52671 + Interview Questions in Chemistry in Class 11 Page 1019 InterviewSolution

50901.	8 grams of NaOH are totally consumed to react with carbondioxide produced by the calcination of sodium bicarbonate. What weight of sodium bicarbonate is to be calcinated?
Answer» ANSWER :06.8g

Discussion

50902.	8 gm O_(2) has the same number of molecules as in
Answer» 11 g `CO_(2)` 22g `CO_(2)` 7g CO 14g CO Solution :`8gm O_(2)=(8)/(32)=(1)/(4)` MOLES same molecules IMPLIES same moles `(11)/(44)=(1)/(4),(7)/(28)=(1)/(4)`

Discussion

50903.	8 gm NaOH dissolve in 250 mL solution. This solution is diluted by using 500 ml water. Find the molartiy in' dilute solution. Also find moles of NaOH ?
Answer» ANSWER :0.4 M and 0.2 MOLE

Discussion

50904.	8 gm H_2 and 256 gm HI in 4 L flask calculate this active mass. (H = 1 g "mol"^(-1), I = 127 g "mol"^(-1))
Answer» SOLUTION :1 MOL `L^(-1) H_2`, 0.5 mol `L^(-1)` HI

Discussion

50905.	8 g of sulphur is burnt to form SO_(2) which is oxidised by chlorine water . The solution is treated with BaCl_(2) solution . The amount of BaSO_(4) precipitated is
Answer» 1 MOLE 0.5 mole 0.25 mole 0.125 mole ANSWER :C

Discussion

50906.	7g of nitrogen occupies a volume of 5 litres under certain conditions. Under the same conditions one mole of a gas, having molecular weight 56, occupies a volume of
Answer» 40L 20L 10L 80L Answer :B

Discussion

50907.	7g of nitrogen contains the same number of molecules as
Answer» 8g of oxygen 16G of oxygen 8g of CARBON monoxide 22G of carbon dioxide Answer :A

Discussion

50908.	7g of N_2 is allowed to react with 2g H_2 in presence of catalyst to form NH_3 Which is the limiting reagent?
Answer» SOLUTION :`N_2`-LIMITING REAGENT,

Discussion

50909.	7g of N_2 is allowed to react with 2g H_2 in presence of catalyst to form NH_3 Calculate the maximum amount of NH_3 that can be formed in the reaction.
Answer» SOLUTION :`NH_3=8.5g`

Discussion

50910.	750 mL of nitrogen are collected over water at 25^@C and 740 mm pressure. If the aqueous tension at this temperature is 23.8 mm Hg, calculate the mass of the dry gas.
Answer» SOLUTION :According to Dalton.s law of partial pressure, `P_(gas) = P_(obs). - " aqueous TENSION " = 740 - 23.8 = 716.2 mm Hg` The given gas thus occupies a volume of 750 mL at `25^@C " and " 716.2 mm Hg`. According to the ideal gas equation, `PV =nRT= m/M RT` `m=(PVM)/(RT)` In the present case, `P =716.2 mm = (716.2)/760 =0.9424` atm `V=750 mL = 0.750 L, "" T=25+273 =298 K` `M = 28 " and " R =0.0821 L " atm " K^(-1) mol^(-1)` `m=(0.9424xx0.750xx28)/(0.0821xx298) = 0.809 g` Hence, the mass of the given gas = 0.809 g

Discussion

50911.	7.5 g of an acid are dissolved per litre of the solution. 20 mL of this acid solution required 25 mL of N//15 NaOH solution for complete neutralisation. Calculate the equivalent mass of the acid.
Answer» Solution :Calculation of the normality of the acid Volume of acid taken = 20 ML volume of base required = 25 mL Normality of base `= N//15` Applying normality equation : `ubrace(N_(1)xxV_(1))_("Acid") equiv ubrace(N_(2) XX V_(2))_("Base")` `N_(1) xx (20 mL) = ((1)/(15) N) xx 25 mL` `N_(1) = ((1)/(15) N) xx ((25 mL))/((20 mL)) = (1)/(12) N` Step II. Calculation of the equivalent mass of the acid Amount of acid dissolved per litre of solution = 7.5 g `:.` Strength of the acid `= 7.5 g L^(-1)` Equivalent mass of the acid `= ("Strength of the acid")/("Normality of acid") = ((7.5 g L^(-1)))/((1//12 N)) = 90.0`

