Explore topic-wise InterviewSolutions in Class 11.

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51901.

0.96 g of HI were, heated to attain equilibrium 2HI hArr H_(2)+I_(2). The reaction mixture on titration requires 15.7 mL of N//10 hypo solution. Calculate the degree of dissociation of HI.

Answer» <html><body><p><br/></p>Solution :`{:(,2HI,hArr,H_(2),+,I_(2)),("moles at",<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>.96/128,,0,,0),(t=0,,,,,),(,=7.5xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>),,0,,0),("moles",(7.5xx10^(-3)-x)x/2,,,,x/2),("at <a href="https://interviewquestions.tuteehub.com/tag/equilibrium-974342" style="font-weight:bold;" target="_blank" title="Click to know more about EQUILIBRIUM">EQUILIBRIUM</a>",,,,,):}` <br/> mEq of `I_(2)` at equilibrium =mEq of hypo used for reaction mixture. <br/> `w_(I_(2))/Exx1000=15.7xx1/10` <br/> `:. (w/E)` of `I_(2)=1.57xx10^(-3)` <br/> `:.` moles of `I_(2)` formed `=(1.57xx10^(-3))/2=0.785xx10^(-3)` <br/> `x/2=0.785xx10^(-3) rArr x=1.57xx10^(-3)` <br/> `:.` Degree of dissociation of HI `("or " alpha_(HI))` <br/> `=("moles dissociated")/("moles <a href="https://interviewquestions.tuteehub.com/tag/taken-659096" style="font-weight:bold;" target="_blank" title="Click to know more about TAKEN">TAKEN</a>")=(1.57xx10^(-3))/(7.5xx10^(-3))` <br/> `alpha_(HI)=0.209` or `20.9%`</body></html>
Discussion
51902.

0.9367 g of cadmium combine with chlorine to form 1.5276 g of CdCl_2- Find the equivalent mass of cadmium.

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 cadmium taken = 0.9367 g <br/> Mass of `CdCl_2` formed = 1.5276 g <br/> Mass of chlorine that combines with 0.9367 g of cadmium = 1.5276 - 0.9367 = 0.5909 g <br/> <a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a>, The <a href="https://interviewquestions.tuteehub.com/tag/equivalent-446407" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENT">EQUIVALENT</a> weight of cadmium `=("Mass of the metal")/("Mass of chlorine") xx 35.5` <br/> `=0.9367/0.5909 xx 35.5 = 56.27`</body></html>
Discussion
51903.

0.92 g of an organic compound was analysed by combustion method. The mass of the U-tube increased by 1.08 g. what is the percentage of hydrogen in the compound?

Answer» <html><body><p>0.1304<br/>0.5217<br/>0.6521<br/>0.113</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/increase-1040383" style="font-weight:bold;" target="_blank" title="Click to know more about INCREASE">INCREASE</a> in mass of U-tube =1.08g mass of <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a> formed=1.08g <br/> 18g `H_(2)O-=2` g <a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a> <br/> `1.08" g "H_(2)O=(2)/(<a href="https://interviewquestions.tuteehub.com/tag/18-278913" style="font-weight:bold;" target="_blank" title="Click to know more about 18">18</a>)xx1.08` <br/> % of hydrogen=`(2)/(18)xx(1.08)/(0.92)xx100=13.04%`</body></html>
Discussion
51904.

0.9031 g of a mixture of NaCI and KCI on treatment with H_2SO_4 yielded 1.0784 g of a mixture of Na_2SO_4 and K_2SO_4.Calculate the percent composition of the mixture.

Answer» <html><body><p><br/></p>Solution :Suppose, the <a href="https://interviewquestions.tuteehub.com/tag/amount-374803" style="font-weight:bold;" target="_blank" title="Click to know more about AMOUNT">AMOUNT</a> of NaCI in the mixture is <a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a> g. .-. The amount of <a href="https://interviewquestions.tuteehub.com/tag/kci-527832" style="font-weight:bold;" target="_blank" title="Click to know more about KCI">KCI</a> in the mixture = 0.9031 - x g The corresponding <a href="https://interviewquestions.tuteehub.com/tag/chemical-914796" style="font-weight:bold;" target="_blank" title="Click to know more about CHEMICAL">CHEMICAL</a> equations are<br/> `underset(2 xx 58.44 g)(2NaCl) + H_(2)SO_(4) to underset(142.04 g) + 2HCl` <br/> `underset(2 xx 74.55 g)(2KCl) + H_(2)SO_(4) to underset(174.26 g)(K_(2)SO_(4)) + 2HCl` <br/> `therefore 2 xx 58.44` of NaCl give `Na_(2)SO_(4) = 142.04 g` <br/> `therefore x g` of NaCl will give `Na_(2)SO_(4) =(142.04)/(2 xx 58.44) xx x` <br/> Similarly, the amount of `K_2So_4` formed by (0.9031 x) g of KCl `=(174.26)/(2 xx 74.55) xx (0.9031 -x)` <br/> `therefore` The amount of the resulting mixture of sulphates =1.0784 g <br/> `therefore 142.04/(2 xx 58.44 xx x) + [174.26/(2 xx 74.55) xx (0.9031 -x)] = 1.0784` <br/> which gives x=0.4829 g <br/> `therefore` Percentage of NaCl in the given mixture <br/> `=0.4829/0.9031 xx 100 = 53.47` <br/> and the percentage of KCI in the mixture `=100-53.47 = 46.53`</body></html>
Discussion
51905.

0.900g of a solute was dissolved in 100 ml of benzene at 25^(@)C when its density is 0.879 g/ml. This solution boiled 0.250^(@)C higher than the boiling point of benzene. Molal elevation constant for benzene is "2.52 K.Kg.mol"^(-1). Calculate the molecular weight of the solute.

