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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.
| 51151. |
35ml of dry H_2 gas were collected at 6^@C and 758 mm pressure. What is the volume in ml of H_2 at STP? |
| Answer» <html><body><p>32<br/>24.6<br/>34.16 <br/>30.16 </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> | |
| 51152. |
350 cm^3 of oxygen and 275cm^3 of another gas 'A' diffused in same time under similar conditions. Find the molecualr mass of the gas 'A'. |
| 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> :51.86 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>.`<a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a>^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)`</body></html> | |
| 51153. |
350 cm^3 of CH4and 175cm^3 of an unknown gas 'A' diffused in the same time under similar conditions. The molecular mass of gas A is |
| Answer» <html><body><p>32<br/>64<br/>30<br/>71</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 51154. |
35 mL sample lof hydrogen peroxide gives off 494 mL of O_(2) at 27^(@) C and 1 atm pressure. Volume strength of H_(2)O_(2) sample will be : |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> V<br/>13 V<br/>11 V<br/>12 V</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 51155. |
3.5 grams aluminium bronze was heated in aqueous hydroxhloric solution. The hydrogen liberated measured 4.15L at STP. What is the percentage of copper in the given alloy? |
| 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> :4.72</body></html> | |
| 51156. |
3.45 g of a metallic carbonate were mixed with 240 mL of N//4 HCl. The excess acid was neutralised by 50 mL of N//5 KOH solution. 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> :Equivalent <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 metal carbonate <br/> `=(3.45xx4xx1000)/(<a href="https://interviewquestions.tuteehub.com/tag/200-288914" style="font-weight:bold;" target="_blank" title="Click to know more about 200">200</a>)=69` <br/> Equivalent mass of metal =69- <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a>. mass of carbonate <br/> =(69-30)=39</body></html> | |
| 51157. |
34.95 ml of phosphorus vapour weighs 0.0625gat 0.1 bar pressure. What is the molar mass of phosphorus ? |
| Answer» <html><body><p></p>Solution :`PV=nRT` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>.1"bar" xx(34.*05)/(<a href="https://interviewquestions.tuteehub.com/tag/10000l-1771129" style="font-weight:bold;" target="_blank" title="Click to know more about 10000L">10000L</a>)=(0*0625)/(M) xx0*083xx819K` <br/> `M=(0*0625xx0*083xx819Kxx1000)/(34*05xx0*1)=(4248*5625)/(3*<a href="https://interviewquestions.tuteehub.com/tag/405-1874941" style="font-weight:bold;" target="_blank" title="Click to know more about 405">405</a>)=1247*7 "<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> | |
| 51158. |
343K = ........""^(@)F. |
| Answer» <html><body><p><br/> <br/> </p>Solution :`T_(<a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>)=t_(<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>)^(@)+273` <br/> `t_(C)^(@)=T_(K)-273 = 343 -273=<a href="https://interviewquestions.tuteehub.com/tag/70-333629" style="font-weight:bold;" target="_blank" title="Click to know more about 70">70</a>""^(@)C` <br/> `""^(@)<a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a>=9/5 (""^(@)C)+32=9/5(70)+32=158 ""^(@)F`</body></html> | |
| 51159. |
3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution are |
| Answer» <html><body><p>`3.67xx10^(26)`<br/>`6.6 xx 10^(23)`<br/>`3.67xx10^(<a href="https://interviewquestions.tuteehub.com/tag/24-295400" style="font-weight:bold;" target="_blank" title="Click to know more about 24">24</a>)`<br/>`6.0 xx10^(<a href="https://interviewquestions.tuteehub.com/tag/22-294057" style="font-weight:bold;" target="_blank" title="Click to know more about 22">22</a>)`</p>Solution :Number 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> atoms in 34.2 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> of sucrose <br/> `=(6.02xx10^(23)xx11xx34.2)/(342) = 6.62xx10^(23)` <br/> Number of oxygen atoms in 9 g of `H_2O` <br/> `=(6.02xx10^(23)xx90)/(<a href="https://interviewquestions.tuteehub.com/tag/18-278913" style="font-weight:bold;" target="_blank" title="Click to know more about 18">18</a>) =30.1xx10^(23)` <br/> Total number of oxygen atoms `(30.1 +6.62)10^(23)` <br/> `=36.72xx10^(23) = 3.672xx10^(24)`</body></html> | |
| 51160. |
