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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.
| 51051. |
5 g of a sample of magnesium carbonate on treatment with excess of dilute hydrochloric acid gave 1.12 L of CO_(2) at STP . The percentage of magnesium carbonate in the mixture is |
| Answer» <html><body><p>42<br/>40<br/>84<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> | |
| 51052. |
5 g of a sample of brass were dissolved in 1 litre dil. H_(2)SO_(4). 20 mL of this solution were mixed with KI and liberated iodine required 20 mL of 0.0327 N hypo solution for titration. Calculate the amount of copper in the alloy. |
| Answer» <html><body><p></p>Solution :When brass is extracted with concentrated `H_(2)SO_(4)`, it <a href="https://interviewquestions.tuteehub.com/tag/gives-1007647" style="font-weight:bold;" target="_blank" title="Click to know more about GIVES">GIVES</a> copper sulphate. <br/> `2[Cu+2H_(2)SO_(4)to CuSO_(4)+SO_(2)+2H_(2)O]` <br/> `2CuSO_(4)+4KI to 2K_(2)SO_(4)+2CuI_(2)` <br/> `2CuI_(2) to Cu_(2)I_(2)+I_(2)` <br/> `2Na_(2)S_(2)O_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)+I_(2) to 2NaI+Na_(2)S_(4)O_(6)` <br/> 2moles `Cu -= 1 " mole" I_(2) 0-2` mole <a href="https://interviewquestions.tuteehub.com/tag/hypo-1034629" style="font-weight:bold;" target="_blank" title="Click to know more about HYPO">HYPO</a> <br/> 20 mL of solution reacts with 20 mL of 0.0327 <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> hypo <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> 1000 mL` of solution will react with 1000 mL of 0.0327 N hypo <br/> No. of moles of hypo used `=("Mass")/("Molecular mass" (158))` <br/> `=(Exx NxxV)/(1000xx158)` <br/> where, E=158, N=0.327 given, V=1000 mL <br/> `therefore` No. of moles of hypo used `=(158xx0.0327xx1000)/(1000xx158)=0.0327` <br/> No. of moles of Cu=No. of moles of hypo =0.0327 mole <br/> Mass of copper in brass `=0.0327xx63.5=2.07645` <br/> % of copper in brass `=(2.07645)/(5)xx100=41.529%`</body></html> | |
| 51053. |
4Mg+10HNO_(3)to4Mg(NO_(3))_(2)+NH_(4)NO_(3)+3H_(2)O. In this reaction |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/96-342541" style="font-weight:bold;" target="_blank" title="Click to know more about 96">96</a> gms magnesium can reduce one mole of `HNO_(3)`<br/><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 reduced `HNO_(3)` is `(1)/(8)th` of its molecular weight <br/>Entire `HNO_(3)` <a href="https://interviewquestions.tuteehub.com/tag/involved-7257329" style="font-weight:bold;" target="_blank" title="Click to know more about INVOLVED">INVOLVED</a> in the reaction is reduced <br/>`HNO_(3)` is reduced to the best possible extent</p>Solution :Out of 10 <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 `HNO_(3)`, only 1 mole gets reduced by 4 moles of Mg. <br/> n - f of reduced `HNO_(3)=(+5-(-3))=8`</body></html> | |
| 51054. |
4H_3BO_3 + X oversetDeltato Na_2B_4O_7 + 6H_2O + Y In this reaction, X and Y are ...... respectively. |
| Answer» <html><body><p>`NaBO_2 , CO_2`<br/>`Na_2CO_3 , CO_2`<br/>`NaHCO_3 , NaBO_2`<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> , CO_2`</p>Solution :According to following <a href="https://interviewquestions.tuteehub.com/tag/formula-464310" style="font-weight:bold;" target="_blank" title="Click to know more about FORMULA">FORMULA</a>, on neutralization of Boric <a href="https://interviewquestions.tuteehub.com/tag/acid-847491" style="font-weight:bold;" target="_blank" title="Click to know more about ACID">ACID</a> `(H_3BO_3)` by `Na_2CO_3`(Sodium carbonate) is borax `(Na_2B_4O_7)` obtained <br/>`underset"Boric acid"(4H_3BO_3) + underset"Sodium carbonate"underset"(<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>)"(Na_2CO_3) oversetDeltato underset"Borax"(Na_2B_4O_7)+ 6H_2O + underset"(Y)"(CO_2)`</body></html> | |
| 51055. |
4g of hydrogen is burnt with 4g of oxygen. How many grams of water can be formed? |
| Answer» <html><body><p>0.5<br/>2.5<br/>4.5<br/>8</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> | |
| 51056. |
4 g of copper chloride on analysis was founded to contain 1.890 g of copper (Cu) and 2.110 g of chlorine (Cl). What is the empirical formula of copper chloride? [At. Mass of Cu= 63.5 u, Cl = 35.5 u]. |
| 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> : <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/SPH_BSR_CHE_QB_XI_MQP_01_E02_019_S01.png" width="80%"/></body></html> | |
| 51057. |
