Explore topic-wise InterviewSolutions in Class 11.

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50951.

60 mL of 0.3 M KOH and 40 mL of 0.2 M HCl are mixed. What is the pH of the mixture ?

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> :Milliequivalents of <a href="https://interviewquestions.tuteehub.com/tag/base-892693" style="font-weight:bold;" target="_blank" title="Click to know more about BASE">BASE</a> =` 60 xx 0.3 = 18`. Milliequivalents of acid `= 40xx 0.2 = 8`<br/> <a href="https://interviewquestions.tuteehub.com/tag/resultant-1187362" style="font-weight:bold;" target="_blank" title="Click to know more about RESULTANT">RESULTANT</a> hydroxyl ion concentration,<br/>`[OH^(-)] = (18 -8)/( 60+40 )= (10)/(100)=(1)/(10) = 0.1`<br/> ` pOH =- <a href="https://interviewquestions.tuteehub.com/tag/log-543719" style="font-weight:bold;" target="_blank" title="Click to know more about LOG">LOG</a>(0.1 )=1` <br/> `[<a href="https://interviewquestions.tuteehub.com/tag/ph-1145128" style="font-weight:bold;" target="_blank" title="Click to know more about PH">PH</a>]=14 -pH` <br/> ` pH =14-1=13.`</body></html>
Discussion
50952.

60 gm of an organic compound containing C,H and O atoms on complete combustion gave 88gm CO_(2) and 36 gm H_(2)O. The empirical formula of the organic compound is:

Answer» <html><body><p>`C_(2)<a href="https://interviewquestions.tuteehub.com/tag/ho-1028234" style="font-weight:bold;" target="_blank" title="Click to know more about HO">HO</a>`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/cho-408392" style="font-weight:bold;" target="_blank" title="Click to know more about CHO">CHO</a>`<br/>`CH_(2)O`<br/>`CHO_(2)`</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html>
Discussion
50953.

6 ml of a standard soap solution (1ml = 0.001g of CaCO_3 ) were required in titrating 50ml of water to produce a good lather. Calculate the degree of hardness.

Answer» <html><body><p></p>Solution :50 ml of hard <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a> sample <a href="https://interviewquestions.tuteehub.com/tag/requires-7264857" style="font-weight:bold;" target="_blank" title="Click to know more about REQUIRES">REQUIRES</a> 6 ml of soap solution.<br/>Volume of soap solution required for `10^6`ml of hardwater to get good lather ` = (10^6 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 6)/(50)= 0.12 xx 10^6 ml` <br/>Given that if <a href="https://interviewquestions.tuteehub.com/tag/1ml-1803748" style="font-weight:bold;" target="_blank" title="Click to know more about 1ML">1ML</a> soap is required for the water sample to get good lather that sample contains 0.001g of `CaCO_3`<br/>1ml soap `to 10^3 g CaCO_3` <br/>`0.12 xx 10^6` ml soap `to`<br/>Weight equivalent of `CaCO_3` = 120 g<br/>Degree of <a href="https://interviewquestions.tuteehub.com/tag/hardness-14971" style="font-weight:bold;" target="_blank" title="Click to know more about HARDNESS">HARDNESS</a> of water sample = 120ppm</body></html>
Discussion
50954.

6 grams of magnesite mineral on heating liberated carbondioxide which measures 1.12 L at STP. What is the percentage purity of mineral?

Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/chemical-914796" style="font-weight:bold;" target="_blank" title="Click to know more about CHEMICAL">CHEMICAL</a> constituent of magnesite is magnesium carbonate. On heating it <a href="https://interviewquestions.tuteehub.com/tag/liberates-1072974" style="font-weight:bold;" target="_blank" title="Click to know more about LIBERATES">LIBERATES</a> carbondioxide. <br/> `MgCO_(3) to MgO+CO_(2)` <br/> 1 mole of `CO_(2)="1 mole of "MgCO_(3)` <br/> 22.4 L of `CO_(2)` at STP=84 grams of `MgCO_(3)` <br/> 1.12 L of `CO_(2)` at STP=? <br/> The <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 pure `MgCO_(3)` to be decomposed <br/> `(1.12)/(22.4) xx 84 =4.2"grams"` <br/> Percent purity `=(4.2)/(6) xx 100=70%`</body></html>
Discussion
50955.

6 g. of Urea is dissolved in 90 g. of water. The mole fraction of solute is

Answer» <html><body><p>`1//5`<br/>`1//50`<br/>`1//51`<br/>`1//501`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html>
Discussion
50956.

6 g of the organic compound on heating with NaOH gave NH_(3) which is neutralised by 200 mL of 1 NHCl. Percentage of nitrogen is

Answer» <html><body><p>0.12<br/>0.6<br/>0.4667<br/>0.2667</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html>
Discussion
50957.

5M H_(2)SO_(4) 1 lit a solution is diluted by adding 10 lit water. State the normality of solution ?

Answer» <html><body><p> <br/> <br/> <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> :`5M H_(2)SO_(<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/10n-267183" style="font-weight:bold;" target="_blank" title="Click to know more about 10N">10N</a> H_(2)SO_(4)` <br/> `N_(1)V_(1) = N_(2)V_(2)` <br/> `:.10xx1=N_(2) xx 10` <br/> `:.N_(2)=1`</body></html>
Discussion
50958.

