94420 + Interview Questions in Chemistry in Class 12

1.	One litre of a gas at S.T.P. weighs 1.16 g. It can possibly be
Answer» `C_2H_2` `CO` `O_2` `CH_4` SOLUTION :1L of gas weighs= 1.16 g 22.4 L of gas weigh` = 1.16 xx 22.4 = 26` MOLECULAR weight of gas = 26 it can be POSSIBLY `C_2H_2`

Discussion

2.	One litre of 2 Macetic acid and one litre of 3 M ethyl alcohol are mixed to form ester according to the given equation CH_(3)COOH+C_(2)H_(5)OHrarrCH_(3)COOC_(2)H_(5)+H_(2)O If each solution is diluted by adding equal volume (1 litre) of water, by how many times the initial forward rate is reduced?
Answer» 4times 2times `0.5`TIMES `0.25` times Answer :C

Discussion

3.	One litre of 2 M acetic acid is mixed with one litre of 2 M ethyl alcohol to form ester as : CH_(3) COOH+C_(2)H_(5)Ohrarr CH_(3)COOC_(2)H_(5)+H_(2)O The decrease in initial reaction rate if each solution is diluted by an equal volume of water would be :
Answer» `(1)/(2) ` time `(1)/(4)` TIMES 2 times 4 times SOLUTION :(B) Concentration of both the reactants after dilution WOULD be REDUCED to one half of the INITIAL concentration.

Discussion

4.	One litre of a gas at NTP weights 1.97g. Find the molecular mass of gas.
Answer» ANSWER :44.128

Discussion

5.	One litre of 16 M H_2SO_4 has been diluted to 100L. The normality of the resulting solution is :
Answer» 0.16 N 0.08 N 0.32 N 0.64 N . SOLUTION :`M_1V_1=M_2V_2` `16 M xx 1L = M_2 xx 100 L` `M_2 = (16)/(100) = 0.16 M ` Normality `= 0.16 xx 2 = 0.32 N`

Discussion

6.	One litre of 1 M CH_3 COOH solution is thoroughly agitated with 6gms of charcoal. After the processthe solution shows 0.9M concentration. What is 'x/m' of adsorption.
Answer» Solution :`M = (wt)/(GMt) cdot 1000/V , 1 = (wt)/(60) cdot 1000/1000 = 60 ` after `0.9 = (wt)/(60).1000/1000 = 54` absorbed `60 - 54 = 6 gr rarr on 6` gr of characoal `? Rarr 1 gr , 1 gr`.

Discussion

7.	One litre each of a Buffer solution is taken in two beakers connected through a salt bridge. Platinum electrode is immersed in each beaker and electrysis is carried out for 10,000 sec. using a current of 9.65 Ampere. The Buffer contain (1)/(3) mole each of Na_(2)HPO_(4) and NaH_(2)PO_(4) Initially. using the given data answer the following. Given: ltbr. pK_(a) (H_(3)PO_(4)) = 4, pK_(a) (H_(2)PO_(4)^(-)) = 7 , pK_(a) (HPO_(4)^(2-)) = 10 log 2 = 0.3, log3 = 0.48 2.303 (RT)/(F) = 0.06 at 298K What is pH of solution in which reduction taken place duringelectrolysis.
Answer» `12.59` `11.91` `13.01` `11.51` Solution :In cathode: `K_(eq) = (K_(w))/(K_(3)) = 10^(-4)` `PO_(4)^(3-) +H_(2)O hArr HPO_(4)^(2-) +OH^(-)` `RARR pOH = (1)/(2) pK_(b) - (1)/(2)logC rArr pH = 11.91`

Discussion

8.	One litre each of a Buffer solution is taken in two beakers connected through a salt bridge. Platinum electrode is immersed in each beaker and electrysis is carried out for 10,000 sec. using a current of 9.65 Ampere. The Buffer contain (1)/(3) mole each of Na_(2)HPO_(4) and NaH_(2)PO_(4) Initially. using the given data answer the following. Given: ltbr. pK_(a) (H_(3)PO_(4)) = 4, pK_(a) (H_(2)PO_(4)^(-)) = 7 , pK_(a) (HPO_(4)^(2-)) = 10 log 2 = 0.3, log3 = 0.48 2.303 (RT)/(F) = 0.06 at 298K After electrolysis if H_(2) gas is bubbled at 1 bar pressure on each platinum electrode, what is the EMF of resulting Galvenic cell.
Answer» `1.09V` `0.29V` `0.59V` `0.39V` SOLUTION :`EMF = E_("cathode") - E_("Anode")` `E_(H^(+)//H_(2)) = E_(H^(+)//H_(2))^(@) -(0.0591)/(2)log.(1)/([H^(+)]^(2))` `E_(H^(+)//H_(2)) =- 0.059 PH` `EMF =- 0.0591 pH_("cathode") -(-0.0591 pH_("anode"))` `= (pH_("anode") -pH_("cathode")) 0.06` `= (11.91 -2.09) 0.06` `= 0.59`