4g of a hydrocarbon on complete combustion gave 12.571g of CO_(2) and 5.143 g of water. The compound may be "i) "C_(2)H_(4)"ii) " CH_(4)"iii) "C_(3)H_(8)"iv) "C_(4)H_(8) The correct combination is |
| Answer» <html><body><p>Only (i) is correct <br/>Only (ii) is correct <br/>Only (ii) & (iii) are correct <br/>(i) & (<a href="https://interviewquestions.tuteehub.com/tag/iv-501699" style="font-weight:bold;" target="_blank" title="Click to know more about IV">IV</a>) are correct</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> | |
| 51058. |
4g NH_(4)NO_(3) were dissolved in 100 g water in bomb calorimeter with heat capacity of calorimether system 150JK^(-1).the temperature dropped by 1.5 K Enthalpy of solution of NH_(4)NO_(3)is : |
| Answer» <html><body><p>`450KJmol^(-<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/>`-450KJmol^(-1)`<br/>`4.5KJmol^(-1)`<br/>`-4.5KJ <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)`</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> | |
| 51059. |
4g H_(2),32g" "O_(2),14 g N_(2) and 11g CO_(2) are taken in a bulb of 500ml. Which one of these has maximum active mass ? |
| Answer» <html><body><p>`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/>`O_(2)`<br/>`N_(2)` <br/>`CO_(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</body></html> | |
| 51060. |
4Fe(s)+3O_(2(g)) to 2Fe_2O_3(g). The value of DeltaS is -550 JK^(-1) and the value of DH is -1650 kJ at 298 K. Does the process is spontaneous or not? |
| Answer» <html><body><p></p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/delta-947703" style="font-weight:bold;" target="_blank" title="Click to know more about DELTA">DELTA</a> S _("<a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a>") = + 4980 JK ^(-1) <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> ^(-1)` <a href="https://interviewquestions.tuteehub.com/tag/spontaneous-632196" style="font-weight:bold;" target="_blank" title="Click to know more about SPONTANEOUS">SPONTANEOUS</a></body></html> | |
| 51061. |
4ClO_(3(aq))^(-)to3ClO_(4(aq))^(-)+Cl_((aq))^(-) is an example of |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/oxidation-588780" style="font-weight:bold;" target="_blank" title="Click to know more about OXIDATION">OXIDATION</a> reaction<br/>Reduction reaction<br/>Decomposition reaction<br/>Disproportionation reaction</p>Answer :D</body></html> | |
| 51062. |
4.9 gm/lit of H_(2)SO_(4) is given ........... Is normality |
| Answer» <html><body><p>0.2<br/>20<br/>10<br/>0.1</p>Solution :`H_(2)SO_(4)` equivalent weight `=(98)/(2) = <a href="https://interviewquestions.tuteehub.com/tag/49-318440" style="font-weight:bold;" target="_blank" title="Click to know more about 49">49</a> gm/"equt"^(-1)` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>=("gm/litre")/("equivalent")=(4.9)/(49)=0.1`</body></html> | |
| 51063. |
4.9 g of KClO_(3) when heated produced 1.92 g of oxygen and the residue of KCl left was found to weigh 2.96 g. Show that the data illustrates the law of conservation of mass. |
| Answer» <html><body><p></p>Solution :`2KClO_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)rarr2KCl+3O_(<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/> <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 reactants =4.9 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> , Mass of products`= (1.92 + 2.96) = 4.88 g` <br/> As the mass of the products is nearly same as that of the reactants, this illustrates the Law of conservation of mass.</body></html> | |
| 51064. |
4.9 g of H_2SO_4 decompedes x.g of NaCl to give 5g of sodium hydrogen sulphate and 1.825g of hydrochloric acid. Then .x. is |
| Answer» <html><body><p>6.92<br/>4.65<br/>2.925<br/>1.41</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> | |
| 51065. |
48 g of Mg contains the same number of atoms as 160 g of another element . The atomic mass of the element is |
| Answer» <html><body><p>24<br/>320<br/>80<br/>40</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> | |
| 51066. |
4.6 g of sodium is treated with excess of water. Calcium the volume of hydrogen evolved at N.T.P |
| Answer» <html><body><p><br/></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>(2xx23=46g)(2Na+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>)O)<a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> underset(22400mL)(2NaOH+H_(2))` <br/> 46 g of Na evolve hydrogen at N.T.P = 22400 mL <br/> 4.6 g of Na evolve hydrogen at N.T.P `= (22400xx4.6)/(46)=2240mL`.</body></html> | |
| 51067. |
46 g of ethanol contains |