Discussion

50912.	7.5 g of a gas occupies a volume of 5.6 litres at 0^(@)C and 1 atm pressure. The gas is
Answer» NO `N_(2)O` `CO` `CO_(2)` Solution :7.5 g of gas occupies a volume of 5.6 liters at 273 K and 1 atm pressure. THEREFORE, the mass of gas that occupies a volume of 22.4 litres `(7.5g)/(5.6L) xx22.4L=30G` Molar mass of NO (14+16) = 30g

Discussion

50913.	7.5 g of a gas occupies 5.6 litres as STP. The gas is
Answer» `NO` `N_(2)O` `CO` `CO_(2)` ANSWER :A

Discussion

50914.	74.5 g metal chloride contain 35.5 g chlorine. Then the equlvalent weight of metal is.....
Answer» 39 74.5 78 19.5 Solution :`74.5 - 35.5 = 39 G`

Discussion

50915.	7.38 g of a sample of a metal oxide is quantitatively reduced to 6.84 g of pure metal. If the specific heat of the metal is 0.0332 cal/g, calculate the valency and the accurate atomic weight of the metal.
Answer» Solution :Let the formula of the oxide be `M_(2)O_(x)` , where x is the valency of the METAL M. We have, therefore, `x xx `moles of M = `2 xx` moles of O. We know that, atomic weight `xx` specific HEAT `~~` 6.4 (Dulong and Petit.s law) `:.` approximate atomic weight `= (6.4)/(0.0332) = 193 ` From EQN. `(1), x xx (6.64)/(193) =2xx ((7.38-6.84))/(16)` or x= 1.9. But valency is always a whole number and so the valency of the metal is 2. Now, to calculate the accurate value of the atomic weight of the metal, substitute the value of x as 2 in equation (1) again, or `2xx(6.84)/("atomic WT.") = 2xx((7.38 - 6.84))/(16)` Atomic weight (accurate) = 202.67.

Discussion

50916.	723 KJ mol""^(-1) heat is released when 1 molemethanol combustions in present of O_(2). What amount of heat is enlisted of one mole of O_(2) is used?
Answer» 241 kJ 723 kJ 482 kJ 924 KJ Solution :Combusion reaction of methanol, `CH_(3) OH_((l)) + (3)/(2) O_(2(g)) = CO_(2(g)) + 2H_(2) O_((l)) , DELTA H= -723` kJ 1 mole `(3)/(2)` mole therefore `(3)/(2)` moles `O_2` is used in this reaction. If 1 mole `O_2` is used. then free ENERGY `= (2)/(3) xx` combustion energy 723 `= 482` kJ

Discussion

50917.	720cc of methane diffused through a porous membrane in 30min. Under identical conditions 240cc of gas 'X'diffused in 20 min. Calculate the molecular weight of 'X'.
Answer» Solution :Rate of diffusion of methane ` = 720//30 = 24 "cc" m^(-1)` Rate of diffusion of gas .X.`240//20 = 12 "cc" m^(-1)` Ratio of rates of diffusion is GIVEN as `(r_(1))/(r_(2))= sqrt((M_(2))/(M_(1))) = (24)/(12) = 12` Molecular weight of .X. ` = 2^(2) XX 16 = 64`

Discussion

50918.	7.18 g iron displaces 2.04 g copper from copper sulphate solution. If the equivalent weight of copper is 31.7, calculate the equivalent weight of iron.
Answer» Solution :In the present case, MASS of IRON taken = 1.8 g Mass of copper displaced = 2.04 g Equivalent mass of copper = 31.7 Using the relation `=("Mass of iron")/("Mass of copper") = ("Equivalent weight of iron")/("Equivalent weight of copper") ` and substituting the VALUES, we have `1.8/2.04 = ("Equivalent weight of iron")/31.7` or Equivalent weight of iron `=(1.8 XX 31.7)/2.04= 27.97`

Discussion

50919.	70 g of a sample of magnesite on treatment with excess of HCl gave 11.2 L of CO_(2) at STP. The percentage purify of the sample
Answer» 80 70 60 50 Answer :C

