(88)/(0.3)=(24)/(0.3)` <br/> `""%` of carbon `=80%` <br/> (ii) <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_CHE_XI_SP_10_E01_041_S01.png" width="80%"/></body></html>
Discussion
52026.

0.30 g of a substance gives 0.88 g of carbon dioxide and 0.54 g water. Calculate the percentage of carbon and hydrogen in it.

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/weight-1451304" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHT">WEIGHT</a> of organic compound = 0.30 g <br/> Weight of carbon dioxide = 0.88 g <br/> Weight of water = 0.54 g <br/> Percentage of carbon : <br/> 44 g if carbondioxide contains, carbon = 12 g <br/> 0.88 g of carbon dioxide contains, <br/> carbon `= (12xx0.88)/(44)g` <br/> 0.30 g substance contains, carbon `= (12xx0.88)/(44)g` <br/> 100 g substance contains <br/> `= (12xx0.88)/(44)<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)/(0.30)=<a href="https://interviewquestions.tuteehub.com/tag/80-337972" style="font-weight:bold;" target="_blank" title="Click to know more about 80">80</a> g` of carbon <br/> Percentage of carbon = 80 <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/> 18 g of water contains, hydrogen = 2g <br/> 0.54 g of water contains, hydrogen `= (2xx0.54)/(18)g` <br/> 0.30 g of substance contains hydrogen `= (2xx0.54)/(18xx0.30)g` <br/> 100 g of substance contains `=(2xx0.54)/(18xx0.30)xx100 g` <br/> = 20 g of hydrogen <br/> Percentage of hydrogen = 20</body></html>
Discussion
52027.

0.3 g of an organic compound on Kjeldahl's analysis gave enough ammonia to just neutralise 30 mL of 0.IN H_(2)SO_(4) . Calculate the percentage of nitrogen in the compound.

Answer» <html><body><p></p>Solution :Weight of organic compound (w) = 0.3 g <br/> Strength of <a href="https://interviewquestions.tuteehub.com/tag/sulphuric-655420" style="font-weight:bold;" target="_blank" title="Click to know more about SULPHURIC">SULPHURIC</a> acid used (<a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>) = 0.1 N <br/> Volume of sulphuric acid used (<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>) = 30 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> <br/> 30 ml of 0.1 N sulphuric acid = 30 ml of 0.1 N ammonia <br/> `:. "Percentage of nitrogen" =((14 xx NV)/(1000 xx w)) xx 100=((14 xx 0.1 xx 30)/(1000 xx 0.3)) xx 100=14%`</body></html>
Discussion
52028.

0.3 g of KI is dissolved in 25 " mL of " water. After adding to this solution double its volume of concentration HCl, a solution of KIO_(3) is graduaaly added with stirring. Iodine is liberated as first but redissolved. It is observed that 24.1 " mL of " iodate solution is just sufficient to dissolve the iodine. If the iodate solution contains 0.8 g per 100 mL formulate the reaction that has taken place.

Answer» <html><body><p></p>Solution :0.3 of KI <a href="https://interviewquestions.tuteehub.com/tag/contains-11473" style="font-weight:bold;" target="_blank" title="Click to know more about CONTAINS">CONTAINS</a> `(0.3)/(166)=1.807xx10^(-3)` " mol of "iodide ion, 24.1 " mL of " iodate <a href="https://interviewquestions.tuteehub.com/tag/salution-3004965" style="font-weight:bold;" target="_blank" title="Click to know more about SALUTION">SALUTION</a> contains `(24.1)/(100)xx(0.8)/(214)=9.01xx10^(-4)` mol. It is clear from these figures that 2 moles of `I^(ɵ)` ions react with 1 <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> of iodate ion. In thus <a href="https://interviewquestions.tuteehub.com/tag/process-11618" style="font-weight:bold;" target="_blank" title="Click to know more about PROCESS">PROCESS</a> `I^(ɵ)` is oxidised and `IO_(3)^(ɵ)` is reduced to form some common species in which the oxidation number of iondine is x (say). <br/> Thus. <br/> `2(x-(-1)=(5-x)` <br/> i.e., `2x+2=5-x` <br/> `therefore3x=3impliesx=+1` <br/> In presence of excess HCl, this would correspond to the formation of <a href="https://interviewquestions.tuteehub.com/tag/icl-496585" style="font-weight:bold;" target="_blank" title="Click to know more about ICL">ICL</a>. `therefore` the reaction is formulated as follows, <br/> `2I^(ɵ)+IO_(3)^(ɵ)+3Cl^(ɵ)+6H^(o+)to3ICl+3H_(2)O`</body></html>
Discussion
52029.