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`"Weight of benzene "=100xx0.879=87.9g` <br/> Molality of solution, m `=(0.900//M_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))/(87.9)xx1000` <br/> `=(900)/(87.9M_(2))` <br/> `DeltaT_(b)=K_(b)m" (or)0.52"<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>(900)/(87.9M_(2))` <br/> `therefore""M_(2)=(900xx2.52)/(87.9xx0.250)` <br/> `M_(2)="103.2g/mole"` <br/> `therefore""{:("Molecular weight"),("of the <a href="https://interviewquestions.tuteehub.com/tag/solute-1217068" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTE">SOLUTE</a>"):}}="103.2 g/mole"`</body></html>
Discussion
51906.

0.84g iron are containing X per cent of iron was takenin a solution containg all the iron in ferrous state. The solution required X ml of a porassium dichromate soltuionfor oxidation of ironcontent to ferric state. Calculate the strength of potassiumdichromate 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> :`7.35 <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/litre-1075864" style="font-weight:bold;" target="_blank" title="Click to know more about LITRE">LITRE</a>"`</body></html>
Discussion
51907.

0.80 g of a substance was digested with sulphuric acid and then distilled with an excess of caustic soda. The ammonia gas evolved was passed through 100 ml of 1N H_(2)SO_(4). The excess of the acid required 80 ml of 1N caustic soda solution for its complete neutralisation. Calculate the percentage of nitrogen in the organic compound.

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`<a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> = <a href="https://interviewquestions.tuteehub.com/tag/35-308504" style="font-weight:bold;" target="_blank" title="Click to know more about 35">35</a>%`</body></html>
Discussion
51908.

0.79 g of iron unite directly with 0.4 g sulphur to form ferrous sulphate. If 2.8 g of iron are dissolved in dilute HCl and excess of sodium sulphide solution is added, 4.4 g iron sulphite is precipitated. Show that the data is according to Law of constant composition.

Answer» <html><body><p></p>Solution :In the first sample : <br/> Ratio by mass of <a href="https://interviewquestions.tuteehub.com/tag/iron-1051344" style="font-weight:bold;" target="_blank" title="Click to know more about IRON">IRON</a> and sulphur is : 0.<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a> : 0.4 or 7 : 4 <br/> In the second sample <br/> Mass of iron dissolved = 2.8 <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/> Mass of iron sulphide formed = 4.4 g <br/> Mass of sulphuric reacted `= (4.4 - 2.8) = 1.6 g` <br/> Ratio of mass of iron and sulphur is : 2.8 : 1.6 or 7 : 4 <br/> Since the ratio by mass of Fe and S in the <a href="https://interviewquestions.tuteehub.com/tag/two-714195" style="font-weight:bold;" target="_blank" title="Click to know more about TWO">TWO</a> samples is the same, this <a href="https://interviewquestions.tuteehub.com/tag/verifies-1444819" style="font-weight:bold;" target="_blank" title="Click to know more about VERIFIES">VERIFIES</a> the Law of constant composition.</body></html>
Discussion
51909.

At 400 K 1.5 g of an unknown substance is dissolved in solvent and the solution is made to 1.5 L. Its osmotic pressure is found to be 0.3 bar. Calculate the molar mass of the unknown substance.

Answer» <html><body><p></p>Solution :`DeltaT_(b) = K_(b) m = K_(b) <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> W_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) xx 1000//M_(2) xx W_(1)` <br/> `M _(2) = K_(b)xx W_(2) xx 1000//Delta T _(b) xx W_(1)` <br/> `= 7.5 xx 0.75 xx 1000//0.15 xx 200 = 187.5 g mol ^(-1)`</body></html>
Discussion
51910.

0.759 g of silver salt of a dibasic acid was ignited to form 0.463 g of the silver residue. Calculate the molecular mass of the acid.

Answer» <html><body><p></p>Solution :Step 1: <a href="https://interviewquestions.tuteehub.com/tag/equivalent-446407" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENT">EQUIVALENT</a> <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of silver salt of the acid. <br/> Equivalent mass of silver salt of the acid `= ("Mass of silver salt")/("Mass of silver residue") xx <a href="https://interviewquestions.tuteehub.com/tag/108-266746" style="font-weight:bold;" target="_blank" title="Click to know more about 108">108</a>` <br/> `= (0.759)/(0.463) xx 108 = 177` <br/> Step II. Equivalent mass of acid <br/> Equivalent mass of acid `=` Equivalent mass of silver salt - Equivalent mass of Ag + Equivalent mass of H <br/> `= 177 - 108 + 1 = 70` <br/> Step III. Molecular mass of acid <br/> Molecular mass of acid = Equivalent mass `xx` basicity <br/> `= 70 xx 2 = <a href="https://interviewquestions.tuteehub.com/tag/140-273392" style="font-weight:bold;" target="_blank" title="Click to know more about 140">140</a>`</body></html>
Discussion
51911.

0.759 g of a silver salt of a dibasic organic acid on ignition left 0.463 g metallic silver. The equivalent mass of the acid is :

Answer» <html><body><p>70<br/>108<br/>60<br/>50</p>Answer :A</body></html>
Discussion
51912.

0.75 mole of solid A_(4)and 2 moles of gaseous O_(2)are heated in a sealed vessel, completely using up the reactants and producing only one compound. It is found that when the temperature is reduced to the initial temperature, the contents of the vessel exhibit a pressure equal to half the original pressure. What conclusions can be drawn from these data about the product of the reaction?