Discussion

50920.	7.00 g of a gas occupies a volume of 4.1 L at 300 K and 1 atm pressure. What is the molecular mass of the gas?
Answer» Solution :In the PRESENT case, `P= 1 " atm, " V= 4.1 L, T = 300 K, m= 7.0 g, " and " R=0.0821 L " atm " K^(-1) mol^(-1)` `:."" PV = m/M RT` `:."" M=(mRT)/(PV)= (7.0 xx0.0821 xx 300)/(1 x 4.1)= 42` AMU and `:. ""d = m/V` `:. "" d=(7.0)/(4.1)=1.7g L^(-1)` Hence, the molecular mass of the given gas is 42 amu and its density in the given conditions is`1.7 g L^(-1)`.

Discussion

50921.	70 gms of a metal oxide on reduction produced 54 gms of metal. The atomic weight of the metal is 81. What is its valency?
Answer» Solution :wt. of metal oxide = 70 G wt. of metal = 54 g `therefore` wt of oxygen = `70-54=16g` `(W_(1))/(W_(2))=(E_(1))/(E_(2)),("wt of metal")/("wt of oxygen")=("EQ wt of metal")/("EQ wt of oxygen")` `(54)/(16)=(E)/(8)impliesE=27` At. Wt = EQ wt `xx` valency `81=27xx` valency implies Valency = 3

Discussion

50922.	7 grams of nitrogen is present at 127^@C and 16 grams of oxygen at 27^@C. Calculate (a) ratio of kinetic energy and (b) ratio of average kinetic energy of nitrogen and oxygen.
Answer» Solution :`(K.E) = 3/2 NRT ` For two gases ratio of KINETIC energies = ` ((K.E )_1)/((K.E)_2) = (n_1 T_1)/(n_2 T_2)= (7)/(28 ) xx 400 xx(32)/(16) xx (1)/(300 ) = 2:3`

Discussion

50923.	7 g of a sample of sodium chloride on treatment with excess of silver nitrate gave 14.35 g of AgCl. The percentage of NaCl in the sample is
Answer» 80 50 65.8 83.5 Answer :D

Discussion

50924.	7-Bromo-1,3,5-cycloheptatriene exists as an ion whereas 5-bromo-1,3-cyclopentadiene does not form an ion even in presence of Ag^+. Explain
Answer» Solution :7-Bromo-1,3,5-cycloheptatriene, on ionization, gives tropylium ion. Since , tropylium ion contains `6 PI`-electrons which are completely delocalized, therefore, according to Huckel rule, it is AROMATIC and HENCE stable .Being highly stable , it is easily FORMED . In contrast,5-bromo-1,3-cyclopentadiene, on ionization, will give 1,3-cyclopentadienyl cation which contains `4pi`-electrons and hence in antiaromatic. Being antiaromatic, it is highly unstable and hence is not formed even in the presence of `Ag^+` ion which otherwise facilitates ionization.

Discussion

50925.	6xx10^(22) gas molecules each of mass 10^(-24)kg are taken in a vessel of 10 litre. What is the pressure exerted by gas molecules? The average velocity of the gas molecules is 92.62m/sec.
Answer» `2XX10^(5)PA` 20Pa `2xx10^(6)Pa` `2xx10^(4)Pa` Answer :d

Discussion

50926.	6L H_2O_2 contains 440 gm solute than how much litre O_2 gas will produce from this solution at STP ?
Answer» 24.16 22.8 30.16 25.8 SOLUTION :Molarity=`"Weight of solute GM"/"Molecular massx Volume of solution in litre"` `=440/(34xx6)=2.156 M H_2O_2` `V(O)_2`= M x 11.2 = 2.156 x 11.2 =24.16 L `O_2` GAS

Discussion

50927.	6g. of Urea is dissolved in 90g. of water. The mole fraction of solute is
Answer» `1//5` `1//50` `1//51` `1//501` ANSWER :C

Discussion

50928.	6g of NaOH are dissolved in 200 cm^3 of water. What is the relation between molarity and normality of the solution thus obtained ?
Answer» ANSWER :MOLARITY and NORMALITY are the same

Discussion

50929.	6g of Mg reacts with excess of an acid. The amount of hydrogen produced would be
Answer» 0.5g 1g 2g 4g Answer :C