34.95 mole of phosphorus vapour weighs 0.0625gat 0.1 bar pressure. What is the molar mass of phosphorus ? |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/according-366619" style="font-weight:bold;" target="_blank" title="Click to know more about ACCORDING">ACCORDING</a> to ideal gas equation. <br/> `Pv= nRT` <br/> Now, `n=(W)/(M):. Pv=(WRT)/(M) or M=(WRT)/(PV)` <br/> Substituting the values of from data <br/> `M=(<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>*0625gxx0*83 "<a href="https://interviewquestions.tuteehub.com/tag/bar-892478" style="font-weight:bold;" target="_blank" title="Click to know more about BAR">BAR</a>" L <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>^(-1) "mol"^(-1) xx819*15K)/(1"bar" xx34*04xx10^(-<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/> `:. M=124*8 g "mol"^(-1)`</body></html> | |
| 51161. |
34.05 mL of phosphorus vapour weights 0.0625 g at 546^(@)C and 0.1 bar pressure. What is the molar mass of phosphorus ? |
| Answer» <html><body><p></p>Solution :According to ideal <a href="https://interviewquestions.tuteehub.com/tag/gas-1003521" style="font-weight:bold;" target="_blank" title="Click to know more about GAS">GAS</a> law,<br/>`pV=nRT=(w)/(M)RT`<br/>`therefore M=(wRT)/(pV)`<br/>where,<br/>Molecular <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 phosphorus, (w) = 0.0625 g<br/> Gas Constant `(R )=8.314xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)` <a href="https://interviewquestions.tuteehub.com/tag/bar-892478" style="font-weight:bold;" target="_blank" title="Click to know more about BAR">BAR</a> `L mol^(-1)K^(-1)` <br/> Pressure of phosphorus vapour (p) = 1.0 bar<br/>Absolute <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a> `(T)=(546+273)K=819 K`<br/>Volume of Phosphorus vapour (V) = 34.05 mL<br/>= 0.03405 L<br/>`therefore M =((0.0625 g)(8.314xx10^(-2)"bar L mol"^(-1)K^(-1))(819K))/(("1 bar")(0.03405 L))`<br/>`= 124.98 = 125 g mol^(-1)`<br/>Note : If 0.1 bar is corrct for according to text book then answer will be 1249.8. It is not real because,<br/>`P_(4)=31xx4=124`</body></html> | |
| 51162. |
34 . 05 mL of phosphorus vapours weigh 0.0625 g at 546 .^(@)C and 0.1 bar pressure . What is the molart mass of phosphorus ? |
| 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>Solution :<a href="https://interviewquestions.tuteehub.com/tag/according-366619" style="font-weight:bold;" target="_blank" title="Click to know more about ACCORDING">ACCORDING</a> to gas <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a>, <br/> `PV = nRT` <br/>or `PV =w/M RT` <br/> Given : `P = 0.1 " bar," V =340.5 mL = 340.5 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>) dm^3, W = 0.0625 g, R = 0.0831 " bar " dm^3 K^(-1) mol^(-1)` <br/> `T = 546^@C = 819 K, M` = ? <br/> `:. M=(W.R.T)/(P.V)= (0.0625 xx0.0831 xx 819)/(0.1 xx 340.5 xx 10^(-3))` <br/> = `124.92 g mol^(-1)` <br/>Hence, 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 phosphorus is `124.92 g mol^(-1)`.</body></html> | |
| 51163. |
34.05 mL of phosphorus vapour weigh 0.0625 g at 546^(@)C and 1.0 bar pressure. What is the molar mass of phosphorus ? |
| 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>Solution :Step 1. Calculation of <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> at `0^(@)C` and 1 <a href="https://interviewquestions.tuteehub.com/tag/bar-892478" style="font-weight:bold;" target="_blank" title="Click to know more about BAR">BAR</a> pressure <br/> `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)), i.e., (1xx34.05)/(546+273)=(1xxV_(2))/(273)"or" V_(2)=11.35 mL` <br/> 11.35 mL of vapour at `0^(@)C` and 1 bar pressure weigh `=0.0625" <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/> Step 2. Calculation of mass of 22700 mL of `0^(@)C` and 1 bar pressure <br/> `:. 22700" mL"` of vapour at `0^(@)C` and 1 bar pressure will weigh `=(0.0625)/(11.35)xx22700=125" g"` <br/> `:. """Molar mass" =125" g "mol^(-1)` <br/>Alternatively, using <br/> `R=0.083" bar "<a href="https://interviewquestions.tuteehub.com/tag/dm-432223" style="font-weight:bold;" target="_blank" title="Click to know more about DM">DM</a>^(3)K^(-1)mol^(-1)` <br/> PV=nRT, i.e., `n=(PV)/(RT)=(1.0" bar"xx(34.05xx10^(-3)dm^(3))/(0.083" bar " dm^(3)K^(-1)xx819 K)=5xx10^(-4)mol` <br/> `:.`Mass of 1 mole `"=(0.0625)/(5xx10^(-4))"g"=125 g` <br/> `:. ""` Molar mass`=125 g mol^(-1)`</body></html> | |
| 51164. |
3.4 gm of H_(2)O_(2) decomposes, the weight of oxygen liberated from it is |
