52671 + Interview Questions in Chemistry in Class 11 Page 1020 InterviewSolution

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» SOLUTION :Milliequivalents of BASE =` 60 xx 0.3 = 18`. Milliequivalents of acid `= 40xx 0.2 = 8` RESULTANT hydroxyl ion concentration, `[OH^(-)] = (18 -8)/( 60+40 )= (10)/(100)=(1)/(10) = 0.1` ` pOH =- LOG(0.1 )=1` `[PH]=14 -pH` ` pH =14-1=13.`

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» `C_(2)HO` `CHO` `CH_(2)O` `CHO_(2)` Answer :C

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» Solution :50 ml of hard WATER sample REQUIRES 6 ml of soap solution. Volume of soap solution required for `10^6`ml of hardwater to get good lather ` = (10^6 XX 6)/(50)= 0.12 xx 10^6 ml` Given that if 1ML soap is required for the water sample to get good lather that sample contains 0.001g of `CaCO_3` 1ml soap `to 10^3 g CaCO_3` `0.12 xx 10^6` ml soap `to` Weight equivalent of `CaCO_3` = 120 g Degree of HARDNESS of water sample = 120ppm

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» Solution :CHEMICAL constituent of magnesite is magnesium carbonate. On heating it LIBERATES carbondioxide. `MgCO_(3) to MgO+CO_(2)` 1 mole of `CO_(2)="1 mole of "MgCO_(3)` 22.4 L of `CO_(2)` at STP=84 grams of `MgCO_(3)` 1.12 L of `CO_(2)` at STP=? The WEIGHT of pure `MgCO_(3)` to be decomposed `(1.12)/(22.4) xx 84 =4.2"grams"` Percent purity `=(4.2)/(6) xx 100=70%`

Discussion

50955.	6 g. of Urea is dissolved in 90 g. of water. The mole fraction of solute is
Answer» `1//5` `1//50` `1//51` `1//501` ANSWER :C

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» 0.12 0.6 0.4667 0.2667 Answer :C

Discussion

50957.	5M H_(2)SO_(4) 1 lit a solution is diluted by adding 10 lit water. State the normality of solution ?
Answer» SOLUTION :`5M H_(2)SO_(4)=10N H_(2)SO_(4)` `N_(1)V_(1) = N_(2)V_(2)` `:.10xx1=N_(2) xx 10` `:.N_(2)=1`

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» Solution :Let the FORMULA of the hydrocarbonbe `C_(x)H_(y)`. Its COMBUSTION can be shows by the following equation: `{:(C_(x)H_(y)+,(x+(y)/(4))O_(2)rarr,xCO_(2)+(y)/(2)H_(2)O),(1vol,(x+(y)/(4))mL,xvol),(5mL,5(x+(y)/(4))mL,5xvol):}` Volume of carbon dioxide produced `= (25 - 15) = 10mL` `5x = 10` or `x = 2` Volume of oxygen used `= (30 - 15) = 15mL` `5[x+(y)/(4)] = 15` or `x +(y)/(4) = (15)/(5) = 3` or `2 +(y)/(4) = 3` or `(y)/(4) = 3 - 2 = 1` or `y = 4` THUS, the molecular formula of the hydrocarbon `= C_(2)H_(4)`

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» 4,4 and 5 2,2 and 4 4,2 and 5 2,2 and 5 Answer :D

Discussion

50960.	5Cu + 2HNO_(3)rarr 5CuO + H_(2)O + N_(2), mark out the correct statement(s)
Answer» a. Equivalent wt. of Cu = 31.75 b. Equivalent wt. of `HNO_3 = 12.6` c. Equivalent wt. of `N_2 = 2.8` d. BALANCED chemical reaction suggest that 10 EQUIVALENTS of Cu combines with 10 equivalents of `HNO_3`in order to produce CuO and 10 equivalents of `N_2` SOLUTION :`overset(0)(N_2) rarr 2Hoverset(+5)(NO_3)` Equivalent weight of nitrogen`=(28)/10 = 2.8`