Discussion

9.	One litre CO_2 is passed over hot coke . The volume becomes 1.4 litre . The percent composition of products is :
Answer» 0.6 LITRE CO 0.8 litre `CO_2` 0.6 litre `CO_2` and 0.8 litre CO None Answer :C

Discussion

10.	One litre aqueous solution of sucrose (molar mass = "342 g mol"^(-1)) weighing 1015 g is found to record an osmotic pressure of 4.82 atm at 292 K. What is the molality of the sucrose solution? (R = 0.0821 atm "mol"^(-1)"K"^(-1)).
Answer» Solution :`PI="CRT gives C = 0.2 mol L"^(-2)="0.2 mol in 1015 g solution."` `"Hence, mass of solvent "=1015-(0.2xx342)=946.6g.` `"MOLALITY "=("0.2mol")/("946.6g")xx"1000 g kg"^(-1)=0.2112m.`

Discussion

11.	One industrial processs for the synthesis of 1-butanol begins with ethanol. Show how this synthesis might be carried out.
Answer» Solution :ETHANOL can be converted to an ALDOL via an aldol addition. Then, DEHYDRATION would PRODUCE 2-butenal, which can be hydrogenated to FURNISH 1-butanol.

Discussion

12.	One hundred millilitres of 0.8 M copper sulphate is electrolysed for 30 minutes by passing a current of 5 amp. Calculate the amount of copper sulphate in grams in the solutions.
Answer» ANSWER :5.32 G

Discussion

13.	One hundred grams of a food product was found to contain 6.0 g of nitrogen. If protein contains 16% nitrogen. What is the percentage of protein in food?
Answer» 0.19 `37.5%` `1.9%` `9.5%` SOLUTION :Protein percentage`= (6 xx 100)/(16) = 37.5%`

Discussion

14.	One hundred gram of a 5 m urea solutionare cooled to- 6^(@)C .Whatamountof urea will seprate out ? (K_(f) = 1.86 K m^(-1)) ?
Answer» Solution :Molalityof solution at ` - 6^(@)C` `Delta T_(f) = k_(f) m` or `"" m = (Delta T_(f))/(K_(f)) = (6)/(1.86) = 3.23 MOL kg ^(-1)` 5 m solution of UREA means that 5 MOLES or `5 xx 60` gof urea are present in 1000 g of solventor `(1000+5xx6)`g of solution ,so that . `(100+300)` g of solutioncontain urea = 300 g . 100g of solutioncontain urea ` = (300)/(1300) xx 100` ` =23.08 g` Mass of water in the solution`= 100- 23.08` ` = 76.92 g` Whenurea separates out , mass of waterdoes notchange.At ` - 6^(@)C`, the molalityis `3.23 mol kg^(-1)` . Letus calculatethe amountof urea to makethe MOLALITY`3.23 mol kg^(-1)` . 1000 g of watercontain urea ` = 60 xx 3.23 g` `76.92` g of watercontain urea ` = (60 xx 3.23 xx 76.92)/(1000)` ` = 14.91 g` Mass ofurea SEPARATED ` = 23.08- 14.91` ` = 8.17g`

Discussion

15.	One gram sample of ammonium nitrate is decomposed in a bomb calorimeter. The temperature of calorimeter increases by 6.12 K. The heat capacity of the system is 1.23 kJ//g//deg. What is the molar heat of decomposition of NH_(4)NO_(3)
Answer» `- 7.53 KJ mol^(-1)` `- 398.1 kJ mol^(-1)` `- 16.1 kJ mol^(-1)` `- 602 kJ mol^(-1)` ANSWER :C

Discussion

16.	One having high vapour pressure at temperature below its melting point:
Answer» BENZOIC ACID SALICYLIC acid Citric acid All of these Answer :A

Discussion

17.	One gram sample of NH_(4)NO_(3) is decomposed in a bomb calorimeter. The temperature of the calorimeter increases by 6.12 K the heat capacity of the system is 1.23 kJ/g/deg. What is the molar heat of decomposition for NH_(4)NO_(3)
Answer» `-7.53` kJ/mol `-398.1 kJ//mol` `-16.1` kJ/mol `-602` kJ/mol Solution :Molecular WEIGHT of `NH_(4)NO_(3)=80` Heat EVOLVED = `1.23xx6.12` `therefore`Molar heat CAPACITY = `1.23xx6.12xx80` `=-602 kJ//mol`.