| Answer» <html><body><p>1.6 <a href="https://interviewquestions.tuteehub.com/tag/gm-1008640" style="font-weight:bold;" target="_blank" title="Click to know more about GM">GM</a><br/>2.24 gm <br/>1.16 gm <br/>3.2 gm </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> | |
| 51165. |
338 mL clear saturated solution of AgBrO_(3) requires just 30.4mL of H_(2)S_((g)) at 23^(@)C and 748 mm Hg to precipitate all the Ag^(+) ions Ag_(2)S. What will be K_(SP)of AgBrO_(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> :`<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>.29xx10^(-5);`</body></html> | |
| 51166. |
3,3-Dimethyl-2-butanol on reaction with HCl yields mainly |
| Answer» <html><body><p>2-Chloro-2,3-dimethlybutane<br/>1-Chloro-2,3-dimethylbutane<br/>2-Chloro-3,3-dimethylbutane<br/>1-Chloro-3,3-dimethylbutane</p>Solution :`CH_(3)-<a href="https://interviewquestions.tuteehub.com/tag/underset-3243992" style="font-weight:bold;" target="_blank" title="Click to know more about UNDERSET">UNDERSET</a>(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-underset(<a href="https://interviewquestions.tuteehub.com/tag/oh-585115" style="font-weight:bold;" target="_blank" title="Click to know more about OH">OH</a>)underset(|)(CH)-CH_(3)underset(-H_(2)O)overset(<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</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> CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-overset(+)(C)H-CH_(3) overset("1,2-Methyl <a href="https://interviewquestions.tuteehub.com/tag/shift-1205367" style="font-weight:bold;" target="_blank" title="Click to know more about SHIFT">SHIFT</a>")rarr CH_(3)-underset(+)overset(CH_(3))overset(|)(C)-underset(CH_(3))underset(|)(CH)-CH_(3)overset(+Cl^(-))rarr CH_(3)-underset(Cl)underset(|)overset(CH_(3))overset(|)(C)-underset(CH_(3))underset(|)(CH)-CH_(3)`</body></html> | |
| 51167. |
3.2g of a gas occupies 550 cc of volume at 22^@C and 770 mm of Hg pressure. Find the molecular mass of the gas. |
| 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/139-1786326" style="font-weight:bold;" target="_blank" title="Click to know more about 139">139</a> <a href="https://interviewquestions.tuteehub.com/tag/gm-1008640" style="font-weight:bold;" target="_blank" title="Click to know more about GM">GM</a></body></html> | |
| 51168. |
3.2g of oxygen (At.wt =16) and 0.2g of hydrogen (At.wt = 1) are placed in a 1.12 litre flask at 0^@C. The total pressure of the gas mixture will be atm |
| 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>Solution :`n_1 = 3./32 = 0.1 , n_2 = (0.2)/2 = 0.1 , n_("<a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a>") = 0.2` <br/> `P = (0.2 xx 0.0821 xx <a href="https://interviewquestions.tuteehub.com/tag/273-1832932" style="font-weight:bold;" target="_blank" title="Click to know more about 273">273</a>)/(1.12) = 4`.</body></html> | |
| 51169. |
32.2 gm of an organic compound containing C,H and O when completely combusted produces 61.6gm of CO_(2) and 37.8 gm of H_(2)O. Select the correct option : |
| Answer» <html><body><p>% 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> in the <a href="https://interviewquestions.tuteehub.com/tag/organic-1138713" style="font-weight:bold;" target="_blank" title="Click to know more about ORGANIC">ORGANIC</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> is <a href="https://interviewquestions.tuteehub.com/tag/40-314386" style="font-weight:bold;" target="_blank" title="Click to know more about 40">40</a>%.<br/>% of H in the organic compound is 13.04%<br/>% of O in the organic compound is 17.04 %<br/>Data given is insufficient.</p>Answer :B</body></html> | |
| 51170. |
3200gm sulphur is present in |
| Answer» <html><body><p>9800 <a href="https://interviewquestions.tuteehub.com/tag/gm-1008640" style="font-weight:bold;" target="_blank" title="Click to know more about GM">GM</a> `H_(2)SO_(4)`<br/><a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a> mole `H_(2)SO_(4)`<br/>100 moles `H_(2)SO_(4)`<br/>6400 g `SO_(2)`</p>Solution :3200 gm = 100 moles <br/> a) 100 moles `H_(2)SO_(4)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>) 20 moles `H_(2)SO_(4)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>) 100 moles `H_(2)SO_(4)` <br/> d) 100 moles `SO_(2)`</body></html> | |
| 51171. |