5mL of a gas containing only carbon and hydrogen were mixed with an excess of oxygen (30mL) and the mixture exploded by means of an electric spark. After the explosion, the volume of the mixed gases remaining was 25mL. On adding a concentrated solution of potassium hydroxide, the volume further diminished to 15mL, the residual gas being pure oxygen. All volumes have been reduced to NTP. Calculate the molecular formula of the hydrocarbon gas.

Answer» <html><body><p></p>Solution :Let the <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 hydrocarbonbe `C_(x)H_(y)`. Its <a href="https://interviewquestions.tuteehub.com/tag/combustion-13975" style="font-weight:bold;" target="_blank" title="Click to know more about COMBUSTION">COMBUSTION</a> can be shows by the following equation: <br/> `{:(C_(x)H_(y)+,(x+(y)/(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>))O_(2)rarr,xCO_(2)+(y)/(2)H_(2)O),(1vol,(x+(y)/(4))mL,xvol),(5mL,5(x+(y)/(4))mL,5xvol):}` <br/> Volume of carbon dioxide produced `= (25 - 15) = 10mL` <br/> `5x = 10` <br/> or `x = 2` <br/> Volume of oxygen used `= (30 - 15) = 15mL` <br/> `5[x+(y)/(4)] = 15` or `x +(y)/(4) = (15)/(5) = <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/> or `2 +(y)/(4) = 3` <br/> or `(y)/(4) = 3 - 2 = 1` <br/> or `y = 4` <br/> <a href="https://interviewquestions.tuteehub.com/tag/thus-2307358" style="font-weight:bold;" target="_blank" title="Click to know more about THUS">THUS</a>, the molecular formula of the hydrocarbon `= C_(2)H_(4)`</body></html>
Discussion
50959.

5H_(2)O_(2)+xCl_(2)O_(2)+y OH^(-) to xCl^(-)+z O_(2)+6H_(2)O. In this reaction the coefficient x,y and z are respectively.

Answer» <html><body><p>4,4 and 5<br/>2,2 and 4<br/>4,2 and 5<br/>2,2 and 5</p>Answer :D</body></html>
Discussion
50960.

5Cu + 2HNO_(3)rarr 5CuO + H_(2)O + N_(2), mark out the correct statement(s)

Answer» <html><body><p> a. Equivalent wt. of Cu = 31.75<br/> b. Equivalent wt. of `HNO_3 = 12.6`<br/> c. Equivalent wt. of `N_2 = 2.8`<br/> d. <a href="https://interviewquestions.tuteehub.com/tag/balanced-389334" style="font-weight:bold;" target="_blank" title="Click to know more about BALANCED">BALANCED</a> chemical reaction suggest that 10 <a href="https://interviewquestions.tuteehub.com/tag/equivalents-974752" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENTS">EQUIVALENTS</a> of Cu combines with 10 equivalents of `HNO_3`in order to produce CuO and 10 equivalents of `N_2` </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> :`overset(<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>)(N_2) rarr 2Hoverset(+5)(NO_3)` <br/> Equivalent weight of nitrogen`=(28)/10 = 2.8`</body></html>
Discussion
50961.

5.85 g of NaCl is dissolved in water and the solution was made upto 500 ml using a standard flask. The strength of the solution in formality is ……………

Answer» <html><body><p></p>Solution :`0.2F` <br/> <a href="https://interviewquestions.tuteehub.com/tag/formality-996498" style="font-weight:bold;" target="_blank" title="Click to know more about FORMALITY">FORMALITY</a> `= ("Number of <a href="https://interviewquestions.tuteehub.com/tag/formula-464310" style="font-weight:bold;" target="_blank" title="Click to know more about FORMULA">FORMULA</a> weight of solute")/(<a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> of solution in <a href="https://interviewquestions.tuteehub.com/tag/litre-1075864" style="font-weight:bold;" target="_blank" title="Click to know more about LITRE">LITRE</a>") = (5.85)/(58.5 xx 0.5) =0.2F`</body></html>
Discussion
50962.

5.85 g of NaCl are dissolved in 500 cm^3 of water. Calculate the formality of the 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> :0.2 <a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a></body></html>
Discussion
50963.

5.85 g NaCl dissolved in 1 L solution. Find the concentration of solution. (molecular mass of NaCl : 58.5g//"mol"^(-1))

Answer» <html><body><p> 0.1 M<br/>2 M<br/>1 M<br/>0.5 M</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/molarity-1100268" style="font-weight:bold;" target="_blank" title="Click to know more about MOLARITY">MOLARITY</a> `=(5.85)/(58.5xx1)= 0.1M`</body></html>
Discussion
50964.

5.85 g of NaCl is dissolved in water and the solution was made up to 500 mL using a standard flask. The strengh of the solution in molarity is …………..

Answer» <html><body><p></p>Solution :`0.2M` <br/> Molarity `= ("Number of <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> of <a href="https://interviewquestions.tuteehub.com/tag/solute-1217068" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTE">SOLUTE</a>")/(" <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 the solvent in Litre") = ((5.845)/(58.45))/(0.5) = (0.1)/(0.5) =0.2M`</body></html>
Discussion
50965.