Discussion

50930.	6g of hydrogen at 1.5 atm and 273^@C occupies a volume of
Answer» 44.8L 89.6L 67.2L 11.2L ANSWER :B

Discussion

50931.	""_(6)C^(12) and ""_(1)T^(3) are formed in nature due to the nuclear reaction of neutron with
Answer» `""_(7)N^(14)` `""_(6)C^(13)` `""_(2)He^(4)` `""_(3)Li(6)` SOLUTION :`""_(7)^(14)N+_(0)^(1)n to ""_(6)^(12)C+""_(1)^(3)T`

Discussion

50932.	6.9g N_2O_4 is taken 0.5 L closed vessel at 400 K temperature. The equilibrium N_2O_(4(g)) hArr 2NO_(2(g)) total pressure at equilibrium is 9.15 atm calculate K_c, K_p and partial pressure of each component.
Answer» SOLUTION :`p_(N_2O_4)`= 0.69 atm , `p_(NO_2)`= 8.46 atm , `K_p`=103.73 atm , `K_c=3.162 "MOL L"^(-1)`

Discussion

50933.	6.8 gm of H_2O_2 is present in 500 ml of an aqueous solution. When 10 ml of this solution is decomposed, what is the S.T.P. volume of oxygen obtained.
Answer» 560 ML 11.2 ml 44.8 ml 22.4 ml ANSWER :C

Discussion

50934.	66 gm sample of an oxalate salt Al_(x)H_(y)(C_(2)O_(4))_(z).nH_(2)O is dissolved in water to form 500mL solution. 50mL solution requires 60mL of 0.5M Ba(OH)_(2) and 240mL 0.1 M KMnO_(4) in acidic medium separately. If in salt x,y,z and n are present in simplest ratio, then select the correct statement(s).
Answer» MOLES of oxalate salt in original sample is 0.2 Ratio of y/z is EQUAL is 1 Value of (x+y+z)-n=5 Number of oxalate ION per molecule of oxalate salt is 2 Answer :a,b,c

Discussion

50935.	6.54 g of zinc are treated with 11.5 g of H_2SO_4. Calculate the volume of hydrogen evolved at S.T.P. How much H_2SO_4 will be left in excess?
Answer» Solution :The CORRESPONDING CHEMICAL equation is: `underset(65.38 g)(Zn) + underset(2.016 + 32.06 + 64.0= 98.1 g)(H_(2)SO_(4)) to ZnSO_(4) + underset(22.4 "L at S.T.P")(H_(2))` From the equation, it is clear that 6.54 g of zinc will react with 9.81 g of `H_2SO_4`. Thus, the amount of `H_2SO_4` TAKEN is in excess. Hence, Zn will get consumed completely and this is the limiting reagent. `therefore 65.38 g` of zinc EVOLVE `H_(2) = 22.4 L` at S.T.P. `therefore 6.54 g` of zinc will evolve `H_(2) = 22.4/(65.38) xx 6.54 = 2.24` L at S.T.P. Further, `therefore 65.38` g of zinc react with `=H_(2)SO_(4) = 98.1 g` `therefore H_(2)SO_(4)` left in excess `=11.5- 98.1 = 1.69 g` Hence, 2.24 L hydrogen will be evolved at S.T.R and 1.69 g of `H_2SO_4` will be left in excess.

Discussion

50936.	.^(64)Cu (hlaf lifre = 12.8h)deacays by beta^(-) emission (38%), beta^(+)- emission (19%) and electron capute (43%). Write the decay products and calculatepartial half-live for each of the decay processes.
Answer» Solution :We know, Given `lambda_(av) = (0.693)/(12.8) hr^(-1)` `:. lambda_(1) +lambda_(2) +lambda_(3) =lambda_(av) = (0.693)/(12.8) = 5.41xx10^(-2) hr^(-1)` .....(1) ALSO for parallel PATH DECAY `lambda_(1)` = Fractionlal yield of `._(30)^(64)Zn xx lambda_(av)` `lambda_(2)` = Fractionalyeild of `._(28)^(64) Ni xx lambda_(av)` `lambda_(3)` = Fractionlal yield of `._(28)^(64) Ni xx lambda_(av)` `:. (lambda_(1))/(lambda_(2)) = (38)/(19)` ....(2) `(lambda_(1))/(lambda_(3)) = (38)/(43)` ....(3) From eqs. (1), (2) and (3) `lambda_(1) = 2.056xx10^(-2) hr^(-1)` `lambda_(2) = 1.028xx10^(-2) hr^(-1)` `lambda_(3) = 2.327xx10^(-2) hr^(-1)` `:. t_(1//2)` for `beta^(-)` emission `= (0.693)/(2.056xx10^(-2)) = 33.70 hr` `t_(1//2)` for `beta^(+)` emission `= (0.693)/(1.028xx10^(-2)) = 67.41 hr` `t_(1//2)` for electron capture `= (0.693)/(2.327xx10^(-2)) = 29.28 hr`