3.2 g of a mixture of calcium carbonate and sodium chloride was dissolved in 100 mL of 1.02 N HCl. After the reaction the solution was filtered and after separating the precipitate the volume was raised to 200 mL . 20 mL of this solution required 25 mL N//5 caustic soda solution for neutralisation. Find out the percentage of calcium carbonate in the mixture. |
| 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> | |
| 51172. |
3.15 g of oxalic acid dihydrate is dissolved in water and the solution was made up to 100 ml using a standard flask. The strength of the solution in normality is …………… |
| Answer» <html><body><p></p>Solution :`0.5N` <br/> Normality `= ("Number of gram equivalents")/("Volume of the solution in L")` <br/> `= ("Mass of oxalic <a href="https://interviewquestions.tuteehub.com/tag/acid-847491" style="font-weight:bold;" target="_blank" title="Click to know more about ACID">ACID</a>")/(" <a href="https://interviewquestions.tuteehub.com/tag/equivalent-446407" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENT">EQUIVALENT</a> mass of oxalic acid") + ` Volume of the solution in L <br/> `= ((3.15)/(63))/(0.1) = (0.05)/(0.1) =0.5N`</body></html> | |
| 51173. |
314.000 has how many significant figures? |
| Answer» <html><body><p>6<br/>3<br/>5<br/>4</p>Answer :A</body></html> | |
| 51174. |
30ml of sample of hydrogen peroxide solution after acidification with dil. H_2SO_4required 30ml of 0.02M KMnO_4solution for complete oxidation. Calculate molarity and percentage strength of the solution. |
| Answer» <html><body><p></p>Solution :`KMnO_(4)+5H_(2)O_(2)+3H_(2)SO_(4)toK_(2)SO_(4)+2MnSO_(4)+8H_(2)O+5O_(2)` <br/> 2 moles of `KMnO_(4)` = 5 moles `H_(2)O_(2)` <br/> The equation for titration `(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>)V_(1))/(n_(1))=(M_(2)V_(2))/(n_(2))=(30xx0.02)/(2)=(M_(H_(2)O_(2))xx30)/(5)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/molarity-1100268" style="font-weight:bold;" target="_blank" title="Click to know more about MOLARITY">MOLARITY</a> of `H_(2)O_(2)` solution `M_(H_(2)O_(2))=(5)/(2)xx0.02M=0.05M` <br/> (w/v)% of `H_(2)O_(2)` solution`=Mxx"G.M.W//10=0.05xx34//10=0.17%</body></html> | |
| 51175. |
30mL of mixture of methane and ethane in x : y ratio of volumes on combustioin gave 40mL of carbon dioxide. If 30mL of mixture is taken in y:x ratio, what is the volume of carbondioxide obtained under similar conditions? |
| Answer» <html><body><p></p>Solution :The <a href="https://interviewquestions.tuteehub.com/tag/combustion-13975" style="font-weight:bold;" target="_blank" title="Click to know more about COMBUSTION">COMBUSTION</a> of methane and ethane is given by the equations <br/> `CH_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>) overset(O_(2))to CO_(2)+2H_(2)O`: <br/> `C_(2)H_(6) overset(O_(2))to 2CO_(2)+3H_(2)O` <br/> 1 volume of `CH_(4)="1 volume of "CO_(2)` <br/> 1 volume of `C_(2)H_(6)="2 volume of "CO_(2)` <br/> Let, Volume of `CH_(4)` in the given mixture=x <br/> Let, volume of `C_(2)H_(6)` in the given mixture=y <br/> Then, x+y=30 `""` x+2y=40 <br/> `"y=10"x=30-10=20` <br/> If y mL of `CH_(4)` is taken <br/> the volume of `CO_(2)` obtained =y=10mL <br/> If x mL of `C_(2)H_(6)` is taken <br/> the volume of `CO_(2)` obtained =2x=2 x 20=40mL <br/> The volume of <a href="https://interviewquestions.tuteehub.com/tag/carbondioxide-413336" style="font-weight:bold;" target="_blank" title="Click to know more about CARBONDIOXIDE">CARBONDIOXIDE</a> obtained by the combustion of 30mL of y:x volumes of methane and ethane =10+40=50mL</body></html> | |
| 51176. |
30ml of hydrogen peroxide solution was added with added with excess potasium iodide. The iodide liberated was titrated using starch indicator with 20ml of 0.3N sodium thiosulphate solution.Calculate the normality of hydrogen peroxide solution. |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/normality-1124423" style="font-weight:bold;" target="_blank" title="Click to know more about NORMALITY">NORMALITY</a> equation for iodometry is given as ,`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>)N_(1)=V_(2)N_(2)` <br/> `N_(1)=(V_(2)N_(2))/(V_(1))=(20xx0.3)/(<a href="https://interviewquestions.tuteehub.com/tag/30-304807" style="font-weight:bold;" target="_blank" title="Click to know more about 30">30</a>)=0.2` <br/> Normality `H_(2)O_(2)` solution = 0.2 eq `L^(-1)`</body></html> | |
| 51177. |
30mL of a soltuion containg 9.15g//"litre" of an oxalte K_(X) H_(Y) (C_(2)O_(4))_(Z).nH_(2)O are required for titrating 27mL of 0.12N NaOH and 36mL of 0.12N KMnO_(4) separalty. Calculalte X,Y,Z are in the simple ratio of g atoms. |