| Answer» <html><body><p><br/></p>Solution :`C_(<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_(<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)OH`= <a href="https://interviewquestions.tuteehub.com/tag/molecular-562994" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULAR">MOLECULAR</a> mass =`(12xx2)+(1xx6)+(1xx16)` <br/>`=24+6+16=46` <br/>2 Carbon atoms are present <br/>`:.2xx6.023xx10^(<a href="https://interviewquestions.tuteehub.com/tag/23-294845" style="font-weight:bold;" target="_blank" title="Click to know more about 23">23</a>)`C atoms is correct.</body></html> | |
| 51068. |
45.6 L of dinitrogen reacted with 22.7 L of dioxygen and 45.4 L of nitrous oxide was formed. The reaction is given below : 2N_(2(g))+O_(2(g)) rarr 2N_(2)O_((g)) Which law is being obeyed in this experiment ? Write the statement of the law ? |
| Answer» <html><body><p> </p>Solution :For the reaction,`{:(2N_(2(g)),+,O_(2(g)),rarr,2N_(2)O_((g)),,),(<a href="https://interviewquestions.tuteehub.com/tag/2v-300496" style="font-weight:bold;" target="_blank" title="Click to know more about 2V">2V</a>,,<a href="https://interviewquestions.tuteehub.com/tag/1v-282855" style="font-weight:bold;" target="_blank" title="Click to know more about 1V">1V</a>,,2V,,):}` <br/> `(45.4)/(22.7)=2 (22.7)/(22.7)= 1 (45.4)/(22.7)=2` <br/> Hence, the ratio between the volumes of the reactants and the product in the given question is <a href="https://interviewquestions.tuteehub.com/tag/simple-1208262" style="font-weight:bold;" target="_blank" title="Click to know more about SIMPLE">SIMPLE</a> i.e., `2:1:2`. It follows the Gay-Lussac.s law of gaseous solution. <br/> Note : Gay-Lussac.s law of gaseous volumes, when gases combine or are produced in a <a href="https://interviewquestions.tuteehub.com/tag/chemical-914796" style="font-weight:bold;" target="_blank" title="Click to know more about CHEMICAL">CHEMICAL</a> reaction, they do so in a simple ratio by volume provided all gases are at same temperature and <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a>.</body></html> | |
| 51069. |
4.5g of urea (molar mass = 60g "mol"^(-1)) are dissolved in water and solution is made to 100 ml in a volumetric flask. Calculate the molarity of solution. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Mass of Urea = 4.5 <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/> Moles of Urea `=("Mass ")/( "Molar Mass ")=(4.5g)/(60 g " <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> "^(-1))` <br/> `=0.075` mol <br/> Volume of Solution = <a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a> ml =`(100)/(1000)L` <br/> `=0.1 L` <br/> Molarity of Solution =`(" Mass of solute in g")/("Volume of Solution in litres ")` <br/> `=(0.075)/(0.12) " mol " = 0.75`M</body></html> | |
| 51070. |
4.5 mol each of H_(2) and I_(2) are present in a 10 lit vessel . At equilibrium the mixture contains 3 mole . HI , Kp for H_(2) + I_(2) hArr 2 HI and mass of I_(2) at equilibrium (in g) respectively are |
| 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> , 0.3`<br/>`1, 381`<br/>`1 , 762`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a> , 762`</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> | |
| 51071. |
4.5 moles, each of hydrogen and iodine was heated in a sealed 10 L vessel. At equilibrium, 3 moles of HI were found. The equilibrium constant for H_(2) (g) + I_(2) (g) rarr 2HI (g), is |
| Answer» <html><body><p><br/></p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_B_C04_E01_073_S01.png" width="80%"/> <br/> at <a href="https://interviewquestions.tuteehub.com/tag/equlibrium-452655" style="font-weight:bold;" target="_blank" title="Click to know more about EQULIBRIUM">EQULIBRIUM</a> no. 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 HI = 3=<a href="https://interviewquestions.tuteehub.com/tag/2x-301182" style="font-weight:bold;" target="_blank" title="Click to know more about 2X">2X</a> `implies` <a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>=1.5 <br/> `K_(c)=((2x)^(2))/((4.5 - x)^(2))=1`</body></html> | |
| 51072. |
4.4g of an oxide of nitrogen gives 2.24L of nitrogen and 60g of another oxide of nitrogen gives 22.4L of nitrogen at S.T.P. The data illustrates |