5.82 g of a silver coin were dissolved in strong nitric acid and excess of NaCI solution was added. The silver chloride precipitated was dried and weighed 7.20 g. Calculate the percentage of silver in the coin

Answer» <html><body><p><br/></p>Solution :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("one mole" 107.9 g) (2HNO_(3)) to underset("one mole")(AgNO_(3)) + NO_(2) + H_(2)O` <br/>`underset("One mole")(AgNO_(3)) + NaCl to underset("one mole" <a href="https://interviewquestions.tuteehub.com/tag/143-273618" style="font-weight:bold;" target="_blank" title="Click to know more about 143">143</a>.35 g)(<a href="https://interviewquestions.tuteehub.com/tag/agcl-368919" style="font-weight:bold;" target="_blank" title="Click to know more about AGCL">AGCL</a>) <a href="https://interviewquestions.tuteehub.com/tag/darr-943899" style="font-weight:bold;" target="_blank" title="Click to know more about DARR">DARR</a> + NaNO_(3)` <br/> From the equations, it is clear that one mole (107.9 g) of Ag gives one mole (143. 35 g) of AgCI. <br/> `therefore` The mass of Ag which gives 7.20 g of AgCI<br/> `=107.9/143.35 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 7.20 = 5.42 g` <br/> `therefore` Percentage of Ag in the given coin <br/> `=5.42/5.82 xx 100 = 93.1`</body></html>
Discussion
50966.

5.84 grams of hydrogen chloride and 5.22 grams of manganese dioxide are reacted. Calculate the volume of chlorine gas liberated at STP?

Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :0.996Lit</body></html>
Discussion
50967.

A silver coin weighing 11.34 g was dissolved in nitric acid When sodium chloride was added to the solution all the silver (present as AgNO_(3)) precipitated as silver chloride. The mass of the precipitated silver chloride was 14.35 g. Calculate the percentage of silver in the coin.

Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :0.931</body></html>
Discussion
50968.

5.82 g of a silver coin is dissolved in nitric acid. One adding solution, all the silver present gets precipitated as AgCl. The weight of the precipitate is found to be 7.20 g. What is the percentage of silver in the coin ?

Answer» <html><body><p><br/></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 AgCl <a href="https://interviewquestions.tuteehub.com/tag/precipitated-2948214" style="font-weight:bold;" target="_blank" title="Click to know more about PRECIPITATED">PRECIPITATED</a> = 7.20 g <br/> Weight of silver in the preciptate `= ((<a href="https://interviewquestions.tuteehub.com/tag/108g-1773491" style="font-weight:bold;" target="_blank" title="Click to know more about 108G">108G</a>))/((143.50g))xx(7.20g)=5.42g` <br/> <a href="https://interviewquestions.tuteehub.com/tag/percentage-13406" style="font-weight:bold;" target="_blank" title="Click to know more about PERCENTAGE">PERCENTAGE</a> of silver in the <a href="https://interviewquestions.tuteehub.com/tag/coin-921316" style="font-weight:bold;" target="_blank" title="Click to know more about COIN">COIN</a> `= ((5.42g))/((5.83g))xx100=93.1%`.</body></html>
Discussion
50969.

5.6g of an organic compound on burning with excess of oxygen gave 17.6g of CO_2 and 7.2g H_2O. The organic compound is

Answer» <html><body><p>`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_(6)`<br/>`C_(4)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_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)H_(8)`<br/>`CH_(3)COOH`</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
50970.

5.6dm^(3) of an unknown gas at S.T.P. requires 52.25 J of heat to raise itstemperature by 10^(@)C at constant volume. Calculate volume. Calculate C_(v) , C_(p) and atomicity of the gas.

Answer» <html><body><p></p>Solution :`22.4dm(3)` of a gas at S.T.P. `=1` mol `:. 5.6 dm^(3)`of the gas at S.T.P. `= (1)/(22.4) xx 5.6 =0.25 mol` <br/> Thus, for `10^(@)` rise , `0.25` mol of the gas at constant volume require heat `= 52.25 <a href="https://interviewquestions.tuteehub.com/tag/j-520843" style="font-weight:bold;" target="_blank" title="Click to know more about J">J</a>` <br/> `:. `For `1^(@)` rise, 1 mol of the gas at constant volume will require heat `= (52.25)/(10 xx 0.25) J = 20.9 J` <br/> `:. C_(<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>) = 20.9 J K^(-1) mol^(-1)` <br/>Now, `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>= 20. 9 J K^(-1) mol^(-1) + 8. 314 J K^(1) mol^(-1)= 29.214 J K^(-1) mol^(-1)` <br/> `:. gamma = (C_(p))/( C_(v))= (29.214)/(20.9) = 1.4`. Hence, the gas is diatomic</body></html>
Discussion
50971.

56.0 g KOH, 138.0 g K_(2)CO_(3) and 100.0 g KHCO_(3) is dissolved in water and the solution is made 1 L. 10 " mL of " this stock solution is titrated with 2.0 M HCl. Which of the following statements is/are correct?