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :First method: As both `A_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)` and `O_(2)` are consumed <br/> `("<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" A_(4))/("moles of" O_(2)) =0.75/2.0` <br/> `=(("moles of A")//4)/(("moles of O")//2) = 0.75/2.0` <br/> `=("moles of A")/("moles of O") = 3/4`<br/> Thus, the empirical <a href="https://interviewquestions.tuteehub.com/tag/formula-464310" style="font-weight:bold;" target="_blank" title="Click to know more about FORMULA">FORMULA</a> of the product is `A_3O_4`. Further, as 2 moles of `O_(2)`give 1 mole of `A_3O_4` (for gases, pressure `prop` mole at constant temperature and <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a>), `A_3O_4` is also the molecular formula of the product.<br/> Second Method :`underset("0.75 mole")(A_(4)s) + underset("2.0 mole")(O_(2) g) to underset("1.0 mole")(A_(x)O_(y)(g))` <br/> Applying POAC for A atoms, <br/> `4 xx` moles of `A_(4) = x xx `moles of `A_(x)O_(y)` <br/> `4 xx 0.75 = x xx 1, x=3` <br/> Applying POAC for O atoms. <br/> `2 xx` moles of `O_(2)= y xx` moles of `A_(x)O_(y)` <br/> `2 xx 2 = y xx 1, y =4`<br/> In the following chapters, we shall apply the principle of atom conservation (POAC) along with the said rules in tackling the various problems encountered in chemical practice.</body></html>
Discussion
51913.

0.75 g platinichloride of monoacid base on ignition gave 0.245 g platinum. The molecular mass of the base is :

Answer» <html><body><p>`75.0`<br/>`93.0`<br/>100<br/>`80.0`</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> :If <a href="https://interviewquestions.tuteehub.com/tag/e-444102" style="font-weight:bold;" target="_blank" title="Click to know more about E">E</a> = Eq.wt.of base<br/>W = weight of platinichloride<br/>w = weight of platinum, then<br/>`(W)/(2E+410)=(<a href="https://interviewquestions.tuteehub.com/tag/omega-585625" style="font-weight:bold;" target="_blank" title="Click to know more about OMEGA">OMEGA</a>)/(195)`,<br/>`(0.75)/(2E+410)=(0.245)/(195)<a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> E = 93.5`</body></html>
Discussion
51914.

0.72 gm of an oxide of a metal M on reduction with H_(2) gave 0.64 g of the metal . The atomic weight of the metal is 64 . The empirical formula of the compound is

Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/mo-548568" style="font-weight:bold;" target="_blank" title="Click to know more about MO">MO</a><br/>`M_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) O`<br/>`MO_(2)`<br/>`M_(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>)`</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>
Discussion
51915.

0.70 g of a sample of Na_(2)CO_(3).xH_(2)O were dissolved in water and the volume was made to 100 ml. 20 ml of this solution required 19.8 ml of (N/10) HCl for complete neutralisation. The value of x is

Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a><br/>1<br/>4<br/>10</p>Solution :Eq. `Na_(2)CO_(3)` = eq. HCl <br/> `20xxN=19.8xx(1)/(10)impliesN=(19.8)/(200)` <br/> `(<a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a>)/(VxxM.wt)xx2=(19.8)/(200)impliesM.wt=141.4impliesx=2`</body></html>
Discussion
51916.

0.7 moles of potassium sulphate is allowed to react with 0.9 moles of barium chloride in aqueous solutions. The number of moles of the substance precipitated in the reaction is

Answer» <html><body><p>1.<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a> <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 potassium chloride <br/>0.7 moles of barium sulphate<br/>1.6 moles of potassium chloride<br/>1.6 moles of barium sulphate</p>Solution :`K_(2)SO_(4) +BaCl_(2) rarr BaSO_(4) + 2KCl 38`<br/> `CaCO_(3) rarr CaO +CO_(2)`</body></html>
Discussion
51917.

0.7 g of Na_(2)CO_(3).xH_(2)O was dissolved in water to make 100 mL solution, 20 mL of this solution required 19.8 mL of 0.1 N HCl for complete neutralisation. The value of x is :

Answer» <html><body><p>5<br/>2<br/>3<br/>4</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>
Discussion
51918.

Combustion of 0.6gm of an organic compound gave 1.17gm of carbondioxide. 0.84 gm of water. Vapour density of the compound is equal to 22.4. The compound contains carbon, hydrogen and nitrogen. Calculate the molecular formula.

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> :`C_(2)H_(7)<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>
Discussion
51919.

0.66 g of a compound gave 112 ml of nitrogen at STP in the Dumas method. The percentage of Nitrogen in the compound is

Answer» <html><body><p>25<br/>41.5<br/>42.4<br/>21.2</p>Answer :D</body></html>
Discussion
51920.

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

Answer» <html><body><p></p>Solution :Mass of the metal taken (`w_(1)) = 0.65` g Mass of 224 <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 hydrogen <a href="https://interviewquestions.tuteehub.com/tag/liberated-2789268" style="font-weight:bold;" target="_blank" title="Click to know more about LIBERATED">LIBERATED</a> <br/> `w_(2) = 2/(22400) <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 224 = 0.02 g` <br/> <a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a>, Equivalent weight of the metal <br/> `=w_(1)/w_(2) =0.65/0.02 = 32.5`</body></html>
Discussion
51921.

0.66 g of H_(3)PO_(2) will require x ml of 0.1M NaOH for complete neutralization. X is

Answer» <html><body><p>100 ml <br/>200 ml <br/><a href="https://interviewquestions.tuteehub.com/tag/300-305868" style="font-weight:bold;" target="_blank" title="Click to know more about 300">300</a> ml <br/>none of these</p>Solution :eq. `H_(3PO_(2)` = eq. NaOH <br/> `(0.<a href="https://interviewquestions.tuteehub.com/tag/66-331311" style="font-weight:bold;" target="_blank" title="Click to know more about 66">66</a>)/(66)<a href="https://interviewquestions.tuteehub.com/tag/xx1-1463705" style="font-weight:bold;" target="_blank" title="Click to know more about XX1">XX1</a>=(x xx0.1)/(<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>)xx1impliesx 100mL`</body></html>
Discussion
51922.