| Answer» <html><body><p>Law of <a href="https://interviewquestions.tuteehub.com/tag/conservation-929810" style="font-weight:bold;" target="_blank" title="Click to know more about CONSERVATION">CONSERVATION</a> of mass<br/>Law of constant proportions<br/>Law of <a href="https://interviewquestions.tuteehub.com/tag/multiple-1105557" style="font-weight:bold;" target="_blank" title="Click to know more about MULTIPLE">MULTIPLE</a> proportions<br/>Law of <a href="https://interviewquestions.tuteehub.com/tag/reciprocal-1180061" style="font-weight:bold;" target="_blank" title="Click to know more about RECIPROCAL">RECIPROCAL</a> proportions </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> | |
| 51073. |
4.48 L of an ideal gas at STP requires 12.0 calories to raise its temperature by 15^(@)Cat constant volume .The C_(p) of the gas is |
| Answer» <html><body><p>3 <a href="https://interviewquestions.tuteehub.com/tag/cal-412083" style="font-weight:bold;" target="_blank" title="Click to know more about CAL">CAL</a><br/>4 cal<br/>7 cal<br/> 6 cal</p>Solution :No. of molesin 4.48 L of ideal gas at STP`= ( 4.48 )/( 22.4) = 0.2` <br/>Thus, to raise the <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a> of0.2 mol of the ideal gas through `15^(@)<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>`, heat absorbed `= 12 cal`. <br/> `:. ` To raise the temperature of 1 mol of the gas through `1^(@)C`, heat absorbed`= ( 12)/( 15) xx ( 1)/( 0.2) = 4 cal` i.e., `C_(v) = 4 cal` <br/> `:. C_(p) = C_(v) + <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>= 4+2 cal = 6 cal`.</body></html> | |
| 51074. |
4.48 L of an ideal gas at STP requires 12 cal to raise the temperature by 15^(0)C at constant volume. The C_(P) of the gas is ____ cal |
| 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><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :At constant volume heat released `= nC_(<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>)dT` <br/> So `12 = (4.48)/(22.4) xx C_(v) xx 15, C_(v) = 4` cal/mole <br/> We know that `C_(p) C_(V) + R = 4 + <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/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>` cal/mole</body></html> | |
| 51075. |
44.8 litres of H_2 and 67.2 litres of O_2are mixed at STP. The weight of the mixture is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/112g-1777508" style="font-weight:bold;" target="_blank" title="Click to know more about 112G">112G</a> <br/>22.4g <br/><a href="https://interviewquestions.tuteehub.com/tag/100g-265940" style="font-weight:bold;" target="_blank" title="Click to know more about 100G">100G</a> <br/>44.8g </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> | |
| 51076. |
448 mL of SO_2 at NTP is passed through 100 mL. of a 0.2 N solution of NaOH. Find the weight of the salt formed. |
| Answer» <html><body><p></p>Solution :`SO_2 + <a href="https://interviewquestions.tuteehub.com/tag/naoh-572531" style="font-weight:bold;" target="_blank" title="Click to know more about NAOH">NAOH</a> to NaHSO_3` <br/> m.e. of `NaOH = 0.2 xx 100 = 20. "...(Eqn. 1, Chapter 7)"` <br/> <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 NaOH `= ( 20)/( <a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>) = 0.02 ". .. (Eqn. 3, Chapter 7)"` <br/> Since 1 equivalent of NaOH combines with 1 mole of `SO_2` according to the above <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a>, <br/> `therefore` for `SO_2`, 1 mole = 1 equivalent, i.e., 1 equivalent of `SO_2` will occupy 22.4 litres at NTP. <br/> Equivalent of `SO_2 = ( 448)/( 22400)` <br/> `= 0.02` <br/> We see that number of equivalents of `SO_2` and that of NaOH are equal. Number of eq. of ` NaHSO_3` willalso be 0.02 by the law of <a href="https://interviewquestions.tuteehub.com/tag/equivalence-452629" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENCE">EQUIVALENCE</a>. <br/>`therefore` weight of `NaHSO_3` = equivalent of `SO_2xx` equivalent weight <br/> `=0.02 xx 104` <br/> `=2.08`g. <br/> (According to the given reaction, eq. wt. of `NaHSO_3=` mol. wt. `=104.` )</body></html> | |
| 51077. |