Discussion

50937.	6.46gm BaSO_(4) is obtain from 4.81gm compound in estimation of sulphur by Carius method. Calculat ethe % of sulphur
Answer» ANSWER :0.185

Discussion

50938.	64 gm of CH_(4) "and" 68gm "of" H_(2)S was placed in an close container and heated up to 727^(@)C following equilibrium is established in gaseous phase reaction is: CH_(4)(g)+2H_(2)S(g)hArrCS_(2)(g)+4H_(2)(g) The total pressure at equilibrium is 1.6 atm and partial pressure of H_(2) "is" 0.8 atm. Then
Answer» Total moles at equilibrium `4.8` `K_(P)=K_(C)(RT)^(2)` Mole fraction `H_(2)` at equilibrium`=0.5` On increasing moles of `H_(2)S` equilibrium constant increases. Solution :`CH_(4)(g)+2H_(2)S(g)hArrCS_(2)(g)+4H_(2)(g)` `{:(t=0,64gm,68gm),("Moles",4"mole",2"mole"):}` `4-x 2-2x x 4x` Total moles at equilibrium`=4-x+2-2x+x+4x=(6-2x)` ACC. To `PV=nRT,T=727+273=1000K` `1.84xxV=(6-2x)xxRxx1000 `....(1) For `H_(2) "gas" V,P_(H_(2))=n_(H_(2))RTto0.8xxV=4xxRxx1000` ....(2) From EQ (1) and(2) `(1.6)/(0.8)=((6-2x))/(4x)` `x=0.6`

Discussion

50939.	6.3 g of oxalic acid dihydrate have been dissolved in water to obtain a 250 ml solution. How much volume of 0.1 N NaOH would be required to neutralize 10 mL of this solution ?
Answer» 40 mL 20 mL 10 mL 4 mL Solution :EQ. wt of `overset(COOH)underset(COOH)(\|).2H_(2)O = 63` Normality of the solution prepared `N=(w xx 1000)/(E xx V) = (6.3 xx 1000)/(63 xx 250) = 0.4 N` `underset(NaOH)(N_(1)V_(1)) = underset("OXALIC acid")(N_(2)V_(2))` `therefore V_(1) = (N_(2)V_(2))/N_(1) =(0.4 xx 10)/0.1 = 40 mL`

Discussion

50940.	60mL of pure ozone is heated and then colled. The increase in the volume is 10mL. Determine the percentage volume of ozone decomposed into oxygen.
Answer» Solution :The balanced equations for the decomposed of ozone is `2O_(3)(g)to3O_(2)(g)` `2 vol "" 3vol"Gay-Lussac.s COEFFICIENTS"` `60mL "" "0mL at start"` `60-2x "" 3x "after the REACTION"` Total volume of the mixture after the reaction =60-2x+3x=60+10 SOLVING, x=10 Volume of `O_(3)` decomposed =20mL Percentage volume of `O_(3)` decomposed `=(20/(60)xx100=33.3%)`