| 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> :`X = <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>, Y = <a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>, Z = <a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>, n = 2`</body></html> | |
| 51178. |
30.4 KJ is required to melt one mole of sodium chloride. The entropy change during melting is 28.4 JK^(-1) mol^(-1). Calculate the melting point of sodium chloride. |
| 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> : <br/> `delta H_(f) (NaCl) = 30.4kJ = 30400 J <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)` <br/> `delta S_(f) (NaCl) = 28.4 JK^(-1) mol^(-1)` <br/> `T_(f) = ?` <br/> `DeltaS_(f) = (Delta H_(f))/(Delta T_(f)) , T_(f) = (Delta H_(f))/(Delta S_(f))` <br/> `T_(f) = (30400 J mol^(-1))/(28.4 J <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>^(-1) mol^(-1)) = 1070.4K`</body></html> | |
| 51179. |
30.4 kJ is required to melt one mole of sodium chloride. The entropy change during melting is 28.4 JK^(-1)mol^(-1) . Calculate the melting point of sodium chloride. |
| 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> : `DeltaH_f (NaCl)=30.4 kJ =30400 "<a href="https://interviewquestions.tuteehub.com/tag/j-520843" style="font-weight:bold;" target="_blank" title="Click to know more about J">J</a> <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a>"^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)` <br/> `DeltaS_f (NaCl)=28.4 <a href="https://interviewquestions.tuteehub.com/tag/jk-521675" style="font-weight:bold;" target="_blank" title="Click to know more about JK">JK</a>^(-1) mol^(-1)` <br/> `T_f`=? <br/>`DeltaS_f =(DeltaH_f)/(DeltaT_f)` <br/> `T_f =(DeltaH_f)/(DeltaS_f)` <br/> `T_f=(30400 "J mol"^(-1))/(28.4 J K^(-1) "mol"^(-1))` <br/> `T_f` =1070.4 K</body></html> | |
| 51180. |
30.4 KJ is required to melt one mole of sodium chloride . The entropy change during melting is 28.4JK^(-1) mol^(-1) .Calculate the melting point of sodium chloride . |
| Answer» <html><body><p></p>Solution :Heat <a href="https://interviewquestions.tuteehub.com/tag/required-1185621" style="font-weight:bold;" target="_blank" title="Click to know more about REQUIRED">REQUIRED</a> 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> mole of NaCl for <a href="https://interviewquestions.tuteehub.com/tag/melting-1093090" style="font-weight:bold;" target="_blank" title="Click to know more about MELTING">MELTING</a> (q) =30.4 K J <br/> =30.4 x 1000 J <br/> `<a href="https://interviewquestions.tuteehub.com/tag/deltas-947743" style="font-weight:bold;" target="_blank" title="Click to know more about DELTAS">DELTAS</a>` - entropy change =`28.4 J K^(-1) "mol"^(-1)` <br/> Melting <a href="https://interviewquestions.tuteehub.com/tag/point-1157106" style="font-weight:bold;" target="_blank" title="Click to know more about POINT">POINT</a> =`T_m`=?<br/> `DeltaS=q/T_m "" therefore T_m =q/(DeltaS)` <br/>`therefore T_m =(30.4xx1000)/28.4`=1070.4 K <br/> Melting point of NaCl =1070.4 K</body></html> | |
| 51181. |
3.01xx10^(23)Ca^(+2) and CO_(3)^(-2)ions are present in CaCO_(3).The mass of sample is..... |
| Answer» <html><body><p> <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/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a><br/> <a href="https://interviewquestions.tuteehub.com/tag/50-322056" style="font-weight:bold;" target="_blank" title="Click to know more about 50">50</a> g<br/>60 g<br/>70 g</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 51182. |
3.011xx10^(22) molecules are removed from a vessel Containing 1680 cc of nitrogen at STP. How many moles are remaining ? |
| Answer» <html><body><p></p>Solution :Number of <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> of `N_(2)` originally present =`("volume")/("<a href="https://interviewquestions.tuteehub.com/tag/gmv-468789" style="font-weight:bold;" target="_blank" title="Click to know more about GMV">GMV</a>") = (1680)/(22400) = (3)/(40)` <br/> Number of moles of `N_(2)` removed = `("Number of <a href="https://interviewquestions.tuteehub.com/tag/molecules-563030" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULES">MOLECULES</a>")/("Avogadro number") = (3.011xx10^(22))/(6.022xx10^(23)) = (1)/(<a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a>)` <br/> Number of moles remaining `= (3)/(40)-(1)/(20) = (1)/(40) = 2.5xx10^(-2) = 0.025`</body></html> | |
| 51183. |
3.011 xx 10^(22) atoms of an element weighs 1.15 g. The atomic mass of the element is |