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» Solution :`0.2F` FORMALITY `= ("Number of FORMULA weight of solute")/(VOLUME of solution in LITRE") = (5.85)/(58.5 xx 0.5) =0.2F`

Discussion

50962.	5.85 g of NaCl are dissolved in 500 cm^3 of water. Calculate the formality of the solution.
Answer» ANSWER :0.2 F

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» 0.1 M 2 M 1 M 0.5 M Solution :MOLARITY `=(5.85)/(58.5xx1)= 0.1M`

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» Solution :`0.2M` Molarity `= ("Number of MOLES of SOLUTE")/(" VOLUME of the solvent in Litre") = ((5.845)/(58.45))/(0.5) = (0.1)/(0.5) =0.2M`

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» Solution :The corresponding CHEMICAL equations are: `underset("one mole" 107.9 g) (2HNO_(3)) to underset("one mole")(AgNO_(3)) + NO_(2) + H_(2)O` `underset("One mole")(AgNO_(3)) + NaCl to underset("one mole" 143.35 g)(AGCL) DARR + NaNO_(3)` From the equations, it is clear that one mole (107.9 g) of Ag gives one mole (143. 35 g) of AgCI. `therefore` The mass of Ag which gives 7.20 g of AgCI `=107.9/143.35 XX 7.20 = 5.42 g` `therefore` Percentage of Ag in the given coin `=5.42/5.82 xx 100 = 93.1`

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» ANSWER :0.996Lit

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» ANSWER :0.931

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» Solution :WEIGHT of AgCl PRECIPITATED = 7.20 g Weight of silver in the preciptate `= ((108G))/((143.50g))xx(7.20g)=5.42g` PERCENTAGE of silver in the COIN `= ((5.42g))/((5.83g))xx100=93.1%`.

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» `C_(6)H_(6)` `C_(4)H_(8)` `C_(3)H_(8)` `CH_(3)COOH` Answer :B

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» 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` Thus, for `10^(@)` rise , `0.25` mol of the gas at constant volume require heat `= 52.25 J` `:. `For `1^(@)` rise, 1 mol of the gas at constant volume will require heat `= (52.25)/(10 xx 0.25) J = 20.9 J` `:. C_(V) = 20.9 J K^(-1) mol^(-1)` Now, `C_(p) = C_(v) + R= 20. 9 J K^(-1) mol^(-1) + 8. 314 J K^(1) mol^(-1)= 29.214 J K^(-1) mol^(-1)` `:. gamma = (C_(p))/( C_(v))= (29.214)/(20.9) = 1.4`. Hence, the gas is diatomic

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» When phenolphthalein is used as an indicator from the very beginning the titre value of HCl will be 60 mL When phenolphthalein is used as an indicator from the very beginning the titre value of HCl will be 40 mL. When methyl orange is used as an indicator from the very beginning, the titre value of HCl will be 80 mL. When methyl orange is used as an indicator after the first end point the titre value of HCl will be 30 mL. Solution :`m" Eq of "`(KOH)/(L)=(56)/((56)/(1))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 `m" Eq of "(KHCO_(3))/(L)=(100)/((100)/(1))xx10^(3)=1000=(20)/(20)mL` (b). With phenolphthalein: `m" Eq of "KOH+(1)/(2)m" Eq of "K_(2)CO_(3)=m" Eq of "HCl` `20+(40)/(2)=vxx1N(n=1)` `V_(HCl)=40mL` (c). With methyl orange: `m" Eq of "KOH+m" Eq of "K_(2)CO_(3)+m" Eq of "KHCO_(3)=m" Eq of "HCl` LTBRGT `20+40+20=Vxx1N` `V_(HCl)=80mL` (d). With methyl orange after the first end point: `(1)/(2)m" Eq of "K_(2)CO_(3)+m" Eq of "KHCO_(3)=m" Eq of "HCl` `20+20=Vxx1N` `V_(HCl)=40mL`

Discussion

50972.	5.6 litres of oxygen at STP is equivalent to
Answer» 1/4 MOLE 1 mole 1/2 mole 1/8 mole Solution :22.4 LITRES of `O_(2)` = 1 mole `:.` 5.6 litres of `O_(2)` =`1/22.4 XX 5.6` = 0.25 mole = 1/4 mole.