Answer» <html><body><p>When phenolphthalein is used as an indicator from the very beginning the titre value of HCl will be <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<br/>When phenolphthalein is used as an indicator from the very beginning the titre value of HCl will be 40 mL.<br/>When methyl orange is used as an indicator from the very beginning, the titre value of HCl will be 80 mL.<br/>When methyl orange is used as an indicator after the first end point the titre value of HCl will be 30 mL.</p>Solution :`m" Eq of "`(KOH)/(L)=(56)/((56)/(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>))xx10^(3)=100=(20)/(20)mL` solution m" Eq of "`(K_(2)CO_(3))/(L)=(138)/((138)/(2))xx10^(3)=200=(40)/(20)mL` solution <br/> `m" Eq of "(KHCO_(3))/(L)=(100)/((100)/(1))xx10^(3)=<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>=(20)/(20)mL` <br/> (b). With phenolphthalein: <br/> `m" Eq of "KOH+(1)/(2)m" Eq of "K_(2)CO_(3)=m" Eq of "HCl` <br/> `20+(40)/(2)=vxx1N(n=1)` <br/> `V_(HCl)=40mL` <br/> (c). With methyl orange: <br/> `m" Eq of "KOH+m" Eq of "K_(2)CO_(3)+m" Eq of "KHCO_(3)=m" Eq of "HCl` <a href="https://interviewquestions.tuteehub.com/tag/ltbrgt-2804393" style="font-weight:bold;" target="_blank" title="Click to know more about LTBRGT">LTBRGT</a> `20+40+20=Vxx1N` <br/> `V_(HCl)=80mL` <br/> (d). With methyl orange after the first end point: <br/> `(1)/(2)m" Eq of "K_(2)CO_(3)+m" Eq of "KHCO_(3)=m" Eq of "HCl` <br/> `20+20=Vxx1N` <br/> `V_(HCl)=40mL`</body></html>
Discussion
50972.

5.6 litres of oxygen at STP is equivalent to

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>/4 <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a><br/>1 mole<br/>1/<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> mole<br/>1/8 mole</p>Solution :22.4 <a href="https://interviewquestions.tuteehub.com/tag/litres-1075876" style="font-weight:bold;" target="_blank" title="Click to know more about LITRES">LITRES</a> of `O_(2)` = 1 mole <br/>`:.` 5.6 litres of `O_(2)` =`1/22.4 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 5.6` = 0.25 mole = 1/4 mole.</body></html>
Discussion
50973.

5.6 litres of oxygen at NTP is equivalent to

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> <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a><br/>`1/2` mole<br/>`1/4` mole<br/>`1/8` mole </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html>
Discussion
50974.

5.6 L of a gas at STP are found to have mass of 11 g. The molecular mass of the gas is

Answer» <html><body><p>Phosphine<br/>Phosgene<br/>Nitric Oxide<br/>Nitrous Oxide</p>Solution :5.6 L of gas has the <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of <a href="https://interviewquestions.tuteehub.com/tag/11-267621" style="font-weight:bold;" target="_blank" title="Click to know more about 11">11</a> g. <br/>`:.`22.4 L of gas will have the mass `11/5.6xx 22.4 = <a href="https://interviewquestions.tuteehub.com/tag/44-316683" style="font-weight:bold;" target="_blank" title="Click to know more about 44">44</a>`</body></html>
Discussion
50975.

56 g of nitrogen and 96 g of oxygen are mixed isothermally at a total pressure of 10 atm. The partial pressures of oxygen and nitrogen (in atm) are respectively:

Answer» <html><body><p>4, <a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a><br/>5, 5<br/>2, 8<br/>8, 2</p>Solution :`n_(N_(2)) = (56)/(28) = 2 , n_(O_(2)) = (96)/(32) = 3` <br/> `P_(N_(2)) = x_(N_(2)) xx P_("Total")` <br/> `= (2)/(2 + 3) 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> = 4 atm` <br/> `P_(O_(2)) = 10 - 4 = 6 atm`</body></html>
Discussion
50976.

5.6 L of a gas at NTP weighs equal to 8g. The vapour density of gas is

Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/32-307188" style="font-weight:bold;" target="_blank" title="Click to know more about 32">32</a><br/><a href="https://interviewquestions.tuteehub.com/tag/16-276476" style="font-weight:bold;" target="_blank" title="Click to know more about 16">16</a><br/>8<br/>40<br/></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>) `therefore` 5.6 L of gas weighs at NTP = 8g <br/> `therefore` L of gas will weights at NTP = (8 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 22.4)/(5.6) = 32g <br/> `therefore` Molecular weight of gas = 32 <br/> We know that, Molecular weight = <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 vapour density <br/> `therefore` Vapour density = `32/2` = 16</body></html>
Discussion
50977.

56 gm of nitrogen and 4 gm of hydrogen are taken in a closed vessel

Answer» <html><body><p>`0.4`<br/>1<br/>1.3<br/>0.2</p>Answer :D</body></html>
Discussion
50978.

54.5 mg of Na_(3)po_(4) containsp^(32) (15.6% of sample) and P^(31) atoms. Assumingonly P^(32) (15.6% of sample) and P^(31) atoms. Assuming onlyP^(32) atoms radioactive caculate the rate of decay for the given sample of Na_(3)PO_(4) (Half-life period for p^(32) = 14.3 days mol. we of Na_(2) PO_(4) = 161.2)

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> :`1.78xx10^(<a href="https://interviewquestions.tuteehub.com/tag/13-271882" style="font-weight:bold;" target="_blank" title="Click to know more about 13">13</a>) <a href="https://interviewquestions.tuteehub.com/tag/dps-432950" style="font-weight:bold;" target="_blank" title="Click to know more about DPS">DPS</a>`</body></html>
Discussion
50979.

5.4 grams of an element of group IIIA an oxidation gave 10.2 grams of its oxide. Calcute the atomic weight of element.

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/27-298706" style="font-weight:bold;" target="_blank" title="Click to know more about 27">27</a></body></html>
Discussion
50980.