Discussion

50941.	60g CH_(3)COOH and 46g C_(2)H_(5)OH react in 5L flask to form 44g CH_(3)COOC_(2)H_(5) at equilibrium. On taking 120 g CH_(3)COOH and 46g C_(2)H_(5)OH, CH_(3)COOC_(2)H_(5) formed at equilibrium is :
Answer» 44g 20.33g 22g 58.66g Solution :Molar mass of `CH_(3)COOH=60gmol^(-1)` Molar mass of `C_(2)H_(5)OH=46g mol^(-1)` Molar mass of `CH_(3)COOC_(2)H_(5)=88g mol^(-1)` `:. [CH_(3)COOH)_("Initial")=(60)/(60xx5)=0.2 "mol"L^(-1)` `[C_(2)H_(5)OH]_("Initial")=(46)/(46xx5)=0.2"mol" L^(-1)` `[CH_(3)COOC_(2)H_(5)]_(eqm)=(44)/(88)xx(1)/(5)=0.1 "mol" L^(-1)` `CH_(3)COOH+C_(2)H_(5)OH hArr CH_(3)COOC_(2)H_(5)+H_(2)O` Initial `0.2 M ""0.2M` At eqm. `(0.2-0.1)M""(0.2-0.1)M""0.1M""0.1M` `:. K=([CH_(3)COOC_(2)H_(5)][H_(2)O])/([CH_(3)COOH][C_(2)H_(5)OH])` `=(0.1xx0.1)/(0.1xx0.1)=1` In second case, `[CH_(3)COOH]_("Initial")=0.4M` `[C_(2)H_(5)OH]_("Initial")=0.2M` If x is the amount of acid and ALCOHOL reacted `[CH_(3)COOH]_("eqm.")=(0.40-x)M` `[C_(2)H_(5)OH]_("eqm.")=[0.2-x]M` `[CH_(3)COOC_(2)H_(5)]_("eqm.")=[H_(2)O]_("eqm.")=xM` `:. K=(x^(2))/((0.4-x)(0.2-x))=1` `:. x^(2)=(0.4-x)(0.2-x)` `x^(2)=0.08+x^(2)-0.6x` `0.6x=0.08` `x=(8)/(60)M` `:. `Moles of ethyl acetate produced `=(8)/(60)XX5=(2)/(3)` Mass of ethyl acetate produced `=(2)/(3)xx88=58=58.66g`.

Discussion

50942.	60g of a compound on analysis gave C=24g, H=4g and O=32g. Its empirical formula is
Answer» `C_(2)H_(4)O_(2)` `C_(2)H_(2)O_(2)` `CH_(2)O_(2)` `CH_(2)O` ANSWER :D

Discussion

50943.	6.048 g H_(2) and 28 g N_(2)react with each other and produce 34.048 g NH_(3). This reaction is explained by which law ?
Answer» CONSTANT PROPORTION MULTIPLE proportion Combining weights Charles law Answer :A::C

Discussion

50944.	6.023xx10^(22) atoms of Hydrogen can make
Answer» 0.05 MOLES of `H_(2)` molecules 0.1 gms of HYDROGENS ATOMS 0.1 GRAM-molecules of Hydrogen 0.1 gram atoms of Hydrogen Solution :1 mole of compound `-6.023xx10^(23)` molecules Hydrogen `(H_(2))` is diatomic 1 gram atom of any ELEMENT `=6.023xx10^(23)` atoms of that element Atomic weight of hydrogen = 1 Number of gram atoms = `("Weight in grams")/("Gramotomicweight")`

Discussion

50945.	6.023xx10^(20) molecules of urea are present in 100 ml of its solution. The concentration of the solution is
Answer» 0.001 M 0.01 M 0.02 M 0.1 M Solution :100 ml of solution contain `6.023 XX 10^(20)` molecules. `6.023 xx 10^(23)` molecules in 1000 ml = 1 M `:. 6.023 xx 10^(20)` molecules in 1000 ml=`(1xx6.023xx10^(20))/(6.023xx10^(23))=10^(-3)` `10^(-3)` moles of PRESENT in 100 ml `:.` In 1000ml the moles present is=`10^(-3)/cancel(100)xxcancel(1000)=10^(-2)` moles CONCENTRATION = 0.01 M

Discussion

50946.	6.022 xx 10^(22) molecules of N_(2) at NTP will occupy a volume of
Answer» 22.4 LITRES 2.24 litres 6.02 litres 6.02 ML ANSWER :B

Discussion

50947.	6.022 xx 10^(20) molecules of urea are present in 100 mL of its solution. The concentration of urea solution is :
Answer» 0.001 M 0.01 M 0.02 M 0.1 M SOLUTION :No. of MOLES urea `(6.022xx10^(20))/(6.022xx10^(23))=0.001` MOL Concentration of solution `=((10^(-3)"mol"))/(("0.1 L"))xx10^(-2)"molL"^(-1)=0.01 M`.