448 mL of a hydrocarbon, having C = 87.8%, H = 12.19%, weigh 1.64 g at S.T.P Determine the molecular formula of 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> :Calculation of <a href="https://interviewquestions.tuteehub.com/tag/empirical-445984" style="font-weight:bold;" target="_blank" title="Click to know more about EMPIRICAL">EMPIRICAL</a> formula: <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/NTN_HCS_ISC_CHE_XI_P1_C01_E09_018_S01.png" width="80%"/> <br/> The empirical formula of hydrocarbon = `C_3H_5` Calculation of molecular formula : <br/> One gram mole of any gas occupies 22.4 <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a> or 22400 mL at S.T.R <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>` 448 mL at S.T.R of hydrocarbon weigh = 1.64 g<br/> `therefore 22400` mL at S.T.R of hydrocarbon will weigh<br/> `=1.64/448 xx 22400 = <a href="https://interviewquestions.tuteehub.com/tag/82-339053" style="font-weight:bold;" target="_blank" title="Click to know more about 82">82</a>.0 g` <br/> Hence, gram molecular mass of hydrocarbon = 82.0 g Empirical formula mass = `(3 xx 12.01) + (5 xx 1.008) = 41.07` <br/> `n=("gram molecular mass")/("Empirical formula mass") = 82/(41.07) = 2` <br/> `therefore` The molecular formula `=2 xx C_(3)H_(5)= C_(6)H_(10)`</body></html> | |
| 51078. |
4.4 gms of a hydrocarbons one combustion produced 6.72 L of CO_(2) at STP and sufficient water vapour that can generate 0.4gms of H_(2) with excess Na. The hydrocarbon is .x. times heavier than CO_(2). What is .x.? |
| Answer» <html><body><p><br/></p>Solution :`2Na+2H_(2)Orarr2NaOH+H_(2)` <br/> `36g___________2g` <br/> `7.2g___________0.4g` <br/> `22.4LCO_(2)-STP-44g` <br/> `6.72LCO_(2)-STP-13.2g` <br/> `underset("propane")(C_(3)H_(8))+5O_(2)rarr3CO_(2)+4H_(2)O` <br/> `44g_______132g_______72g` <br/> `4.4g_______13.2g_______7.2g` <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=(<a href="https://interviewquestions.tuteehub.com/tag/44-316683" style="font-weight:bold;" target="_blank" title="Click to know more about 44">44</a>)/(44)=1`</body></html> | |
| 51079. |
4.4 grams of carbondioxide present in a vessel at certain temperature exerts a pressure of 1000 kPa. Keeping the temperature constant if the gas is evacuated to have a rediual pressure of 0.01 kPa,how many molecules are left in the vessel ? |
| 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/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a> <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(<a href="https://interviewquestions.tuteehub.com/tag/17-278001" style="font-weight:bold;" target="_blank" title="Click to know more about 17">17</a>)`</body></html> | |
| 51080. |
4.4 g of CO_(2) contains how many litre of CO_(2) at STP ? |
| Answer» <html><body><p>2.4 litre<br/>2.24 litre<br/>44 litre<br/>22.4 litre</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> | |
| 51081. |
4.4 g of an unknown gas occupies 2.24 L of volume under N.T.P. conditions. The gas may be : |
| Answer» <html><body><p>`CO_(<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/>CO<br/>`O_(2)`<br/>`SO_(2)`</p>Solution :2.24 L of gas <a href="https://interviewquestions.tuteehub.com/tag/correspond-935558" style="font-weight:bold;" target="_blank" title="Click to know more about CORRESPOND">CORRESPOND</a> to <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> = 4.4 g <br/> 22.4 L of gas correspond to mass `= (4.4)/(2.24) xx22.4` <br/> `= 44.0 g` <br/> The gas is `CO_(2)`.</body></html> | |
| 51082. |
4.4 g CO_(2)contain ........... oxygen atoms. |
| Answer» <html><body><p>`1.2044xx10^(<a href="https://interviewquestions.tuteehub.com/tag/23-294845" style="font-weight:bold;" target="_blank" title="Click to know more about 23">23</a>)`<br/>`1.2044xx10^(<a href="https://interviewquestions.tuteehub.com/tag/22-294057" style="font-weight:bold;" target="_blank" title="Click to know more about 22">22</a>)`<br/>`1.68xx10^(23)`<br/>`1.68xx10^(22)`</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::<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>::<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>::D</body></html> | |
| 51083. |
4.2g of baking soda on strong ignition in an open container leaves a residue of mass |
| Answer» <html><body><p>2.65 g<br/>3.1 g<br/>2.1g<br/>3.35g</p>Answer :A</body></html> | |
| 51084. |
4.28g of NaOH is dissolved in water and the solution is made to 250 cc. what will be the molarity of the solution ? |
| Answer» <html><body><p>0.615 mol `<a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</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/>0.428 mol `L^(-1)`<br/>0.99 mol `L^(-1)` <br/>0.301 mol `L^(-1)`<br/></p>Solution :No. moles of `<a href="https://interviewquestions.tuteehub.com/tag/naoh-572531" style="font-weight:bold;" target="_blank" title="Click to know more about NAOH">NAOH</a> =(4.28)/(40)=0.107` ltbr. <a href="https://interviewquestions.tuteehub.com/tag/volumeof-3265347" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUMEOF">VOLUMEOF</a> solution `=250cm^3` <br/> `M=n/(V("in L"))=(0.107)/(250)xx1000=0.428mol L^(-1)`</body></html> | |