Discussion

50973.	5.6 litres of oxygen at NTP is equivalent to
Answer» 1 MOLE `1/2` mole `1/4` mole `1/8` mole ANSWER :C

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» Phosphine Phosgene Nitric Oxide Nitrous Oxide Solution :5.6 L of gas has the MASS of 11 g. `:.`22.4 L of gas will have the mass `11/5.6xx 22.4 = 44`

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» 4, 6 5, 5 2, 8 8, 2 Solution :`n_(N_(2)) = (56)/(28) = 2 , n_(O_(2)) = (96)/(32) = 3` `P_(N_(2)) = x_(N_(2)) xx P_("Total")` `= (2)/(2 + 3) xx 10 = 4 atm` `P_(O_(2)) = 10 - 4 = 6 atm`

Discussion

50976.	5.6 L of a gas at NTP weighs equal to 8g. The vapour density of gas is
Answer» 32 16 8 40 Solution :B) `therefore` 5.6 L of gas weighs at NTP = 8g `therefore` L of gas will weights at NTP = (8 XX 22.4)/(5.6) = 32g `therefore` Molecular weight of gas = 32 We know that, Molecular weight = 2 xx vapour density `therefore` Vapour density = `32/2` = 16

Discussion

50977.	56 gm of nitrogen and 4 gm of hydrogen are taken in a closed vessel
Answer» `0.4` 1 1.3 0.2 Answer :D

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» ANSWER :`1.78xx10^(13) DPS`

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» ANSWER :27

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» SOLUTION :0.6 gram HYDROGEN is produced by 5.4g metal 1 gram hydrogen is produced by `(1xx5.4)/(0.6)=9g` metal `therefore` EQ WT of metal = 9

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» Solution :MOLARITY of solution (M) `= (("Mass of "Na_(2)CO_(3))/("Molecular mass of "Na_(2)CO_(3)))/("Volume of solution in mL"/(1000))=((5.3g))/(("106 g MOL"^(-1))XX("0.5 L"))` `="0.1 mol L"^(-1)=0.1M`

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» Solution :The reaction of sodium CARBONATE with acid is given as `Na_(2)CO_(3)+2HClrarr 2NaCl+H_(2)O+CO_(2)` 1 MOLE of `Na_(2)CO_(3) = 1` mole of `CO_(2)` 106 grams of `Na_(2)CO_(3) = 44` grams of `CO_(2)` 5.3 grams of `Na_(2)CO_(3)=?` The AMOUNT of `CO_(2)` produced `=(5.3)/(106)xx44=2.2` grams Oxidation of graphite is given as `C+O_(2)rarr CO_(2)` 1 mole of `CO_(2) = 1` mole of graphite 44 grams of `CO_(2) = 6.022 XX 10^(23)` atoms 2.2 grams of `CO_(2) = ?` Number of atoms of .C. to be oxidised `= (2.2xx6.022x10^(23))/(44)= 3.011xx10^(22)`

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» 0.1 M 0.02 M 0.04 M none SOLUTION :`M_(1)=(5.3)/(106xx0.25)=0.2M,M_(1)V_(1)=M_(2)V_(2)` `0.2xx10=M_(2)xx50impliesM_(2)=0.04M`

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» ANSWER :1.7 ( w/v)% ,5.6v

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» SOLUTION :1 mol of `H_(2)` will be produced from 1 mol of `Li[(CH_(3))_(3)CO]_(3)AlH`. 1 mol of `Li[(CH_(3))_(3)O]_(3)` AlH will be produced from 1 mol of `LiAlH_(4)`. From second EQUATION, 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 REACT with excess of `CH_(4)` to produce 52.5 millimon of `H_(2)`.