0.635 g of a metal gives on oxidation 0.795g g of its oxide. Calculate the equivalent mass of the metal.

Answer» <html><body><p></p>Solution :Mass of the <a href="https://interviewquestions.tuteehub.com/tag/metal-1094457" style="font-weight:bold;" target="_blank" title="Click to know more about METAL">METAL</a> oxide = 0.795g <br/>Mass of the metal= 0.635g <br/>Mass of oxygen= 0.795 - 0.635 = 0.16g<br/> 0.16 g of oxygen has <a href="https://interviewquestions.tuteehub.com/tag/combined-409631" style="font-weight:bold;" target="_blank" title="Click to know more about COMBINED">COMBINED</a> with 0.635 g of a metal<br/> `:.` 8g of oxygen will combine with <br/> ` (8 xx 0.635)/(0.16)=31.75g " equiv"^(-<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>
Discussion
51923.

0.63 grams of a dioic acid required 50mL of 0.2M NaOH for neutralisation. Calculate the molarmass of acid .

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/123-271064" style="font-weight:bold;" target="_blank" title="Click to know more about 123">123</a></body></html>
Discussion
51924.

0.6% solution of urea and 1.8% solution containing a solute (A) are isotonic with each other. Calculate the molecular weight of the solute (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> :`<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>=0.6% y=<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.8%, M_(x)="<a href="https://interviewquestions.tuteehub.com/tag/urea-1440871" style="font-weight:bold;" target="_blank" title="Click to know more about UREA">UREA</a>" (60 g "mol"^(-1)),M_(y)=?` <br/> `(x)/(100)xx(100)/(M_(x))=(y)/(100)xx(100)/(M_(y))=(x)/(M_(x))=(y)/(M_(y))` <br/> `M_(y)=(y xx M_(x))/(x)=(1.8xx60 g "mol"^(-1))/(0.6)` <br/> `=180 g "mol"^(-1)`</body></html>
Discussion
51925.

0.6 moles of K_(2)Cr_(2)O_(7) can oxidise-

Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>.6 mol of `FeSO_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)` to `Fe_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)(SO_(4))_(3)`<br/>0.1 mol of `FeSO_(4)` to `Fe_(2)(SO_(4))_(3)`<br/>0.05 mol of `Sn^(+2)` to `Sn^(+4)`<br/>1.8 mol of `Sn^(+2)` to `Sn^(+4)`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A::D</body></html>
Discussion
51926.

0.6 mol K_(2)Cr_(2)O_(7), in acid medium can oxidise :

Answer» <html><body><p>3.6 mol `FeSO_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>) " to " Fe_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)(SO_(4))_(3)`<br/>0.1 mol `FeSO_(4) " to " Fe_(2)(SO_(4))_(3)`<br/>0.05 mol of `Sn^(2+) " to" Sn^(4+)`<br/>1.8 mol of `Sn^(2+) " to " Sn^(4+)`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A::D</body></html>
Discussion
51927.

0.6 g of an organic compound was Kjeldhalised and NH_(3) evolved was absorbed into 50 ml of semi-normal solution of H_(2)SO_(4) . The residual acid solution was diluted with distilled water and the volume made up to 150 ml. 20 mL of this diluted solution required 35 mL of (N)/(20) NaOH solution for complete neutralisation. Calculate the % of N in the compound.

Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/weight-1451304" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHT">WEIGHT</a> of Organic compound = 0.6 g <br/> Volume of <a href="https://interviewquestions.tuteehub.com/tag/sulphuric-655420" style="font-weight:bold;" target="_blank" title="Click to know more about SULPHURIC">SULPHURIC</a> acid taken = 50 mL <br/> Strength of sulphuric acid taken=0.5 N <br/> 20 ml of diluted solution of unreacted sulphuric acid was neutralised by 35 ml of 0.05 N Sodium laydroxide <br/> `"Strength of the diluted sulphuric acid"=(35 xx 0.05)/(20) = 0.0875 N` <br/> Volume of the sulphuric acid remaining after reaction with ammonia= `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>)` mL <br/> Strength of `H_(2)SO_(4)` = 0.5 N <br/> Volume of the diluted `H_(2)SO_(4)` = <a href="https://interviewquestions.tuteehub.com/tag/150-275254" style="font-weight:bold;" target="_blank" title="Click to know more about 150">150</a> mL <br/> Strength of the diluted sulphuric acid = 0.0875 N <br/> `V_(1)= (150 xx 0.087)/(0.5) = 26.25 ml` <br/> Volume of `H_(2)SO_(4)` consumed by ammonia = 50 - 26.25 = 23.75 mL <br/> 23.75 mL of 0.5 N `H_(2)SO_(4)` - 23.75 mL of 0.5 N `NH_(3)` <br/> The amount of <a href="https://interviewquestions.tuteehub.com/tag/nitrogen-1118291" style="font-weight:bold;" target="_blank" title="Click to know more about NITROGEN">NITROGEN</a> present in the 0.6 g of organic compound <br/> `=(14 g)/(1000 mL xx 1 N) xx 23.75 xx 0.5 N=0.166 g` <br/> `"Percentage of nitrogen"=(0.166)/(0.6) xx 100=27.66%`</body></html>
Discussion
51928.

0.6 g of a metal gives on oxidation 1 g of its oxide. Calculate its equivalent mass.