| 51085. |
4.215 g of a metallic carbonate was heated in a hard glass tube and the CO_2 evolved was found to measure 1336 mL at 27°C and 700 mm pressure. What is the equivalent weight of the metal? |
| Answer» <html><body><p></p>Solution :Volume of `CO_2` at NTP `= ( 1336 xx 273)/(300) xx (700)/( 760)` <br/> `= <a href="https://interviewquestions.tuteehub.com/tag/tildephi-fltildephi-ck-1120-268501" style="font-weight:bold;" target="_blank" title="Click to know more about 1120">1120</a>` mL. <br/> Suppose the equivalent weight of the metal is E. <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>` equivalent weight of the metal carbonate `= (E + 30)` <br/> `(because "eq. wt. of"CO_(3)^(2-) = ( 60)/( 2) = 30)`. <br/> Now, equivalent of metallic carbonate = equivalent of `CO_2` <br/> `(4.215)/( E+30) = ( 1120("<a href="https://interviewquestions.tuteehub.com/tag/vol-723961" style="font-weight:bold;" target="_blank" title="Click to know more about VOL">VOL</a>. at NTP") )/( 11200 ("vol. of 1 eq. at NTP") )` <br/> `therefore E=12.15`.</body></html> | |
| 51086. |
4.215 g a metallic carbonate was heated in a hard glass tube and CO_2 evolved was found to measure 1336 mL at 27°C and 700 mm pressure. What is 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> :`("Mass of metal <a href="https://interviewquestions.tuteehub.com/tag/carbonate-909380" style="font-weight:bold;" target="_blank" title="Click to know more about CARBONATE">CARBONATE</a>")/("Mass of metal <a href="https://interviewquestions.tuteehub.com/tag/oxide-1144484" style="font-weight:bold;" target="_blank" title="Click to know more about OXIDE">OXIDE</a>") = ("Eq. mass of metal + Eq. mass of" CO_(3)^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>-))/("Eq. mass of metal + Eq. mass of " O^(2-))` <br/> and Mass of metal oxide = Mass of metal carbonate - Mass of `CO_(2)`</body></html> | |
| 51087. |
4.184g of benzoic acid was burnt in bomb calorimeter. The rise in the temperature is 10^@. The heat capacity of calorimeter and its contents is 2.644 kcal K^(-1). Calculate the heat of combustion of benzoic acid. |
| 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> :The heat of combustion = `-(h <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> q xx M)//m` <br/>Rise in <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a> of calorimeter and its <a href="https://interviewquestions.tuteehub.com/tag/contents-25449" style="font-weight:bold;" target="_blank" title="Click to know more about CONTENTS">CONTENTS</a>, `theta = 10^@ ` <br/> The heat capacity of calorimeter and its contents, `h = 2.644 xx 4.184 J K^(-1)`. The inass of combustible <a href="https://interviewquestions.tuteehub.com/tag/substance-1231528" style="font-weight:bold;" target="_blank" title="Click to know more about SUBSTANCE">SUBSTANCE</a>, (benzoic acid), `m = 4.184g`. The molar mass of benzoic acid, `M = 122g mol^(-1)`. <br/> The heat of combustion of benzoic acid <br/> `= (2.644 xx 4.184 xx 10 xx122)/(4.184) = -3226 kJ mol^(-1)`</body></html> | |
| 51088. |
40 mL of 0.05 M solution of sodium sesquicarbionate (Na_(2)CO_(3). NaHCO_(3).2H_(2)O) is titrated against 0.05 M HCl solution. X ml of HCl solution is used when phenophthalein is the indicator and y ml of HCl is used when methyl orange is the indicator in two separate titrations Hence (y-x) is |
| Answer» <html><body><p>80 mL<br/>30 mL<br/>120 mL<br/>none of these</p>Solution :In the first <a href="https://interviewquestions.tuteehub.com/tag/titration-710389" style="font-weight:bold;" target="_blank" title="Click to know more about TITRATION">TITRATION</a> <br/> <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a>. <a href="https://interviewquestions.tuteehub.com/tag/hcl-479502" style="font-weight:bold;" target="_blank" title="Click to know more about HCL">HCL</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>)/(2)` eq `NC_(2)CO_(3)` <br/> `(x xx0.05)/(1000)=(1)/(2)<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>(0.050xx40xx2)/(1000)impliesx=40ml` <br/> In the second titration <br/> Eq of HCl = Eq `NC_(2)CO_(3)+` Eq `NKHCO_(3)` <br/> `(yxx0.05)/(1000)=(40xx0.05xx2)/(1000)+(40xx0.05xx1)/(1000)` <br/> `y=120impliesy-x=80`</body></html> | |
| 51089. |
40mL of 0.2M oxalic acid can completely decolourise what volume of 0.1M acidified permanganate solution? |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :32mL</body></html> | |
| 51090. |