Discussion

50986.	52% water is in mixture of alochol and water state mole fraction ethanol.
Answer» Solution :`H_(2)O=54` gmand ETHANOL `= 46 ` gm Moles of `H_(2)O = (54)/(18)=3` Moles of Ethanol `=(46)/(46)=1` `:.` MOLE FRACTION of ethanol `= 1/4 = 0.25`

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» `DeltaT_(1)= 2DeltaT_(1)` `DeltaT_(2)= 2DeltaT_(1)` `DeltaT_(2)= DeltaT_(1)` `DeltaT_(2)= 4DeltaT_(1)` Answer :C

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» `-29kcal` `-21kcal` `29 KCAL` `21 kcal` ANSWER :B

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» ANSWER :`T_(1)=T_(2)`

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» `1.0` `1//9` `4//9` `8//9` SOLUTION :`K_(c) =( 1.5 xx 1.5)/(1.5 xx 1.5) = 1 .` On CHANGING CONCENTRATIONS , equilibrium constant does not change .

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» `X CAL` 2 x cal zero `x// 2 cal` ANSWER :C

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» SOLUTION :`(500a+500b)/(2000)xx1.5impliesa+b=6` GIVEN `a=2bimplies a=4`

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» SOLUTION :N//A

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» `0.05N` `0.1N` `0.2N` REDUCED to ZERO SOLUTION :`[NO_(3)^(-)]=(0.1xx500)/(1000)=0.05N`

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» Solution :(i) Millimoles of `CH_(3)CO OH = 500 xx 0.2 = 100` Millimoles of HCl `= 500xx 0.2=100` Final volume after mixing `= 500 + 500 = 1000 ` mL `:. [ CH_(3)CO OH]=(100)/(1000) = 0.1 M,"" [HCl ] = (100)/(1000) = 0.1 M` `{:(,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-X),," "x,,(0.1+x),,):}` `K_(a) = (x(0.1+))/((0.1+x))` 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, `K_(a) = (x(0.1))/(0.1) = x = 1.75xx10^(-5) `MOL `L^(-1) ` (Given) `:.` Degree of dissociation `= (x)/(0.1) = (1.75xx10^(-5))/(0.1) = 1.75xx10^(-4)= 0.00175 %` Further, `[H^(+)]= 0.1 + x ~~ 0.1 :. pH = - log 0.1 = 1 ` (ii) 6g of NAOH = `(6)/(40) ` mole = 0.15 mole Hence, now the equilibrium will be `{:(,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):}` 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. `pH = - log K_(a) + log. (["Salt"])/(["Acid"]) = - log (1.75xx10^(-5)) + log (0.05)/(0.05) = 4. 757`

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» `[K^(+)] = 0.04, [AG^(+)]= 0.04, {Na^(+)]=0.002[Cl^(-)] = 0.04 , [NO_(3^(-)] = 0.042` `[K^(+)]=0.04, [Na^(+)] = 0.00166, [NO^(-_(3)]]= 0.0433` `[K^(+)]= 0.04, [Ag^(+)] = 0.05, [Na^(+)] = 0.0025, [Cl^(-)] = 0.05, [NO^(-_(3)] = 0.0525]` `[K^(+)] = 0.05, [Na^(+)]= 0.0025 [Cl^(-)]= 0.05, [NO^(-_(3)] = 0.0525` ANSWER :B

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» Solution :i. TOTAL mEq of acid `= 500 xx 2 + 100 xx 2 xx 2` (`n` factor) `= 1000 + 400 = 1400` Total mEq of monoacid alkali = 1000 Acid left after neutralisation with alkali `= 1400 - 1000 = 400 mEq` `N` fo acid left `= (400 mEq)/((500 + 100)mL) = (400)/(600) = 0.6 N` mEq of acid `-= mEq "of"Na_(2)CO_(3)` `30 mL xx 0.66 N = 20 mL xx N Na_(2)CO_(3)` Strength of `Na_(2)CO_(3) x H_(2)O = 143 g L^(-1)` `M_(Na_(2)CO_(3).xH_(2)O) = ("Strenght")/(Mw) = ((143)/(106 + 18x))` `M_(Na_(2)CO_(3).xH_(2)O) = ((2 xx 143)/(106 + 18 x))` (`n` factor = 2) Substitution the value of `N` of `Na_(2)CO_(3).xH_(2)O` in EQUATION (i), we get `30 xx 0.66 x = 20 xx ((2 xx 143)/(106 + 18 x))` solve for `x` `x = 10.16 ~~ 10` So WATER of crystallisation of `Na_(2)CO_(3).xH_(2) O = 10` formula : `Na_(2) CO_(3). 10 H_(2) O`