| Answer» <html><body><p>23<br/>10<br/>16<br/>35.5</p>Answer :A</body></html> | |
| 51184. |
300ml of O_2 diffused though a porous pot in 50 Sec. How long will it take for 500ml of CO_2 to diffuse under similar conditions. |
| Answer» <html><body><p>97.72 <a href="https://interviewquestions.tuteehub.com/tag/seconds-1198593" style="font-weight:bold;" target="_blank" title="Click to know more about SECONDS">SECONDS</a> <br/>5.117 <a href="https://interviewquestions.tuteehub.com/tag/sec-1197209" style="font-weight:bold;" target="_blank" title="Click to know more about SEC">SEC</a> <br/>26.18 sec <br/>9.772 sec </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> | |
| 51185. |
3*00 " mol of " PCl_(5) kept in 1 L closed reaction vessel was allowed to attain equilibrium at 380 K. Calculate the composititon of the mixture at equilibrium . K_(c) = 1* 80 |
| Answer» <html><body><p></p>Solution :` {:(,PCl_(5)(g),hArr,PCl_(3)(g),+,Cl_(2)(g)),("Initial",3*<a href="https://interviewquestions.tuteehub.com/tag/00-254995" style="font-weight:bold;" target="_blank" title="Click to know more about 00">00</a> " mol ",,0,,0),("At. eqm.",(3*0-<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>)"mol",,x " mol",,x" mol "),("Molar conc",(3*0-x),,x,,x):}` (as volume of <a href="https://interviewquestions.tuteehub.com/tag/vessel-1445872" style="font-weight:bold;" target="_blank" title="Click to know more about VESSEL">VESSEL</a> = 1 L) <br/> ` K_(c) = ([PC_(3)] [Cl_(2)])/([PC_(5)])`<br/> ` 1*80 = (x xxx)/(3-x)=x^(2)/(3-x)or x^(2) = 5*<a href="https://interviewquestions.tuteehub.com/tag/40-314386" style="font-weight:bold;" target="_blank" title="Click to know more about 40">40</a> - 1*80 x or x^(2) + 1*80 x - 5* 40=0` <br/>` :. x= (-bpm sqrt(b^(2) - 4ac))/(2a)=(-1*80 pm sqrt((1*80)^(2) - 4 (-5*40)))/2=1*59`(neglecting - ve value) <br/> `:. " AT. equilibrium",[PCl_(5)]=3- 1*59 = 1*41 M` <br/> ` [PCl_(3)] = [Cl_(2)] = 1*59 M `</body></html> | |
| 51186. |
3.00 mol of PCl_5 kept in 1 L closed reaction vessel was allowed to attain equilibrium at 380 K. Calculate composition of the mixture at equilibrium. K_c=1.80 PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g)) |
| Answer» <html><body><p></p>Solution :`{:("Equilibrium reaction :",PCl_(5(g)) hArr , PCl_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>(g)) + , Cl_(<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))),("Initial mol :", 3.0,0,0),("Change of mol :", -x,x,+x),("Mol at equilibrium :",(3-x),x,x),("mol" L^(-1) "at equilibrium :",(3-x)/1=(3-x),x/1=x, x/1=x):}` <br/> where , x=(<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 `PCl_5`) = (Mole of `PCl_3` and `Cl_2` ) <br/> So, `K_c=([PCl_3][Cl_2])/[[PCl_5]]` <br/> `therefore 1.80 = ((x)(x))/(3-x)=x^2/(3-x)` <br/> `therefore x^2`=1.80 (3-x)=5.4 -1.80x <br/> `therefore x^2+1.80x-5.4=0` <br/> `therefore` a=1 |b=1.80|c=-5.4<br/> `x=(-bpmsqrt(b^2-Delta4ac))/(<a href="https://interviewquestions.tuteehub.com/tag/2a-300249" style="font-weight:bold;" target="_blank" title="Click to know more about 2A">2A</a>)` <br/> `=(-1.80 pm sqrt((-1.8)^2-4(1)(-5.4)))/(2(1))` <br/> `=(-1.80 pm sqrt(3.24+21.6))/2` <br/>`=(-1.80 pm sqrt(24.84))/2=(-1.8 pm 4.984)/2`<br/> `therefore x=3.184/2` OR `(-6.084)/2` <br/>=1.592 M OR -3.042 M <br/> So, x=1.592and -3.42 is <a href="https://interviewquestions.tuteehub.com/tag/impossible-498943" style="font-weight:bold;" target="_blank" title="Click to know more about IMPOSSIBLE">IMPOSSIBLE</a> <br/> At equilibrium `(PCl_5)`=(3-x)=(3-1.592) =1.408 M<br/> `approx` 1.41 M<br/> equilibrium `[PCl_3]=[Cl_2]`= x= 1.592 M `approx` 1.59 M</body></html> | |
| 51187. |
300 litres of ammonia gas at 20^(@)C and 20 atmosphere pressure are allowed to expend in a space of 600litres capacity and to a pressure of one atmosphere. Calculate the drop in temperature. |
| Answer» <html><body><p><br/></p>Solution :`(P_(1)V_(1))/(T_(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>)V_(2))/(T_(2)), i.e., (20 atmxx300 L)/(<a href="https://interviewquestions.tuteehub.com/tag/293-1834823" style="font-weight:bold;" target="_blank" title="Click to know more about 293">293</a> <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>)=(1 atmxx600 L)/(T_(2))"or" T_(2)=29.3 K`<br/> <a href="https://interviewquestions.tuteehub.com/tag/drop-443018" style="font-weight:bold;" target="_blank" title="Click to know more about DROP">DROP</a> intemperature =293-29.3=263.7 K</body></html> | |