Answer» <html><body><p></p>Solution :Mass of <a href="https://interviewquestions.tuteehub.com/tag/metal-1094457" style="font-weight:bold;" target="_blank" title="Click to know more about METAL">METAL</a> = 0.6<br/>Mass of metal oxide = <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> g<br/>`:.` Mass 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> = 1 - 0.6 = 0.4 g<br/>0.4 g of oxygen combines with 0.6 g of metal<br/>`:.` 8 g of oxygen will combine with `0.6/0.4 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 8`<br/>Equivalent mass of the metal = 12 g `eq^(-1)` .</body></html>
Discussion
51929.

0.5g sample of iron containg mineral mainly in the form of CuFeS_(2) was reduced suitably to vonvert all the ferruic ions into ferrous ions (Fe^(3+) rarr Fe^(2+)) and was obtained as solution. In the absenceof any interferringradical, the solution. In the absenceof any interfeerring radical, the solution required 42mL of interferring radical, the solution required 42mL of 0.1M K_(2)Cr_(2)O_(7) for tirtration. Calculate% of CuFeS_(2) in sample.

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> :`92.48%`</body></html>
Discussion
51930.

0.5g of an organic compound gave 62.2ml of N_(2) by Duma's method. Calculate the percentage of N in this compound.

Answer» <html><body><p><br/></p>Solution :% of <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> (By Duma's method) `= (<a href="https://interviewquestions.tuteehub.com/tag/28-299271" style="font-weight:bold;" target="_blank" title="Click to know more about 28">28</a> xx <a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a> xx 100)/(22400 xx <a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a>)` <br/> `= (28 xx 62.2 xx 100)/(22400 xx 0.5) = 15.55%`</body></html>
Discussion
51931.

0.5g of a metal on oxidation gave 0.79g of its oxide. The equivalent mass of the metal is

Answer» <html><body><p>10<br/>14<br/>20<br/>40</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>
Discussion
51932.

0.59g of the silver salt of an organic acid (mol.wt. 210) on ignition gave 0.36 g of pure silver. The basicity of the acid is [AW of Ag = 108]

Answer» <html><body><p>1<br/>2<br/>3<br/>4</p>Solution :`(<a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a>)/(E+107)=(omega)/(108)(0.59)/(E+107)=(0.36)/(108)rAr E = <a href="https://interviewquestions.tuteehub.com/tag/70-333629" style="font-weight:bold;" target="_blank" title="Click to know more about 70">70</a>`<br/><a href="https://interviewquestions.tuteehub.com/tag/basicity-394484" style="font-weight:bold;" target="_blank" title="Click to know more about BASICITY">BASICITY</a> `= ("Mol. wt")/("<a href="https://interviewquestions.tuteehub.com/tag/eg-445433" style="font-weight:bold;" target="_blank" title="Click to know more about EG">EG</a>. wt")=(<a href="https://interviewquestions.tuteehub.com/tag/210-293619" style="font-weight:bold;" target="_blank" title="Click to know more about 210">210</a>)/(70)=3`</body></html>
Discussion
51933.

0.59g of the silver salt of an organic acid (mol. wt. 210) on ignition gave 0.36g of pure silver. The basicity of the acid is [AW of Ag = 108]

Answer» <html><body><p>1<br/>2<br/>3<br/>4</p>Answer :A</body></html>
Discussion
51934.

0.59 g of an organic compound produces 112 mL nitrogen at NTP. The percentage of nitrogen in the compound through Duma's method is

Answer» <html><body><p>23.7<br/>27.5<br/>33.07<br/>16.8</p>Solution :`%` of N through Duma's method is <a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> by <br/> `%` of `=N=(28xx"<a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> of " N_(2)" at <a href="https://interviewquestions.tuteehub.com/tag/stp-633132" style="font-weight:bold;" target="_blank" title="Click to know more about STP">STP</a>")/(22400 xx "Mass of an <a href="https://interviewquestions.tuteehub.com/tag/organic-1138713" style="font-weight:bold;" target="_blank" title="Click to know more about ORGANIC">ORGANIC</a> compound")xx100` <br/> `=(28xx112xx100)/(22400xx0.59)=23.7`.</body></html>
Discussion
51935.

0.548 g of the metal reacts with dilute acid and liberates 0.0198 g of hydrogen at S.T.P. Calculate the equivalent mass of the metal.

Answer» <html><body><p></p>Solution :0.548 g of the <a href="https://interviewquestions.tuteehub.com/tag/metal-1094457" style="font-weight:bold;" target="_blank" title="Click to know more about METAL">METAL</a> <a href="https://interviewquestions.tuteehub.com/tag/displaces-956139" style="font-weight:bold;" target="_blank" title="Click to know more about DISPLACES">DISPLACES</a> 0.0198 g of hydrogen<br/> The mass of the metal which will <a href="https://interviewquestions.tuteehub.com/tag/displace-956065" style="font-weight:bold;" target="_blank" title="Click to know more about DISPLACE">DISPLACE</a><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>.008 g of hydrogen ` = (1.008 xx 0.548)/( 0.0198)g` of metal <br/> The equivalent mass of the metal `=27.90 g"equiv"^(-1)` .</body></html>
Discussion
51936.

0.532 g of the chloroplatinate of a diacid base gave 0.195 g of the platinum residue on ignition. Calaculate the molecular mass of the base.