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» Solution :The balanced chemical equation for the reaction is : `underset("1 MOL")(N_(2)(G))+underset("3 mol")(3H_(2)(g))rarrunderset("2 mol")(2NH_(3)(g))` STEP I. To find the limiting reactant Mass of `N_(2)` taken `= 50.0 KG = 50,000 g` Moles of `N_(2)=("Mass of" N_(2))/("MOLAR mass of "N_(2))=(5000g)/((28"g mol"^(-1)))=1785.7mol` Mass of `H_(2)` taken `= 10.0 kg = 10,000 g` Moles of `H_(2)=(10,000g)/((2.0"g mol"^(-1)))=5000 mol` From the above balanced equation, 1 mol of `N_(2)` reacts with 3 mol of `H_(2)` `:. 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. Step II. To find the amount of `NH_(3)` formed. 3 moles of `H_(2)` from `NH_(3) =2` mol 5000 moles of `H_(2)` form `NH_(3)=5000 xx ((2 mol))/((3 mol))=3333.3 mol` Mass of `NH_(3)` formed `= 3333.3 xx 17 = 56666.6 g = 56.67 kg`.

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» Solution :A BALANCED equation for the above reaction is written as follows, `N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g))`. Mole of `N_(2)=50.0 kg N_(2) xx (1000 g N_(2))/(1 kg N_(2)) xx (1 "mol" N_(2))/(28.09g N_(2))` `= 17.86 xx 10^(2)` mol 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))` `= 4.96xx10^(3)` mol According to the equation ,1 mol `N_(2(g))` required 3 mol `(H_(2))`. For the reaction. Hence, for `17.86xx10^(2)` mol of `N_(2(g))` , the MOLES of `H_(2(g))` required would be : `= 17.86 xx 10^(2)` mol `xx(3 "mol"H_(2(g)))/(1 "mol"N_(2(g)))` `= 5.36xx10^(3)` mol `H_(2(g))` But we have only `4.96xx10^(3)` mol `H_(2)`. Hence diphydrogen is the limiting reagent in this CASE. So, `NH_(3(g))` would be formed only from that amount of available dihydrogen , i.e `4.96xx10^(3)` mol `H_(2(g))`. Since 3 mol `H_(2(g)), 2 mol NH_(3(g))` `4.96xx10^(3) mol H_(2(g)) xx (2 mol NH_(3(g)))/(3 mol H_(2(g)))` `= 3.30xx10^(3) ` mol `NH_(3(g))` `3.30xx10^(3)` mol `NH_(3(g))` is obtained. If they are to be converted to GRAMS, it is done as follows: 1 mol `NH_(3(g))= 17.0 g NH_(3 (g))` `:.3.30xx10^(3) mol NH_(3(g))xx(17.0 g mol NH_(3(g)))/(1 mol NH_(3(g)))` `= 3.30xx10^(3) xx17 g NH_(3(g))` `= 56.1 xx 10^(3) g NH_(3)` `= 56.1 kg NH_(3(g))`

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» SOLUTION :At CONSTANT volume, `DeltaV=0`. Applying `DeltaU= Q+w=q+PDeltaV, ` we get `DeltaU = q= 500J`.

Discussion

Explore topic-wise InterviewSolutions in .

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

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:

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.

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

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

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

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

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.

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

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 ……………

5.85 g of NaCl are dissolved in 500 cm^3 of water. Calculate the formality of the solution.

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

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 …………..

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

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

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.

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 ?

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

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.

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?

5.6 litres of oxygen at STP is equivalent to

5.6 litres of oxygen at NTP is equivalent to

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

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:

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

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

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

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?

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

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?

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.

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.

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

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

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)

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 ?

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

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

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.

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

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:

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

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.

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.

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 ?