5.4 grams of a metal is able to produce 0.6 grams of H_(2) gas with acid action. What is the equivalent weight of that 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> :0.6 gram <a href="https://interviewquestions.tuteehub.com/tag/hydrogen-22331" style="font-weight:bold;" target="_blank" title="Click to know more about HYDROGEN">HYDROGEN</a> is produced by 5.4g metal 1 gram hydrogen is produced by <br/> `(1xx5.4)/(0.6)=9g` metal `therefore` EQ <a href="https://interviewquestions.tuteehub.com/tag/wt-1462289" style="font-weight:bold;" target="_blank" title="Click to know more about WT">WT</a> of metal = <a href="https://interviewquestions.tuteehub.com/tag/9-340408" style="font-weight:bold;" target="_blank" title="Click to know more about 9">9</a></body></html>
Discussion
50981.

5.3 g of sodium carbonate have been dissolved in 500 mL of the solution. What is the molarity of the solution ?

Answer» <html><body><p><br/></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/molarity-1100268" style="font-weight:bold;" target="_blank" title="Click to know more about MOLARITY">MOLARITY</a> of solution (M) `= (("Mass of "Na_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)CO_(3))/("Molecular mass of "Na_(2)CO_(3)))/("Volume of solution in mL"/(<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>))=((5.3g))/(("106 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))<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>("0.5 L"))` <br/> `="0.1 mol L"^(-1)=0.1M`</body></html>
Discussion
50982.

5.3 grams of anhydrous sodium carbonate is treated with dilute hydrochloric acid to give carbondioxide. In order to produce the same amount of gas. How many graphite atoms are to be oxidised?

Answer» <html><body><p></p>Solution :The reaction of sodium <a href="https://interviewquestions.tuteehub.com/tag/carbonate-909380" style="font-weight:bold;" target="_blank" title="Click to know more about CARBONATE">CARBONATE</a> with acid is given as <br/> `Na_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)CO_(3)+2HClrarr 2NaCl+H_(2)O+CO_(2)` <br/> 1 <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> of `Na_(2)CO_(3) = 1` mole of `CO_(2)` <br/> 106 grams of `Na_(2)CO_(3) = 44` grams of `CO_(2)` <br/> 5.3 grams of `Na_(2)CO_(3)=?` <br/> 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 `CO_(2)` produced `=(5.3)/(106)xx44=2.2` grams <br/> Oxidation of graphite is given as `C+O_(2)rarr CO_(2)` <br/> 1 mole of `CO_(2) = 1` mole of graphite <br/> 44 grams of `CO_(2) = 6.022 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10^(23)` atoms <br/> 2.2 grams of `CO_(2) = ?` <br/> Number of atoms of .C. to be oxidised <br/> `= (2.2xx6.022x10^(23))/(44)= 3.011xx10^(22)`</body></html>
Discussion
50983.

5.3 g of Na_(2)CO_(3) taken in a 250 ml flask and water added upto the mark. 10ml of that solution was taken in a 50 ml flask water added upto the mark. Find molarity of that dilute solution.

Answer» <html><body><p>0.<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> M<br/>0.02 M<br/>0.04 M<br/>none </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> :`M_(1)=(5.3)/(106xx0.25)=0.2M,M_(1)V_(1)=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>)V_(2)` <br/> `0.2xx10=M_(2)xx50impliesM_(2)=0.04M`</body></html>
Discussion
50984.

52ml of sample of hydrogen peroxide solution after acidification with dil. H_(2)O_(4) requires 100ml of 0.1M KMnO_(4) solution for complete oxidaitoin. Calculate percenatge strength and volume strength of H_(2)O_(2) 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> :1.7 ( w/v)% ,5.6v</body></html>
Discussion
50985.

52.5 millimol LiAlH_(4) reacts with 15.6g (210 millimol) tert-butyl alcohol. In the following reaction 157.5 millimol H_(2) is produced. LiAlH_(4)+3(CH_(3))_(3)COH to 3H_(2)+LiC(CH_(3))_(3) O]_(3)AlH On adding extra methanol or alchol in the above reaction, displacement of the 4th hydrogen atom LiAlH_(4)will be observed by the following reaction. Li[(CH_(3))_(3)O]_(3)AlH+CH_(3)OH to H_(2)+Li(CH_(3))_(3)O]_(3)[CH_(3)]Al How much H_(2) will evolve on adding methanol?

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> :1 mol of `H_(2)` will be produced from 1 mol of `Li[(CH_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>))_(3)CO]_(3)AlH`. <br/> 1 mol of `Li[(CH_(3))_(3)O]_(3)` AlH will be produced from 1 mol of `LiAlH_(4)`. From second <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a>, 1 mol of `Li[(CH_(3))_(3)CO]_(3)AlH` will produce 1 mol of `H_(2)`. 52.5 millimol of `Li[(CH_(3))_(3)CO]_(3)AlH` will <a href="https://interviewquestions.tuteehub.com/tag/react-613674" style="font-weight:bold;" target="_blank" title="Click to know more about REACT">REACT</a> with excess of `CH_(4)` to produce 52.5 millimon of `H_(2)`.</body></html>
Discussion
50986.

52% water is in mixture of alochol and water state mole fraction ethanol.

Answer» <html><body><p> <br/> <br/> <br/></p>Solution :`H_(2)O=<a href="https://interviewquestions.tuteehub.com/tag/54-324973" style="font-weight:bold;" target="_blank" title="Click to know more about 54">54</a>` gmand <a href="https://interviewquestions.tuteehub.com/tag/ethanol-975823" style="font-weight:bold;" target="_blank" title="Click to know more about ETHANOL">ETHANOL</a> `= 46 ` gm <br/> Moles of `H_(2)O = (54)/(18)=3` <br/> Moles of Ethanol `=(46)/(46)=<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/> `:.` <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> <a href="https://interviewquestions.tuteehub.com/tag/fraction-458259" style="font-weight:bold;" target="_blank" title="Click to know more about FRACTION">FRACTION</a> of ethanol `= 1/4 = 0.25`</body></html>
Discussion
50987.