0.3 g of a gas has a volume of 112 ml at 0^@C and 2atm pressure. Its Molecular weight is

Answer» <html><body><p>60<br/>30<br/>44<br/>28</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
52030.

0.2m aqueous solution of KCl freezes at -0.68^(@)C calculate van't Hoff factor. K _(f) for water is 1.86 K kg mol ^(-1).

Answer» <html><body><p></p>Solution :`I = ("Observed propoerty")/("Theoretical <a href="https://interviewquestions.tuteehub.com/tag/property-1170155" style="font-weight:bold;" target="_blank" title="Click to know more about PROPERTY">PROPERTY</a> (<a href="https://interviewquestions.tuteehub.com/tag/calculated-907694" style="font-weight:bold;" target="_blank" title="Click to know more about CALCULATED">CALCULATED</a>)")` <br/> Given `Delta T_(f) = 0.680 Km ,m=0.2 m, Delta T_(f) ` (observed) `=0.680K,` <br/> `Delta T _(f)` (colculated) `=K _(f). M = <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>. 86 K kg mol ^(-1) <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 0.2 mol kg ^(-1) = 0.372K` <br/> `i = ((Delta T _(f)) "observed")/((Delta T _(f))"calculated") = (0.680K )/(0.372K) =1.82`</body></html>
Discussion
52031.

0.2g of an organic compound on Kjeldahl's analysis gave enough ammonia to just neutralize 20 mL of 0.1 N H_(2)SO_(4). 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> :0.14</body></html>
Discussion
52032.

0.2g of a sample of H_2O_2required 10mL of N KMnO_4in a titration in the presence of H_2SO_4Purity of H_2O_2 is

Answer» <html><body><p>0.25<br/>0.<a href="https://interviewquestions.tuteehub.com/tag/85-339428" style="font-weight:bold;" target="_blank" title="Click to know more about 85">85</a><br/>0.65<br/>0.95</p>Solution :`W/17xx1000=10xx1` <br/> `W = (<a href="https://interviewquestions.tuteehub.com/tag/170-278357" style="font-weight:bold;" target="_blank" title="Click to know more about 170">170</a>)/(1000) = 0.17 ` gm <br/> 0.2 gm `rarr` 0.17 gm <a href="https://interviewquestions.tuteehub.com/tag/pure-1172806" style="font-weight:bold;" target="_blank" title="Click to know more about PURE">PURE</a> <br/> 100 gm `rarr` ? <br/> `=(100xx0.17)/(0.2) =85%`</body></html>
Discussion
52033.

0.29 g of an organic compound were analysed by Liebig's method. The increase in the mass of U-tube and the potash bulbs at the end of the experiment were found to be 0.27g and 0.66g respectively. Calculate the percentage of carbon and hydrogen in it

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> = 62.1 and %<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a> = 11.1</body></html>
Discussion
52034.

0.28g of an organic compound in Dumas method liberated 22.4 ml of nitrogen at STP. The percentage of nitrogen in the compound is

Answer» <html><body><p>20<br/>10<br/>80<br/>50</p>Answer :A</body></html>
Discussion
52035.

0.2828g of iron wire was dissolved in excess dilute H_(2)SO_(4) and the solution was made up to 100 mL. 20mL of this solution required 30mL of N//30 K_(2)Cr_(2)O_(7) doluyion for exact oxidation. Calculate % purity of Fe in wire.

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/99-342952" style="font-weight:bold;" target="_blank" title="Click to know more about 99">99</a>%`</body></html>
Discussion
52036.