Discussion

50948.	6.02 xx 10^(20) molecules of urea are present in 100 mL of its solution. The concentration of urea solution is:
Answer» 0.001 M 0.01 M 0.02 M 0.1 M Solution :No. of MOLES of urea taken `=(6.02 xx 10^(20))/(6.022 xx 10^(23)) = 10^(-3)` `therefore` MOLARITY of solution `=10^(-3)/100 xx 1000 = 0.01 M`

Discussion

50949.	60 mL of methane and 180mL of oxygen are mixed and fixed. The resulting gases are passed throughcaustic soda solution. Determine the composition of gases before and after passing through base solution.
Answer» Solution :The storichiometric equation is `underset((g))(CH_(4))+underset((g))(2O_(2))tounderset((g))(CO_(2))+underset((l))(2H_(2))O` `1 vol 2 vol 1vol ovol """Gay-Lussac.s coefficients"` `60mL 180mL 0mL 0mL """ at start"` `0mL 60mL 60mL 0mL "" "after the reaction" ` Before passing through caustic soda solution unreacted oxygen =60mL Carbondioxide FORMED =80ML After passing through caustic soda solution(since `CO_(2)` is ABSORBED) unreaced oxygen =60mL

Discussion

50950.	60 gm of oleum (labelled as 118%) is mixed with 11.8 gm of water. What will be the composition of final mixture?
Answer» Only `H_(2)SO_(4)`, having MASS 71.8 gm 118 gm of `H_(2)SO_(4)` 70.8 gm `H_(2)SO_(4)` and 1 gm water 32 gm `SO_(3)` and 39.8 gm `H_(2)SO_(4)` Answer :C

Discussion

Explore topic-wise InterviewSolutions in .

8 grams of NaOH are totally consumed to react with carbondioxide produced by the calcination of sodium bicarbonate. What weight of sodium bicarbonate is to be calcinated?

8 gm O_(2) has the same number of molecules as in

8 gm NaOH dissolve in 250 mL solution. This solution is diluted by using 500 ml water. Find the molartiy in' dilute solution. Also find moles of NaOH ?

8 gm H_2 and 256 gm HI in 4 L flask calculate this active mass. (H = 1 g "mol"^(-1), I = 127 g "mol"^(-1))

8 g of sulphur is burnt to form SO_(2) which is oxidised by chlorine water . The solution is treated with BaCl_(2) solution . The amount of BaSO_(4) precipitated is

7g of nitrogen occupies a volume of 5 litres under certain conditions. Under the same conditions one mole of a gas, having molecular weight 56, occupies a volume of

7g of nitrogen contains the same number of molecules as

7g of N_2 is allowed to react with 2g H_2 in presence of catalyst to form NH_3 Which is the limiting reagent?

7g of N_2 is allowed to react with 2g H_2 in presence of catalyst to form NH_3 Calculate the maximum amount of NH_3 that can be formed in the reaction.

750 mL of nitrogen are collected over water at 25^@C and 740 mm pressure. If the aqueous tension at this temperature is 23.8 mm Hg, calculate the mass of the dry gas.

7.5 g of an acid are dissolved per litre of the solution. 20 mL of this acid solution required 25 mL of N//15 NaOH solution for complete neutralisation. Calculate the equivalent mass of the acid.

7.5 g of a gas occupies a volume of 5.6 litres at 0^(@)C and 1 atm pressure. The gas is

7.5 g of a gas occupies 5.6 litres as STP. The gas is

74.5 g metal chloride contain 35.5 g chlorine. Then the equlvalent weight of metal is.....

7.38 g of a sample of a metal oxide is quantitatively reduced to 6.84 g of pure metal. If the specific heat of the metal is 0.0332 cal/g, calculate the valency and the accurate atomic weight of the metal.

723 KJ mol""^(-1) heat is released when 1 molemethanol combustions in present of O_(2). What amount of heat is enlisted of one mole of O_(2) is used?

720cc of methane diffused through a porous membrane in 30min. Under identical conditions 240cc of gas 'X'diffused in 20 min. Calculate the molecular weight of 'X'.