50mL of 0.10 B Ba(OH)_(2) is added to 50 mLof 0.10 M H_(2)SO_(4). The rise in temperature isDeltaT_(1) . If experiment is replaced by taking 100 mL of each solution, the rise in temperature isDeltaT_92). Then

Answer» <html><body><p>`DeltaT_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)= 2DeltaT_(1)`<br/>`DeltaT_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)= 2DeltaT_(1)`<br/>`DeltaT_(2)= DeltaT_(1)`<br/>`DeltaT_(2)= 4DeltaT_(1)`</p>Answer :C</body></html>
Discussion
50988.

50g of an ideal gas is undergoing a process for which heat capacity is 50 cal//mole-K. Ifrise in temperature during the process is 100^(@)C then calculate work involved in process (given molar mass of gas =10 g//"mole", C_(p) of gas =5R)

Answer» <html><body><p>`-29kcal`<br/>`-21kcal`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/29-299821" style="font-weight:bold;" target="_blank" title="Click to know more about 29">29</a> <a href="https://interviewquestions.tuteehub.com/tag/kcal-531581" style="font-weight:bold;" target="_blank" title="Click to know more about KCAL">KCAL</a>`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/21-293276" style="font-weight:bold;" target="_blank" title="Click to know more about 21">21</a> kcal`</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
50989.

500mL of IM HCl and 500mL of 1M NaOH are mixed in Dewar flask. Then rise in temperature is T_1. In another experiment 1000mL 1M HCl and 1000 mL of 1M NaOH are mixed in Dewar flask. Then rise in temperature is T_2. What is the relation between T_1 and T_2 ?

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> :`T_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)=T_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)`</body></html>
Discussion
50990.

500 ml vessel contaions 1.5 M each of A,B,C and D at equilibrium . If 0.5 M each of C and D are taken out, the value of K_(c)"for"A + BhArr C+D will be

Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.0`<br/>`1//9`<br/>`4//9`<br/>`8//9`</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> :`K_(c) =( 1.5 xx 1.5)/(1.5 xx 1.5) = 1 .` On <a href="https://interviewquestions.tuteehub.com/tag/changing-2508325" style="font-weight:bold;" target="_blank" title="Click to know more about CHANGING">CHANGING</a> <a href="https://interviewquestions.tuteehub.com/tag/concentrations-928106" style="font-weight:bold;" target="_blank" title="Click to know more about CONCENTRATIONS">CONCENTRATIONS</a> , equilibrium constant does not change .</body></html>
Discussion
50991.

500 mL of0.1 MH_(2)SO_(4) was added into 1 L of 0.1 M NaOH solution. The heat evolved was x calories. If further 500 mL of H_(2)SO_(4) is added into the solution, now heat evolved will be

Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a> <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/><a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> x cal<br/>zero<br/>`x// 2 cal`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html>
Discussion
50992.

500 ml of aM and 500 ml of bM solution of a solute are mixed and diluted to 2 litre to prepare a solution of 1.5 M. If a and b are in the ratio 2 : 1, then the value of a is.

Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`(500a+500b)/(<a href="https://interviewquestions.tuteehub.com/tag/2000-289566" style="font-weight:bold;" target="_blank" title="Click to know more about 2000">2000</a>)xx1.5impliesa+b=6` <br/> <a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> `a=2bimplies a=4`</body></html>
Discussion
50993.

500 mL of 2 M HCl, 100 mL of 2 M H_(2)SO_(4) and one gram equivalent of monoacidic alkali are mixed together. 30 mL of this solution required 20 mL of Na_(2)CO_(3). xH_(2)O solution obtained by dissolving 143 g Na_(2)CO_(3).xH_(2)O in one litre solution. Calculate the water of crystallisation of Na_(2)CO_(3) .xH_(2)O.

Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :N//A</body></html>
Discussion
50994.

500 ml of a 0.1 N solution of AgNO_(3) added to 500 ml of 0.1 N solution of KCl.The concentration of nitrate ion in the resulting mixture is

Answer» <html><body><p>`0.05N`<br/>`0.1N`<br/>`0.2N`<br/><a href="https://interviewquestions.tuteehub.com/tag/reduced-1181337" style="font-weight:bold;" target="_blank" title="Click to know more about REDUCED">REDUCED</a> to <a href="https://interviewquestions.tuteehub.com/tag/zero-751093" style="font-weight:bold;" target="_blank" title="Click to know more about ZERO">ZERO</a> </p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`[NO_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)^(-)]=(0.1xx500)/(<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>)=0.05N`</body></html>
Discussion
50995.