4.0g of argon gas has pressure P and temperature TK in a vessel. On keeping the vessel at 50^@C higher, 0.8g of argon was given out to maintain the pressure at P. The original temperature was ____xx 10^(-2) K. |
| 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_1 T_1 = n_2T_2 , 0.1 (T) = 0.08(T + <a href="https://interviewquestions.tuteehub.com/tag/50-322056" style="font-weight:bold;" target="_blank" title="Click to know more about 50">50</a>)` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/implies-1037962" style="font-weight:bold;" target="_blank" title="Click to know more about IMPLIES">IMPLIES</a> T = <a href="https://interviewquestions.tuteehub.com/tag/200-288914" style="font-weight:bold;" target="_blank" title="Click to know more about 200">200</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>`</body></html> | |
| 51091. |
40.05 mL of 1.0 M Ce^(4+) are required to titrate 20.0 mL of 1.0 M Sn^(2+) " to "Sn^(4+) . What is the oxidatin state of cerium in the reduction product ? |
| Answer» <html><body><p></p>Solution :The reactions occurring are : <br/> `Ce^(4+)+ "ne"^(-) rarr Ce^((4-n)+)` <br/> `Sn^(2+) rarr Sn^(4+)+2e^(-)` <br/> To balance the equation , (the no . Of electrons <a href="https://interviewquestions.tuteehub.com/tag/lost-537630" style="font-weight:bold;" target="_blank" title="Click to know more about LOST">LOST</a> = no of electrons <a href="https://interviewquestions.tuteehub.com/tag/gained-2665852" style="font-weight:bold;" target="_blank" title="Click to know more about GAINED">GAINED</a> ) multiply eq. (i) by 2 eq (ii) by n and add <br/> `2Ce^(4+)+ nSn^(2+) rarr Ce^((4-n)+)+Sn^(4+)` <br/> Moles of `Ce^(4+)` in 40.05mL of 1.0 M solution , <br/> `=(1.0)/(1000)xx40.05=40.05 xx10^(-3)mol` <br/> Now 2 mol of `Ce^(4+)` will oxidise n mole of `Sn^(2+)` <br/> `40.05 xx10^(-3) " mol of " Ce^(4+) ` will oxidise `Sn^(2+)` <br/> `=n/2 xx40.05xx10^(-3)` mol `=20.02n xx 10^(-3)` mol = `20.02 n xx10^(-3)` mol<br/> But moles of `Sn^(2+)` in 20.0 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> of 1.0 M solution <br/>`=(1.0)/(1000)xx20.0=20.0xx10^(-3)` mol <br/> `:. 20.02n xx10^(-3) mol = 20.0 xx10^(-3)` mol <br/> `:. n = 1` <br/> Hence 1 mol of electrons are required in the <a href="https://interviewquestions.tuteehub.com/tag/reduction-621019" style="font-weight:bold;" target="_blank" title="Click to know more about REDUCTION">REDUCTION</a> of each mol of `Ce^(4+)` ion. <br/> `Ce^(4+)+e^(-) rarr Ce^(3+)` <br/> `Ce^(3+)` is the reduction <a href="https://interviewquestions.tuteehub.com/tag/product-25523" style="font-weight:bold;" target="_blank" title="Click to know more about PRODUCT">PRODUCT</a>.</body></html> | |
| 51092. |
400 moles of van der Waals gas having b=0.02Lmo1^(-1) are contained in a 1000 litre vessel The temperature and pressure of the gas are 400K and 90 atm respectively Calculate the pressure of the gas at 700K . |
| 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> :`99.93 <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a>`</body></html> | |
| 51093. |
400 mL of N_(2) gas at 700 mm and 300 mL of H_(2) gas at 800 mm were introducedinto a vessel of 2 litres at the same temperature. Calculate the final pressure of the gas mixture. |
| 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/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 51094. |
400cc of N_2 at 600mm and 500cc of O_2 at 300mm are quantitatively transferred in to an empty vessel of X 2 L capacity. Calculate the pressure of the mixture and partial pressures. |
| 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/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a> of the final mixture `= P = (P_(1)V_(1) + P_(2)V_(2))/(V)` <br/> ` = (400 xx 600 + <a href="https://interviewquestions.tuteehub.com/tag/500-323288" style="font-weight:bold;" target="_blank" title="Click to know more about 500">500</a> xx 300)/(2000) = 195` mm <br/> <a href="https://interviewquestions.tuteehub.com/tag/partial-596556" style="font-weight:bold;" target="_blank" title="Click to know more about PARTIAL">PARTIAL</a> presssure of nitrogen ` = (P_(1)V_(1))/(V) = (400 xx 600)/(2000) = 120` mm <br/> Partial pressure of oxygen ` = (P_(2)V_(2))/(V) = (500 xx 300)/(2000) = <a href="https://interviewquestions.tuteehub.com/tag/75-334971" style="font-weight:bold;" target="_blank" title="Click to know more about 75">75</a>` mm</body></html> | |
| 51095. |
40 mol of underline(x) M KMnO_4 solution is required to react completely with 200 ml of 0.02 M oxalic acid solution in acidic medium. The value of underline(x) is |