7.18 g iron displaces 2.04 g copper from copper sulphate solution. If the equivalent weight of copper is 31.7, calculate the equivalent weight of iron.

70 g of a sample of magnesite on treatment with excess of HCl gave 11.2 L of CO_(2) at STP. The percentage purify of the sample

7.00 g of a gas occupies a volume of 4.1 L at 300 K and 1 atm pressure. What is the molecular mass of the gas?

70 gms of a metal oxide on reduction produced 54 gms of metal. The atomic weight of the metal is 81. What is its valency?

7 grams of nitrogen is present at 127^@C and 16 grams of oxygen at 27^@C. Calculate (a) ratio of kinetic energy and (b) ratio of average kinetic energy of nitrogen and oxygen.

7 g of a sample of sodium chloride on treatment with excess of silver nitrate gave 14.35 g of AgCl. The percentage of NaCl in the sample is

7-Bromo-1,3,5-cycloheptatriene exists as an ion whereas 5-bromo-1,3-cyclopentadiene does not form an ion even in presence of Ag^+. Explain

6xx10^(22) gas molecules each of mass 10^(-24)kg are taken in a vessel of 10 litre. What is the pressure exerted by gas molecules? The average velocity of the gas molecules is 92.62m/sec.

6L H_2O_2 contains 440 gm solute than how much litre O_2 gas will produce from this solution at STP ?

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

6g of NaOH are dissolved in 200 cm^3 of water. What is the relation between molarity and normality of the solution thus obtained ?

6g of Mg reacts with excess of an acid. The amount of hydrogen produced would be

6g of hydrogen at 1.5 atm and 273^@C occupies a volume of

""_(6)C^(12) and ""_(1)T^(3) are formed in nature due to the nuclear reaction of neutron with

6.9g N_2O_4 is taken 0.5 L closed vessel at 400 K temperature. The equilibrium N_2O_(4(g)) hArr 2NO_(2(g)) total pressure at equilibrium is 9.15 atm calculate K_c, K_p and partial pressure of each component.

6.8 gm of H_2O_2 is present in 500 ml of an aqueous solution. When 10 ml of this solution is decomposed, what is the S.T.P. volume of oxygen obtained.

6.54 g of zinc are treated with 11.5 g of H_2SO_4. Calculate the volume of hydrogen evolved at S.T.P. How much H_2SO_4 will be left in excess?

.^(64)Cu (hlaf lifre = 12.8h)deacays by beta^(-) emission (38%), beta^(+)- emission (19%) and electron capute (43%). Write the decay products and calculatepartial half-live for each of the decay processes.

6.46gm BaSO_(4) is obtain from 4.81gm compound in estimation of sulphur by Carius method. Calculat ethe % of sulphur

6.3 g of oxalic acid dihydrate have been dissolved in water to obtain a 250 ml solution. How much volume of 0.1 N NaOH would be required to neutralize 10 mL of this solution ?

60mL of pure ozone is heated and then colled. The increase in the volume is 10mL. Determine the percentage volume of ozone decomposed into oxygen.

60g CH_(3)COOH and 46g C_(2)H_(5)OH react in 5L flask to form 44g CH_(3)COOC_(2)H_(5) at equilibrium. On taking 120 g CH_(3)COOH and 46g C_(2)H_(5)OH, CH_(3)COOC_(2)H_(5) formed at equilibrium is :

60g of a compound on analysis gave C=24g, H=4g and O=32g. Its empirical formula is

6.048 g H_(2) and 28 g N_(2)react with each other and produce 34.048 g NH_(3). This reaction is explained by which law ?

6.023xx10^(22) atoms of Hydrogen can make

6.023xx10^(20) molecules of urea are present in 100 ml of its solution. The concentration of the solution is

6.022 xx 10^(22) molecules of N_(2) at NTP will occupy a volume of

6.022 xx 10^(20) molecules of urea are present in 100 mL of its solution. The concentration of urea solution is :

6.02 xx 10^(20) molecules of urea are present in 100 mL of its solution. The concentration of urea solution is:

60 mL of methane and 180mL of oxygen are mixed and fixed. The resulting gases are passed throughcaustic soda solution. Determine the composition of gases before and after passing through base solution.

60 gm of oleum (labelled as 118%) is mixed with 11.8 gm of water. What will be the composition of final mixture?