500 mL of 0.2 M aqueous solution of acetic acid is mixed with 500 mL of 0.2 M HCl at 25^(@)C. (i) Calculate the degree of dissociation of acetic acid in the resulting solution and pH of the solution. (ii) If 6 g of NaOH is added to the above solution, determine the final pH [Assume there is no change in volume on mixing : K_(a)of acetic acid is 1.75xx10^(-5) " mol " L^(-1)]

Answer» <html><body><p></p>Solution :(i) Millimoles of `CH_(3)CO OH = <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 0.<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> = 100` <br/> Millimoles of HCl `= 500xx 0.2=100` <br/> Final volume after mixing `= 500 + 500 = 1000 ` mL <br/> `:. [ CH_(3)CO OH]=(100)/(1000) = 0.1 M,"" [HCl ] = (100)/(1000) = 0.1 M` <br/> `{:(,CH_(3)CO OH ,hArr,CH_(3)CO O^(-),+,H^(+),,),("Before dissociation",0.1 M ,," "0,,0.1 M,,),(,,,,,("from HCl"),,),("After dissociation",(0.1-<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>),," "x,,(0.1+x),,):}` <br/> `K_(a) = (x(0.1+))/((0.1+x))` <br/> As in the presence of HCl, dissociation of `CH_(3)CO OH` will be very small (due to common ion effect ),x is very very small. Hence, <br/> `K_(a) = (x(0.1))/(0.1) = x = 1.75xx10^(-5) `<a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> `L^(-1) ` (Given) <br/> `:.` Degree of dissociation `= (x)/(0.1) = (1.75xx10^(-5))/(0.1) = 1.75xx10^(-4)= 0.00175 %` <br/> Further, `[H^(+)]= 0.1 + x ~~ 0.1 :. pH = - log 0.1 = 1 ` <br/> (ii) 6g 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> = `(6)/(40) ` mole = 0.15 mole <br/> Hence, now the equilibrium will be<br/> `{:(,CH_(3)CO OH ,+,HCl,+,NaOH,hArr,CH_(3)CO ONa ,+,NaCl,+,H_(2)O),("Initial",0.1,,0.1,,0.15,,0,,0,,0),("At eqm.",0.05,,0,,0,,0.05,,0,,0):}` <br/> Thus, the solution will now be 0.05 M in `CH_(3)CO OH` and 0.05 M in `CH_(3)CO ONa`, i.e., it is acidic buffer. <br/> `pH = - log K_(a) + log. (["Salt"])/(["Acid"]) = - log (1.75xx10^(-5)) + log (0.05)/(0.05) = 4. 757`</body></html>
Discussion
50996.

500 mL of 0.1 M KCl, 200 ml of 0.01 M NaNO_(3) and 500 ml of 0.1 M AgNO_(3) was mixed. The molarity of K^(+),Ag^(+), Cl^(-), Na^(+), NO^(3-) in the solution would be:

Answer» <html><body><p>`[<a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>^(+)] = 0.04, [<a href="https://interviewquestions.tuteehub.com/tag/ag-362275" style="font-weight:bold;" target="_blank" title="Click to know more about AG">AG</a>^(+)]= 0.04, {Na^(+)]=0.002[Cl^(-)] = 0.04 , [NO_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>^(-)] = 0.042`<br/>`[K^(+)]=0.04, [Na^(+)] = 0.00166, [NO^(-_(3)]]= 0.0433`<br/>`[K^(+)]= 0.04, [Ag^(+)] = 0.05, [Na^(+)] = 0.0025, [Cl^(-)] = 0.05, [NO^(-_(3)] = 0.0525]`<br/>`[K^(+)] = 0.05, [Na^(+)]= 0.0025 [Cl^(-)]= 0.05, [NO^(-_(3)] = 0.0525`</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
50997.

500 mL fo 2 M HCl, 100 mL of 2 M H_(2) SO_(4), and one gram equivalent of a monoacidic alkali are mixed together. 30mL of this solution requried 20 mL of 143 g Na_(2) CO_(3). xH_(2)O in one litre solution. Calculate the water of crystallisation of Na_(2) CO_(3). xH_(2) O

Answer» <html><body><p><br/></p>Solution :i. <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> mEq of acid `= 500 xx 2 + 100 xx 2 xx 2` (`n` factor) <br/> `= 1000 + 400 = 1400` <br/> Total mEq of monoacid alkali = 1000 <br/> Acid left after neutralisation with alkali <br/> `= 1400 - 1000 = 400 mEq` <br/> `N` fo acid left `= (400 mEq)/((500 + 100)mL) = (400)/(600) = 0.6 N` <br/> mEq of acid `-= mEq "of"Na_(2)CO_(<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/> `30 mL xx 0.66 N = 20 mL xx N Na_(2)CO_(3)` <br/> Strength of `Na_(2)CO_(3) x H_(2)O = 143 g L^(-1)` <br/> `M_(Na_(2)CO_(3).xH_(2)O) = ("Strenght")/(Mw) = ((143)/(106 + 18x))` <br/> `M_(Na_(2)CO_(3).xH_(2)O) = ((2 xx 143)/(106 + 18 x))` (`n` factor = 2) <br/> Substitution the value of `N` of `Na_(2)CO_(3).xH_(2)O` in <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a> (i), we get <br/> `30 xx 0.66 x = 20 xx ((2 xx 143)/(106 + 18 x))` <br/> solve for `x` <br/> `x = 10.16 ~~ 10` <br/> So <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a> of crystallisation of `Na_(2)CO_(3).xH_(2) O = 10` <br/> formula : `Na_(2) CO_(3). 10 H_(2) O`</body></html>
Discussion
50998.

50.0 kg of N_(2)(g) and 10.0 kgof H_(2)(g) are mixed to produce NH_(3)(g). Calculate mass of NH_(3)(g) formed. Identify the limiting reagent in the production of NH_(3) in this situation.