Answer» <html><body><p></p>Solution :Mass of the base chloroplatinate taken `(W) = 0.532 <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/> Mass of the platinum residue `(x) = 0.<a href="https://interviewquestions.tuteehub.com/tag/195-281683" style="font-weight:bold;" target="_blank" title="Click to know more about 195">195</a> g` <br/> step I. Equivalent mass of the base <br/> Equivalent mass of the base `= (1)/(2) [(W)/(x) xx 195 - 410] = (1)/(2) [(0.532)/(0.195) xx 195 - 410]` <br/> `= (1)/(2) [532 - 410] = <a href="https://interviewquestions.tuteehub.com/tag/61-330126" style="font-weight:bold;" target="_blank" title="Click to know more about 61">61</a>` <br/> Step <a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>. Molecular mass of the base <br/> Molecular mass of the base = Equivalent mass `xx` <a href="https://interviewquestions.tuteehub.com/tag/acidity-847681" style="font-weight:bold;" target="_blank" title="Click to know more about ACIDITY">ACIDITY</a> `= 61 xx 2 = 122`</body></html>
Discussion
51937.

0.532 g of the chloroplatinate of an organic base (mol.wt 24 D) gave 0.195 g of Pt on ignition. Then the number of nitrogen atoms per molecule of the base is

Answer» <html><body><p>1<br/>2<br/>3<br/>4</p>Solution :`(<a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a>)/(2E+410)=(<a href="https://interviewquestions.tuteehub.com/tag/omega-585625" style="font-weight:bold;" target="_blank" title="Click to know more about OMEGA">OMEGA</a>)/(<a href="https://interviewquestions.tuteehub.com/tag/195-281683" style="font-weight:bold;" target="_blank" title="Click to know more about 195">195</a>)`<br/>`(0.532)/(2E+410)=(0.195)/(195)<a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> E=61`</body></html>
Discussion
51938.

0.530 g of an organic compound gave 0.90 g of BaSO_(4) in carius determination of sulphur. Calculate the percentage of sulphur.

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`S = 23.32%`</body></html>
Discussion
51939.

0.515 g of an organic compound containing phosphorus gave 0.214 g of magnesium pyrophosphate in Carius method for the estimation of phosphorus. Calculate the percentage of phosphorus in the given organic compound.

Answer» <html><body><p>&lt;<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>&gt;</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of organic compound `=0.515 g` <br/> Mass of <a href="https://interviewquestions.tuteehub.com/tag/magnesium-1082716" style="font-weight:bold;" target="_blank" title="Click to know more about MAGNESIUM">MAGNESIUM</a> pyrophosphate, `Mg_(2)P_(2)O_(7)=0.214 g` <br/> `{:(Mg_(2)P_(2)O_(7),equiv,2P),(22g,,62g):}` <br/> 222g of `Mg_(2)P_(2)O_(7)` are obtained from <a href="https://interviewquestions.tuteehub.com/tag/phosphorus-1153299" style="font-weight:bold;" target="_blank" title="Click to know more about PHOSPHORUS">PHOSPHORUS</a> `=62 g` <br/> 0.214 g of `Mg_(2)P_(2)O_(7)` are obtained from phosphorus `=(62xx0.214)/222=0.0597 g` <br/> Percentage of phosphorus `=("Mass of phosphorus")/("Mass of organic compound")xx100=(0.0597xx100)/0.515=11.6`.</body></html>
Discussion
51940.

0.50 g of an organic compound was kjeldahlished. The ammonia evolved was passed in 50 cm^(3) of 1 N H_(2)SO_(4). The residual acid required 60 cm^(3) of N//2 NaOH solution. Calculate the pecentage of nitogen in the compound.

Answer» <html><body><p></p>Solution :Step I. calculation of volume of <a href="https://interviewquestions.tuteehub.com/tag/unused-7277696" style="font-weight:bold;" target="_blank" title="Click to know more about UNUSED">UNUSED</a> acid <br/> Volume of `NaOH` solution required `=<a href="https://interviewquestions.tuteehub.com/tag/60-328817" style="font-weight:bold;" target="_blank" title="Click to know more about 60">60</a> cm^(3)` <br/> Normality of `NaOH` solution `=1//2 N` <br/> Normality of `H_(2)SO_(4)` solution `= 1N` <br/> Volume of unused acid can be calculated by applying normality equation <br/> `ubrace(N_(1)V_(1))_("Acid")=ubrace(N_(1)V_(1))_("Base") <br/> 1xxV=1/2xx60=<a href="https://interviewquestions.tuteehub.com/tag/30-304807" style="font-weight:bold;" target="_blank" title="Click to know more about 30">30</a> cm^(3)` <br/> Step II. Calculation of volume of acid used <br/> Volume of acid added `=50 cm^(3)` <br/> Volume of unused acid `=30 cm^(3)` <br/> Volume of acid used `=(50-30)=20 cm^(3)` <br/> Step III. Calculating of percentage of nitrogen <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 compound `=0.50 g` <br/> Volume of acid used `=20 cm^(3)` <br/> Normality of acid used `= 1 N` <br/> <a href="https://interviewquestions.tuteehub.com/tag/persentage-2923290" style="font-weight:bold;" target="_blank" title="Click to know more about PERSENTAGE">PERSENTAGE</a> of `N=(1.4xx"Volume of acid used"xx"Normality of acid used")/("Mass of the compound")` <br/> `=(1.4xx20xx1)/0.50=56%`</body></html>
Discussion
51941.

0.50 g of an organic compound was Kjeldahlished. The ammonia evolved was passed in 50 cm^(3) of 1N H_(2)SO_(4). The residual acid required 60 cm^(3) of N // 2 NaOH solution. Calculate the percentage of nitrogen in the compound.