0.28 g of nitrogenous compound was subjected to Kjeldahl's process to produce 0.17 g of NH_(3) . The percentage of nitrogen in the organic compound is

Answer» <html><body><p>5<br/>2<br/>50<br/>80</p>Solution :0.17 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> `NH_(3)` will <a href="https://interviewquestions.tuteehub.com/tag/contains-11473" style="font-weight:bold;" target="_blank" title="Click to know more about CONTAINS">CONTAINS</a> `(14/17 xx0.17)` g of nitrogen , i.e., 0.14 g of nitrogen <br/> % Nitrogen =`("<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 nitrogen ")/("Mass of compound ")xx100=0.14/0.28xx100=50%`</body></html>
Discussion
52037.

0.27g of an organic compound gave on combustion 0.396g of CO_(2) and 0.216g of H_(2)O. 0.36g of the same substance gave 48.88 mL of N_(2) at 290 K and 740 mm pressure. Calculate the percentage composition of 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> :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> = <a href="https://interviewquestions.tuteehub.com/tag/40-314386" style="font-weight:bold;" target="_blank" title="Click to know more about 40">40</a>%, <a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a> = 8.89%, <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> = 15.56% and O = 35.55%</body></html>
Discussion
52038.

0.28 g of a commercial sample of K_(2)Cr_(2)O_(7) was dissolved in water. Excess of KI was added to it along with dilute H_(2)SO_(4). Iodine liberated was then titrated against sodium thiosulphate solution containing 24.82 g of Na_(2)S_(2)O_(3). 5H_(2)O per litre. The thiosulphate solution required was 50 mL. Find the percentage purity of the sample of K_(2)Cr_(2)O_(7).

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>
Discussion
52039.

0.262 g of a substance gave, on combustion 0.361 g of CO_(2) and 0.147 g of H_(2)O. What is the empirical formula of the substance

Answer» <html><body><p>`CH_(<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`<br/>`C_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>) H_(6) O`<br/>`C_(3) H_(6) O_(2)`<br/>`C_2 H_(6) O_(2)`</p>Answer :A</body></html>
Discussion
52040.

0.26 g of an organic compundgave 0.039 g of water and 0.245 g of CO_2 on combustion. Calculate the percentage of carbon and hydrogen.

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 compound =0.26 <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 water =0.039 g <br/>Weight of `CO_2`= 0.245 g <br/> Percentage of hydrogen <br/> 18 g of watercontains 2 g of hydrogen <br/> 0.039 mg of water contains`-(2)/(18)xx(0039)/(0.26)`of H <br/> % of hydrogen `-(0.39)/(0.26)xx(2)/(18)xx100=1.66%` <br/> Percentage of csarbon <br/> 44 g of `CO_2` contains 12 g of <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/>0.245 g `CO_2` contains=`(12)/(44)xx(0.245)/(0.26)g` of C <br/> %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(0.245)/(0.26)xx100=25.69%`</body></html>
Discussion
52041.

0.26 g of an organic compound gave 0.039 g of water and 0.245 g of carbon dioxide on combustion. Calculate the percentage of C & H.

Answer» <html><body><p></p>Solution :Weight of organic compound = 0.26 g <br/> Weight 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> = 0.039 g <br/> Weight of `CO_(2)` =0.245 g <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/> 18 g of water contains 2 g of hydrogen <br/> `"0.039 g of water contains" =(2)/(18) xx (0.039)/(0.26) "of H"` <br/> `:. % "of hydrogen"=(0.039)/(0.26) xx (2)/(18) xx 100=1.66%` <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> <br/> 44g of `CO_(2)` contains 12 g of C <br/> `0.245 "g of" CO_(2) "contains"=(12)/(<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 (0.245)/(0.26) "g of C"` <br/> `% "of Carbon"=(12)/(44) xx (0.245)/(0.26) xx 100=25.69%`</body></html>
Discussion
52042.

0.25g of an organic compound an analysis by Duma,s method gave 32cc nitrogen at STP. What is the weight percentage of nitrogen 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> :<a href="https://interviewquestions.tuteehub.com/tag/16-276476" style="font-weight:bold;" target="_blank" title="Click to know more about 16">16</a></body></html>
Discussion
52043.