| 51188. |
300 K temperature at 2.0xx10^(4) Pa pressure volume of 0.09 mole CO_(2) gas is 112.0xx10^(-3)m^(3)then calculate volume at 4.0xx10^(4) Pa ? |
| 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>Solution :`64.2xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)m^(3)`<br/>`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)=p_(2)V_(2)`<br/>`therefore V_(2)=(p_(1)V_(1))/(p_(2))=(2.0xx10^(4)Pa xx112.0xx10^(-3)m^(3))/(4.0xx10^(4)Pa)`<br/>`= 64.2xx10^(-3)m^(3)`<br/>(Note : As the pressure divided the volume become half.)</body></html> | |
| 51189. |
300 पर एक अभिक्रिया के लिए साम्य स्थिरांक 10 है| DeltaG^(ө) का मान क्या होगा ? R=8.314 JK^(-1) "mol"^(-1) |
| Answer» <html><body><p>`-5.74kJ`<br/>`-574kJ`<br/>`+11.48kJ`<br/>`+5.74kJ`</p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/deltag-2053246" style="font-weight:bold;" target="_blank" title="Click to know more about DELTAG">DELTAG</a>= -2.303RT log <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>` <br/> `=-2.303xx8.314xx300 log 10` <br/> `=-5744.1 <a href="https://interviewquestions.tuteehub.com/tag/j-520843" style="font-weight:bold;" target="_blank" title="Click to know more about J">J</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> -5.74 <a href="https://interviewquestions.tuteehub.com/tag/kj-1063034" style="font-weight:bold;" target="_blank" title="Click to know more about KJ">KJ</a>`</body></html> | |
| 51190. |
30 Volumes H_(2)O_(2)means |
| Answer» <html><body><p>`30% H_(2)O_(2)`<br/>`30 cm^(3)` of the <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/contains-11473" style="font-weight:bold;" target="_blank" title="Click to know more about CONTAINS">CONTAINS</a> 1 g of `H_(2)O_(2)`<br/>`1 cm^(3)` of the solution <a href="https://interviewquestions.tuteehub.com/tag/liberates-1072974" style="font-weight:bold;" target="_blank" title="Click to know more about LIBERATES">LIBERATES</a> `30 cm^(3)` of `O_(2)` at STP<br/>`30 cm^(3)` of the solution contain <a href="https://interviewquestions.tuteehub.com/tag/one-585732" style="font-weight:bold;" target="_blank" title="Click to know more about ONE">ONE</a> mole of `H_(2)O_(2)`</p>Answer :C</body></html> | |
| 51191. |
30 volumes H_2O_2 means.... |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/30-304807" style="font-weight:bold;" target="_blank" title="Click to know more about 30">30</a>% `H_2O_2` <br/>30 `"cm"^3` of the <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/contains-11473" style="font-weight:bold;" target="_blank" title="Click to know more about CONTAINS">CONTAINS</a> 1 g of `H_2O_2`<br/>1 `"cm"^3` of the solution <a href="https://interviewquestions.tuteehub.com/tag/liberates-1072974" style="font-weight:bold;" target="_blank" title="Click to know more about LIBERATES">LIBERATES</a> 30 `"cm"^3` of `O_2` at STP<br/>30 `"cm"^3` of the solution liberates 30 `"cm"^3` of `O_2` at STP </p>Answer :B</body></html> | |
| 51192. |
30 volume H_(2)O_(2) means _________. |
| Answer» <html><body><p>30% (w`//`v) `H_(2)O_(2)` solution <br/>`30 <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)` of the solution <a href="https://interviewquestions.tuteehub.com/tag/contains-11473" style="font-weight:bold;" target="_blank" title="Click to know more about CONTAINS">CONTAINS</a> 1g of `H_(2)O_(2)`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> cm^(3)` of the solution <a href="https://interviewquestions.tuteehub.com/tag/liberates-1072974" style="font-weight:bold;" target="_blank" title="Click to know more about LIBERATES">LIBERATES</a> `<a href="https://interviewquestions.tuteehub.com/tag/30cm-306639" style="font-weight:bold;" target="_blank" title="Click to know more about 30CM">30CM</a>^(3)` of `O_(2)` at S.T.P <br/>`30 cm^(3)` of the solution contains one mole of `H_(2)O_(2)`</p>Answer :C</body></html> | |
| 51193. |
30% solution of hydrogen peroxide is well known as |
| Answer» <html><body><p>Perhydrol<br/>10 <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> peroxide<br/>70% w/w `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_(2)`<br/>All of these</p>Answer :A</body></html> | |
| 51194. |
30 mL of N//10 HCl are required to neutralise 50 mL of a sodium carbonate solution. How many mL of water must be added to 30 mL of this solution so that the solution obtained may have a concentration equal to N//50 ? |
| 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> | |
| 51195. |