| Answer» <html><body><p>0.04<br/>0.01<br/>0.03<br/>0.02</p>Answer :A</body></html> | |
| 51096. |
40 mL of mixture of H_(2) & O_(2) was placed in a gas burette at 18^(@) C and 1 atm . A spark was produced so that the formation of water was complete . The remaining pure gas had a volume of 10 mL of 18^(@) C & 1 atm . If the remaining gas was H_(2) what was initial mole % of H_(2) in mixture ? |
| Answer» <html><body><p>0.75<br/>0.25<br/>0.6<br/>0.45</p>Solution :Let the volume of `O_(2)` is the mixture be x mL <br/> `H_(2) + 1//2 O_(2) to H_(2) O (l) ` <br/> <a href="https://interviewquestions.tuteehub.com/tag/initial-499269" style="font-weight:bold;" target="_blank" title="Click to know more about INITIAL">INITIAL</a> vol. in ml `"" (40 - x)"" x` <br/> <a href="https://interviewquestions.tuteehub.com/tag/final-461168" style="font-weight:bold;" target="_blank" title="Click to know more about FINAL">FINAL</a> vol. in ml (40- <a href="https://interviewquestions.tuteehub.com/tag/3x-310805" style="font-weight:bold;" target="_blank" title="Click to know more about 3X">3X</a>) - 0. <br/> Final volume of `H_(2) = 10 mL ""` (<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/> `therefore 40 - 3x = 10 , x = 10 ` mL <br/> `therefore` initial volume of `H_(2) = 30` mL , Initial mole % of `H_(2) = (30)/(40) xx 100 = 75%`</body></html> | |
| 51097. |
40 ml of methane is completely burnt using 80 ml of oxygen at room temperature. The volume of gas left after cooling to room temperature is …………….. . |
| Answer» <html><body><p>40 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> `CO_(2)` gas<br/>40 ml `CO_(2)` gas and 80 ml `H_(2)O` gas<br/>60 ml `CO_(2)` gas and 60 ml `h_(2)O` gas <br/>120 ml `CO_(2)` gas</p>Solution :`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>(g)) + 2O_(2(g)) rarr CO_(2(g)) + 2H_(2)O_((1))` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_CHE_XI_SP_08_E01_001_S01.png" width="80%"/> <br/> <a href="https://interviewquestions.tuteehub.com/tag/since-644476" style="font-weight:bold;" target="_blank" title="Click to know more about SINCE">SINCE</a> the product was <a href="https://interviewquestions.tuteehub.com/tag/cooled-7329693" style="font-weight:bold;" target="_blank" title="Click to know more about COOLED">COOLED</a> to room temperature, water exists mostly as liquid. <a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a>, option</body></html> | |
| 51098. |
40 ml of methane is completely burnt using 80 ml of oxygen at room temperature. The volume of gas left after cooling to room temperature 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> ml of `CO_(2)` <a href="https://interviewquestions.tuteehub.com/tag/gas-1003521" style="font-weight:bold;" target="_blank" title="Click to know more about GAS">GAS</a><br/>40 ml `CO_(2)` gas and 80 ml `H_(2)O` gas<br/><a href="https://interviewquestions.tuteehub.com/tag/60-328817" style="font-weight:bold;" target="_blank" title="Click to know more about 60">60</a> ml of `CO_(2)` and 60 ml `H_(2)O` gas<br/>120 ml `CO_(2)` gas</p>Solution :`CH_(4(g)) + 2O_(2(g)) to CO_(2(g)) + 2H_(2)O_(l)` <br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_CHE_XI_V01_C01_E01_001_S01.png" width="80%"/> <br/>Since the product was cooled to room temperature, water exists <a href="https://interviewquestions.tuteehub.com/tag/mostly-7382870" style="font-weight:bold;" target="_blank" title="Click to know more about MOSTLY">MOSTLY</a> as liquid. Hence, option (a) is correct</body></html> | |
| 51099. |
40 mL of a mixture of Na_(2)CO_(3) and NaOH when titrated against N//10 HCl, the end point with phenolphthalein was reached at 25 mL of HCl and at this stage methyl orange was added, the quantity of acid further required for second end point was 5 mL. Calculate the amount of Na_(2)CO_(3) and NaOH in g//L of the solution. |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :N//A</body></html> | |
| 51100. |
40 ml of a hydrogen undergoes combustion in 260 ml of oxygen and gives 160 ml of carbondioxide. If all gases are measured under similar conditions of temperature and pressure the formula of hydrocarbon is |
| Answer» <html><body><p>`C_(3)H_(<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>)`<br/>`C_(4)H_(8)`<br/>`C_(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)H_(14)`<br/>`C_(3)H_(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)`</p>Solution :`C_(x)H_(y) +(x+(y)/(4)) O_(2) rarr xCO_(2) +(y)/(2)H_(2)O`</body></html> | |