Answer» <html><body><p></p>Solution :Step 1. <a href="https://interviewquestions.tuteehub.com/tag/calculation-907729" style="font-weight:bold;" target="_blank" title="Click to know more about CALCULATION">CALCULATION</a> of volume of unused acid <br/> `"Volume of NaOH solution required" = 60 cm^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)` <br/> `"Normality of NaOH solution" =(1)/(2)N` <br/> `"Normality of" H_(2)SO_(4) "solution"= (1)/(N)` <br/> Volume of unused acid can be calculated by applying normality equation <br/> `underset("Acid")(underbrace(N_(1)V_(1)))=underset("<a href="https://interviewquestions.tuteehub.com/tag/base-892693" style="font-weight:bold;" target="_blank" title="Click to know more about BASE">BASE</a>")(underbrace(N_(1)V_(1)))` <br/> `1 xx V=(1)/(2) xx 60 = 30 cm^(3)` <br/> Step II. Calculation of volume of acid used <br/> `"Volume of acid added" = 50 cm^(3)` <br/> `"Volume of unused acid" = 30 cm^(3)` <br/> `"Volume of acid used"= (50-30) = 20 cm^(3)` <br/> Step III. Calculation of percentage of <a href="https://interviewquestions.tuteehub.com/tag/nitrogen-1118291" style="font-weight:bold;" target="_blank" title="Click to know more about NITROGEN">NITROGEN</a> <br/> Mass of compound = 0.50 g <br/> `"Volume of acid used"=20 cm^(3)` <br/> Normality of acid used = 1 N <br/> `"Percentage of N" = (1.4 xx "Volume of acid used" xx "Normality of acid used")/("Mass of the compound")` <br/> `=(1.4 xx 20 xx 1)/(0.50)=50%`</body></html>
Discussion
51942.

05 moles of NH_(4)HS(s) are taken in a container having air at 1 atm. On warming the closed container to 50^(@)C the pressure attained a constant value of 1.5 atm, with some NH_(4)HS(s) remaining unreacted. The K_(p) of reaction NH_(4)HS(s)hArrNH_(3)(g)+H_(2)S(g) "at" 50^@)C is:

Answer» <html><body><p>`0.25`<br/>`0.625`<br/>`0.025`<br/>`0.0625`</p>Solution :`NH_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)<a href="https://interviewquestions.tuteehub.com/tag/hs-492079" style="font-weight:bold;" target="_blank" title="Click to know more about HS">HS</a>(s)hArrNH_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</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>)+H_(2)S(g)` <br/> `P P` <br/> `2P=1.5-1` <br/> `2P=0.5` <br/> `P=0.25` <br/> `rArrK_(p)=0.25=0.0625`</body></html>
Discussion
51943.

0.5 moles of gaseousnon-metallic X^(-) anions (having positive electron affinity) requires 806.4 kJ energy to get completely converted into gaseous X^(+) ions. Calculate Pauling's electronegativity of the elements X. Use Avogardo's no = 6 xx 10^(23) and 1eV = 1.6 xx 10^(-19)J. [Use the fact that, Pauling's electronegativity =("Mulliken's electronegativity")/(2.8) and Mulliken's electronegativity =("Ionisation energy+ Electron affinity")/(2)]

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> :Let `X rarr X^(+)e^(-)` <br/> `:. <a href="https://interviewquestions.tuteehub.com/tag/ie-496825" style="font-weight:bold;" target="_blank" title="Click to know more about IE">IE</a>` (energy absorbed) = aeV <a href="https://interviewquestions.tuteehub.com/tag/per-590802" style="font-weight:bold;" target="_blank" title="Click to know more about PER">PER</a> atom and `X +e^(-) rarr X^(-)` <br/> `:. EA` (energy <a href="https://interviewquestions.tuteehub.com/tag/released-613817" style="font-weight:bold;" target="_blank" title="Click to know more about RELEASED">RELEASED</a>) =- beV per atom Now `(N)/(2)` of `X^(-)` will lose two elecctrons to give `(N)/(2)X^(+)` <br/> `:. X^(-) rarr X +E^(-)` <br/> `:.` energy absorbed `= + beV` per atom. <br/> `X rarr X^(+) +e^(-)` <br/> `:.` energy absorbed `=+ aeV` per atom. <br/> `:. a xx (N)/(2) + b xx (N)/(2) = (806.4xx 1000)/(1.6 xx <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(-19)) eV` <br/> or `(a+b) = (806.4 xx 1000 xx 2)/(1.6 xx 10^(-19) xx 6 xx 10^(23)) = (806.4 xx 2)/(1.6 xx 60) = 16.8` <br/> `xhi_(m) = (IE +EA)/(2) = (16.8)/(2) = 8.4` <br/> `:.` Mulliken's electronegativity = 8.4 <br/> Pauling's electronegativity `= (8.4)/(2.8) = 3.0`</body></html>
Discussion
51944.

0.5 mole of S_8 and 0.5 mol of P_4 have same number of polyatomic molecules

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/true-713260" style="font-weight:bold;" target="_blank" title="Click to know more about TRUE">TRUE</a></body></html>
Discussion
51945.

0.5 mole of P_(4)IO_(10) contains

Answer» <html><body><p>80 gm oxygen <br/>2 gram atoms <a href="https://interviewquestions.tuteehub.com/tag/phosphorous-599815" style="font-weight:bold;" target="_blank" title="Click to know more about PHOSPHOROUS">PHOSPHOROUS</a> <br/>5 gram atoms oxygen <br/><a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> gram atoms oxygen</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/molecular-562994" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULAR">MOLECULAR</a> weigth of `P_(4)O_(10)=284` <br/> 1 mole `P_(4)O_(10)=284g` <br/> `284 g P_(4)O_(10)` contains <a href="https://interviewquestions.tuteehub.com/tag/124-271143" style="font-weight:bold;" target="_blank" title="Click to know more about 124">124</a> g phosphorous <br/> `284g P_(4)O_(10)` contain 160 g oxygen <br/> 0.5 moles `P_(4)O_(10)-62 g P - 80 g` oxygen <br/> (2 gram atom) (5 gram atom)</body></html>
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51946.