30 mL of mixture of methane and ethane in x:y ratio by volumes on combustion gave 40 mL of carbon dioxide. If 30 mL of mixture is taken in y: x ratio, what is the volume of carbondioxide obtained under similar conditions? |
| Answer» <html><body><p></p>Solution :The combustion of methane and ethane is given by the equations <br/> `CH_(4)overset(O_(2))rarrCO_(2)+2H_(2)O, C_(2)H_(6)overset(O_(2))rarr 2CO_(2)+3H_(2)O` <br/> 1 volume of `CH_(4) = 1` volume of `CO_(2)` <br/> 1 volume of `C_(2)H_(6) = 2` volume of `CO_(2)` <br/> Let, volume of `CH_(4)` in the given mixture `= <a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>` <br/> Let, volume of `C_(2)H_(6)` in the given mixture `= y` <br/> Then, `x + y = 30""x + <a href="https://interviewquestions.tuteehub.com/tag/2y-301307" style="font-weight:bold;" target="_blank" title="Click to know more about 2Y">2Y</a> = 40` <br/> `y = 10""x = 30 - 10 = 20` <br/> If y mL of `CH_(4)` is taken, <br/> the volume of `CO_(2)` obtained = y = 10 mL <br/> If x mL of `C_(2)H_(6)` is taken, <br/> the volume of `CO_(2)` obtained `= 2x = 2 xx 20 = 40 mL` <br/> The volume of carbondioxide obtained by the combustion of 30 mL of `y: x` volumes of methane and ethane `= 10 + 40 = 50 mL`</body></html> | |
| 51196. |
30 mL of K_(2)Cr_(2)O_(7) liberated iodine from KI solution when the iodine was titrated with hypo solution (N//20), the titre value was 45 mL. Find the concentration of K_(2)Cr_(2)O_(7) in g per litre. |
| 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> | |
| 51197. |
30 mL of a H_(2)O_(2) solution after acidification required 30 mL of N/10 KMnO_(4)solution for complete oxidation . Calculate the percentage and volume strength of H_(2)O_(2) solution. |
| Answer» <html><body><p></p>Solution :To determine the normality of `H_(2)O_(2)` solution . From the given <a href="https://interviewquestions.tuteehub.com/tag/data-25577" style="font-weight:bold;" target="_blank" title="Click to know more about DATA">DATA</a>,<br/>for `H_(2)O_(2), 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>)=30 mL, N_(1)=?` <br/>For `KMnO_(4), V_(2)=30 mL, N_(2)=N//10` <br/>Applying normallity equation , `N_(1)V_(1)=N_(2)V_(2) , i.e., 30xxN_(1)=30xx1//10 therefore N_(1)=0.1 N` <br/>Thus, the normality of `H_(2)O_(2)` solution =0.1 N <br/>Step 2. To determine the percentage <a href="https://interviewquestions.tuteehub.com/tag/strength-1229153" style="font-weight:bold;" target="_blank" title="Click to know more about STRENGTH">STRENGTH</a> of `H_(2)O_(2)` solution, <br/>We know that, `H_(2)O_(2) to 2H^(+) + O_(2) + 2e^(-)""therefore"` Eq. wt. of `H_(2)O_(2) =34//2=17` <br/>Hence, strength of `H_(2)O_(2)=(1.7xx100)/(1000)=0.17%` <br/>Step 3. To determinethe <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> strength of `H_(2)O_(2)` solution. <br/>Consider the chemical equation, <br/>`underset(68 g)(2H_(2)O_(2)) to 2H_(2)O + underset(22400 mL at N.T.P.)O_(2)` <br/>Now 68 g of `H_(2)O_(2)` give `O_(2)` at N.T.P. =22400 mL <br/>`therefore 1.7 g ` of `H_(2)O_(2)` will give `O_(2)=(22400)/(68)xx1.7=560` mL <br/> But 1.7 g of `H_(2)O_(2)` are present in 1000 mL of `H_(2)O_(2)` solution <br/>Hence, 1000 mL of `H_(2)O_(2)` solution gives 560 mL of `O_(2)` at N.T.P. <br/>`therefore ` 1 mL of `H_(2)O_(2)` solution will give `=(560)/(1000)=0.56 ` mL of `O_(2)` at N.T.P. <br/>`therefore` Volume strength of `H_(2)O_(2)` solution =0.56</body></html> | |
| 51198. |
30 ml of 20 Vol of H_(2)O_(2) 40 ml of 20Vol H_(2)O_(2) and 30 ml10 Vol of H_(2)O_(2) dissolvedin all proprotios. The normality of the resulting solution 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> :`(V_1 N_1 + V_2 N_2 + V_3N_3)/(V_1 + V_2 + V_3) = <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>`</body></html> | |
| 51199. |
30 mL H_(2)and 20 L O_(2)reacts to form water then what will left at the end of reaction ? |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a> <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> `H_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)`<br/>5 mL `O_(2)`<br/>10 mL `H_(2)` <br/>10 mL `O_(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/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 51200. |
30 lit of oxygen is liberated at STPby the complete decompostion of a sample of 1 lit of H_(2)O_(2). The molarity of that H_(2)O_(2) solution is nearly |
| Answer» <html><body><p>`0.893M`<br/>`1.785M`<br/>`2.679M`<br/>`3.572M`</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> | |