Answer» <html><body><p></p>Solution :The balanced chemical equation for the reaction is : <br/> `underset("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>")(N_(2)(<a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>))+underset("3 mol")(3H_(2)(g))rarrunderset("2 mol")(2NH_(3)(g))` <br/> <a href="https://interviewquestions.tuteehub.com/tag/step-25533" style="font-weight:bold;" target="_blank" title="Click to know more about STEP">STEP</a> I. To find the limiting reactant <br/> Mass of `N_(2)` taken `= 50.0 <a href="https://interviewquestions.tuteehub.com/tag/kg-1063886" style="font-weight:bold;" target="_blank" title="Click to know more about KG">KG</a> = 50,000 g` <br/> Moles of `N_(2)=("Mass of" N_(2))/("<a href="https://interviewquestions.tuteehub.com/tag/molar-562965" style="font-weight:bold;" target="_blank" title="Click to know more about MOLAR">MOLAR</a> mass of "N_(2))=(5000g)/((28"g mol"^(-1)))=1785.7mol` <br/> Mass of `H_(2)` taken `= 10.0 kg = 10,000 g` <br/> Moles of `H_(2)=(10,000g)/((2.0"g mol"^(-1)))=5000 mol` <br/> From the above balanced equation, 1 mol of `N_(2)` reacts with 3 mol of `H_(2)` <br/> `:. 1785.7` mol of `N_(2)` will react with` H_(2)=3xx1785.7 mol = 5355.9 mol` But we have only 5000 mol of `H_(2)`. Therefore, `H_(2)` is the limiting reactant. <br/> Step II. To find the amount of `NH_(3)` formed. <br/> 3 moles of `H_(2)` from `NH_(3) =2` mol <br/> 5000 moles of `H_(2)` form `NH_(3)=5000 xx ((2 mol))/((3 mol))=3333.3 mol` <br/> Mass of `NH_(3)` formed `= 3333.3 xx 17 = 56666.6 g = 56.67 kg`.</body></html>
Discussion
50999.

50.0 kg of N_(2(g)) and 10.0 kg of H_(2(g)) are mixed to produce NH_(3(g)). Calculate the NH_(3(g)) formed. Identify the limiting reagent in the production of NH_(3) in this situation.

Answer» <html><body><p> <br/> <br/> </p>Solution :A <a href="https://interviewquestions.tuteehub.com/tag/balanced-389334" style="font-weight:bold;" target="_blank" title="Click to know more about BALANCED">BALANCED</a> equation for the above reaction is written as follows, <br/> `N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g))`. <br/> Mole of `N_(2)=50.0 kg N_(2) xx (<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a> g N_(2))/(1 kg N_(2)) xx (1 "mol" N_(2))/(28.09g N_(2))` <br/> `= 17.86 xx 10^(2)` mol <br/> Mole of `H_(2) = 10.00 kg H_(2) xx (1000 g H_(2))/(1 kg H_(2)) xx (1 "mol" H_(2))/(2.016 g H_(2))` <br/> `= 4.96xx10^(3)` mol <br/> According to the equation ,1 mol `N_(2(g))` required 3 mol `(H_(2))`. For the reaction. <br/> Hence, for `17.86xx10^(2)` mol of `N_(2(g))` , the <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 `H_(2(g))` required would be : <br/> `= 17.86 xx 10^(2)` mol `xx(3 "mol"H_(2(g)))/(1 "mol"N_(2(g)))` <br/> `= 5.36xx10^(3)` mol `H_(2(g))` <br/> But we have only `4.96xx10^(3)` mol `H_(2)`. Hence diphydrogen is the limiting reagent in this <a href="https://interviewquestions.tuteehub.com/tag/case-910082" style="font-weight:bold;" target="_blank" title="Click to know more about CASE">CASE</a>. So, `NH_(3(g))` would be formed only from that amount of available dihydrogen , i.e `4.96xx10^(3)` mol `H_(2(g))`. <br/> Since 3 mol `H_(2(g)), 2 mol NH_(3(g))` <br/> `4.96xx10^(3) mol H_(2(g)) xx (2 mol NH_(3(g)))/(3 mol H_(2(g)))` <br/> `= 3.30xx10^(3) ` mol `NH_(3(g))` <br/> `3.30xx10^(3)` mol `NH_(3(g))` is obtained. <br/> If they are to be converted to <a href="https://interviewquestions.tuteehub.com/tag/grams-476111" style="font-weight:bold;" target="_blank" title="Click to know more about GRAMS">GRAMS</a>, it is done as follows: <br/> 1 mol `NH_(3(g))= 17.0 g NH_(3 (g))` <br/> `:.3.30xx10^(3) mol NH_(3(g))xx(17.0 g mol NH_(3(g)))/(1 mol NH_(3(g)))` <br/> `= 3.30xx10^(3) xx17 g NH_(3(g))` <br/> `= 56.1 xx 10^(3) g NH_(3)` <br/> `= 56.1 kg NH_(3(g))`</body></html>
Discussion
51000.

500 jouleof heat was suppliedto a system at constant volume. It resulted in the increase of temperature of the system from 20^(@)C to 25^(@)C . What is the change in internal energy of the system ?

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> :At <a href="https://interviewquestions.tuteehub.com/tag/constant-930172" style="font-weight:bold;" target="_blank" title="Click to know more about CONSTANT">CONSTANT</a> volume, `DeltaV=0`. Applying `DeltaU= <a href="https://interviewquestions.tuteehub.com/tag/q-609558" style="font-weight:bold;" target="_blank" title="Click to know more about Q">Q</a>+w=q+PDeltaV, ` we get `DeltaU = q= 500J`.</body></html>
Discussion