0.5 mole of ethanol is mixed with 1.5 mole of water. Then the mole fraction of ethanol and water are ……………..

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> :`0.25, 0.75` <br/> <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> fraction of ethnol `= ("<a href="https://interviewquestions.tuteehub.com/tag/number-582134" style="font-weight:bold;" target="_blank" title="Click to know more about NUMBER">NUMBER</a> of moles of <a href="https://interviewquestions.tuteehub.com/tag/ethanol-975823" style="font-weight:bold;" target="_blank" title="Click to know more about ETHANOL">ETHANOL</a>")/("Total number of moles of ethanol and <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a>")` <br/> `=(0.5)/( 1.5 + 0.5) = (0.5)/(2.0 = 0.25` <br/> Mole fraction of water `= (1.5)/( 2.0) = 0.75`</body></html>
Discussion
51947.

0.5 mole of each of H_(2),SO_(2) and CH_(4) are kept in acontainer. A hole was made in the container. After 3 hours, the order of partial pressures in the container will be

Answer» <html><body><p>`p_(SO_(2)) gt p_(CH_(4)) gt p_(H_(2))`<br/>`p_(H_(2)) gt p_(SO_(2)) gt p_(CH_(4))`<br/>`p_(H_(2)) gt p_(SO_(2)) gt p_(CH_(4))`<br/>`p_(H_(2)) gt p_(CH_(4)) gt p_(SO_(2))`</p>Solution :Equal <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 three gases means equal partial <a href="https://interviewquestions.tuteehub.com/tag/pressures-1164423" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURES">PRESSURES</a> in the beginning. As rate of <a href="https://interviewquestions.tuteehub.com/tag/diffusion-15612" style="font-weight:bold;" target="_blank" title="Click to know more about DIFFUSION">DIFFUSION</a> <br/> `prop sqrt((1)/("Mol mass"))`,therefore, <a href="https://interviewquestions.tuteehub.com/tag/rates-25465" style="font-weight:bold;" target="_blank" title="Click to know more about RATES">RATES</a> of diffusion will he in the <a href="https://interviewquestions.tuteehub.com/tag/order-1138220" style="font-weight:bold;" target="_blank" title="Click to know more about ORDER">ORDER</a> : `H_(2) gt CH_(4) gt SO_(2)`. Hence, amounts left will be in the order `H_(2) lt CH_(4) lt SO_(2)` or `SO_(2) gt CH_(4) gt H_(2)`. Their partial pressures will also be in the same order.</body></html>
Discussion
51948.

0.5 mole of ethanol is mixed with 1.5 mole of water. Then the mole fraction of ethanol and water are ....

Answer» <html><body><p>`0.75 ,0.25`<br/>`0.25,0.75`<br/>`0.5,0.5`<br/>`0.90,0.10`<br/></p>Solution :`0.25 ,0.75` <br/> <a href="https://interviewquestions.tuteehub.com/tag/sol-1216281" style="font-weight:bold;" target="_blank" title="Click to know more about SOL">SOL</a> :molefractionof ethanol<br/> `=( "Numberof <a href="https://interviewquestions.tuteehub.com/tag/molesof-2840420" style="font-weight:bold;" target="_blank" title="Click to know more about MOLESOF">MOLESOF</a> ethanol ")/(" <a href="https://interviewquestions.tuteehub.com/tag/totalnumberof-3225823" style="font-weight:bold;" target="_blank" title="Click to know more about TOTALNUMBEROF">TOTALNUMBEROF</a> molesof ethanoland <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a> ")` <br/>`= ( 0.5 )/( 1.5+0.5 ) = ( 0.5) /( 2.0 ) = 0.25` <br/> Molefractionof water`=( 1.5 )/( 2.0 ) =0.75`</body></html>
Discussion
51949.

0.5 mole of calcium carbonate id decomposed by an aqueous solution containing 25 % JCl by mass. Calculate the mass of the solution consumed.

Answer» <html><body><p><br/></p>Solution :Chemical equation for the <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a> is : <br/> `underset("1 mol")(CaCO_(3))+underset("2 mol")(2HCl)rarrCaCl_(2)+H_(2)O+CO_(2)`<br/> 1 mole of `CaCO_(3)` react with HCl = 2 mol <br/> 0.5 mole of `CaCO_(3)` react with HCl = 1 mol = 36.5 g <br/> Mass of HCl solution <a href="https://interviewquestions.tuteehub.com/tag/required-1185621" style="font-weight:bold;" target="_blank" title="Click to know more about REQUIRED">REQUIRED</a> ot consumed `= 36.5 xx (100)/(<a href="https://interviewquestions.tuteehub.com/tag/25-296426" style="font-weight:bold;" target="_blank" title="Click to know more about 25">25</a>) = 146 g`.</body></html>
Discussion
51950.

0.5 mole each of two ideal gases A (C_(v.m)= (5)/(2)R) and B (C_(v.m)=3R) are taken in a container and expanded reversibly and adiabtically, during this process temperature of gaseous mixture decreases from 350K and 250K. Find DH (in cal/mol) for the process

Answer» <html><body><p>`-100R`<br/>`-137.5R`<br/>`-375R`<br/>none of these</p>Solution :In case of a mixture of gas `C_(P) = (n_(1)Cp_(1) + n_(2)Cp_(2))/(n_(1) n_(2)) = ((0.5) ((<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>)/(2)<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>)+ 0.5 (4R))/(1)` = 3.75R <br/> `<a href="https://interviewquestions.tuteehub.com/tag/delta-947703" style="font-weight:bold;" target="_blank" title="Click to know more about DELTA">DELTA</a> H = nCp Delta T = 1 xx 3.75 xx (-100) = - 375R`</body></html>
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