0.25g of an organic compound gave 30cm^(3) of moist dinitrogen at 288K and 745mm pressure. Calculate the percentage of nitrogen. (Aqueous tension at 288K = 12.7mm)

Answer» <html><body><p>&lt;<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>&gt;</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> :Mass of the substance = 0.25g <br/> Volume of moist dinitrogen = `30cm^(3)` <br/> Temperature = 288K <br/> Pressure `= 745- 12.7 = 732.3mm` <br/> Volume of dinitrogen at S.T.P <br/> `V_(2) = (P_(1)V_(1))/(T_(1)) xx (T_(2))/(P_(2)) = (732.3 xx 30 xx 273)/(288 xx <a href="https://interviewquestions.tuteehub.com/tag/760-335611" style="font-weight:bold;" target="_blank" title="Click to know more about 760">760</a>) = 27.4 <a href="https://interviewquestions.tuteehub.com/tag/cm-919986" style="font-weight:bold;" target="_blank" title="Click to know more about CM">CM</a>^(3)` <br/> `22400cm^(3)` of dinitrogen at STP weight = 28g <br/> `27.4 cm^(3)` of dinitrogen at STP weight `= (28 xx 27.4)/(22400)` <br/> = 0.034g <br/> 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> in organic compound `=(0.034)/(0.25) xx 100 = 13.6`</body></html>
Discussion
52044.

0.25g compound give 0.350g, BaSO_(4). What is 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> :0.1856</body></html>
Discussion
52045.

0.2595 g of organic compound when treated by carius method gave 0.35 g of BaSO_(4). Calculate the percentage of sulphur in the compound.

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/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 ("Mass of "BaSO_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>))/("Mass of compound")xx100=32/233xx0.35/0.2595xx100=18.52%`</body></html>
Discussion
52046.

0.2595 g of an organic substance, when treated by Carius method, gave 0.25 g of BaSO_(4). Calculate the percentage of sulphur 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> :`13.23%`</body></html>
Discussion
52047.

0.257 g of an organic substance was Kjeldahlised and ammonia evolved was absorbed in 50 mL of N/10 HCl which required 23.2 ml of N/10 NaOH for neutralization. Determine the percentage of nitrogen 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/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> = 14.6%`</body></html>
Discussion
52048.

0.25 g of an organic compound 'X' containing carbon, hydrogen and oxygen was analysed by the combustion method. The increase in mass of calciumchloride tube and potash bulbs at the end of the experiment was found to be 0.15 g and 0.1837 g respectively. Calculate the percentage composition of the compound.

Answer» <html><body><p><br/></p>Solution :Wt. 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> = Increase in potash <a href="https://interviewquestions.tuteehub.com/tag/bulbs-905371" style="font-weight:bold;" target="_blank" title="Click to know more about BULBS">BULBS</a> <br/> `= 0.1837` g <br/> Wt. 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> = 0.25<br/> `% C = *(12)/(44) xx (0.1837)/( 0.25) 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>` <br/> `= 20.04%` <br/> Wt. of `H_2 O` produced = Increase in `CaCl_2` <a href="https://interviewquestions.tuteehub.com/tag/tube-1428592" style="font-weight:bold;" target="_blank" title="Click to know more about TUBE">TUBE</a> <br/> `= 0.15` g <br/> `%H= (2)/( 16) xx (0.15)/( 0.25) xx 100` <br/> `= 6.66%` <br/> % of `O - 100 - ( 20. 04 + 6.66)` <br/> `73.30 %`</body></html>
Discussion
52049.

0.25 g of an oxalate salt was dissolved in 100 mL of water. 10 mL of this solution required 8 mL of N//20 KMnO_(4) for its oxidation. Calculate the percentage of oxalate in the salt.

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>
Discussion
52050.

0.25 g of an organic compound was found to produce 0.35 g of AgCl after heating with fuming HNO_(3) and AgNO_(3) in a sealed carius method. Determine 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> = 34.75%`</body></html>
Discussion