Discussion

18.	One gram of silver gets distributed between "10 cm"^(3) of molten zinc and "100 cm"^(3) of molten lead of 800^(@)C. The percentage of silver in the zinc layer is approximately
Answer» 89 91 97 94 Answer :C

Discussion

19.	One gram of mass is equal to:
Answer» `5 xx 10^(10)` ERG `9 xx 10^(2)` erg `7 xx 10^(5)` erg `11 xx 10^(12)` erg Answer :B

Discussion

20.	One gram of hydrogen is found to combine with 80 g of bromine one gram of calcium (valency = 2) combines with 4 g of bromine the equivalent weight of calcium is
Answer» 10 20 40 80 Solution :ONE gram of hydrogen COMBINES with 80 gm of BROMINE. So, EQUIVALENT weight of bromine =80 gm `:'` 4 gm of bromine combines with 1 gm of CA `:.` 80 gm of bromine combines with `=1/4xx80=20`.

Discussion

21.	One gram of fat gives
Answer» Same AMOUNT of ENERGY as one gram of CARBOHYDRATE Same amount of energy as one gram of protein More than twice the energy as onegram of carbohydrate or protein None ANSWER :C

Discussion

22.	One gram of [Cr(H_(2)O)_(6)]Cl_(3) was passed through cationic exchanger and the HCl produced is diluted to 1 L. Find the molarity of acid.
Answer» SOLUTION :0.0112 M

Discussion

23.	One gram of commercial AgNO_3is dissolved in 50 mL of water. It is treated with50 mL of a KI solution. The silver iodide thus precipitated is filtered off. Excess of KI in the filtrate is titrated with (M/10) KIO_3solution in the presence of 6 M HCl till all I ions are converted into ICl. It requires 50 mL of (M/10) KIO_3solution. 20 mL of the same stock solution of KI requires 30 mL of (M/10) KIO_3under similar conditions. Calculate the percentage of AgNO_3in the sample.
Answer» SOLUTION :REACTION : `KIO_3 +2KI + 6HCL= 3ICl + 3H_2O ` 85%

Discussion

24.	One gram of commercial AgNO_(3) is dissolved in 50 mL of water. It is treated with 50 mL of a KI solution. The silver iodide thus precipitated is filtered off. Excess of KI in the filtrate is titrated with (M//10) KIO_(3) solution in presence of 6 M HCl till all I^(-) ions are converted into ICl. It required 50 mL of (M//10) KIO_(3) solution. 20 mL of the same stock solution of KI requires 30 mL of (M//10) KIO_(3) under similar conditions. Calculate the percentage of AgNO_(3) in the sample.Reaction : KIO_(3) + 2KI + 6HCl rarr 3ICl + 3KCl + 3H_(2)O
Answer» Solution :Number of millimoles of `KIO_(3)` in 30 mL of solution = `"MOLARITY" xx"volume in mL" = 1/10 xx 30 = 3` Given reaction : `KIO_(3) + 2KI + 6HCl rarr 3 ICl + 3 KCl + 3H_(2)O` According to the equation given, 1 MOLE of `KIO_(3)` is equivalent to 2 moles of KI Number of millimoles of KI in 20 mL of stock solution = `2 xx 3 = 6` Number of millimoles of KI in 50 mL of the same solution = `6 xx 50/20 = 15 ` Number of millimoles of `KIO_(3)` in 50 mL of solution = 1/10 xx 50 = 5 Number of millimoles of KI used with `AgNO_(3)` = 15 - 10 = 5 `AgNO_(3) + KI rarr AGI + KNO_(3)` 1 mole of `AgNO_(3)` reacts with 1 mole of KI. Therefore, number of millimoles of `AgCl_(3)` is equal to 5 Weight of `AgNO_(3) = 5 xx 10^(-3) xx 170 g = 0.85 g` `% of AgNO_(3) = 0.85 xx 100/1.0 = 85.0 % `

Discussion

25.	One gram of charcoal adsorbs 100 ml of 0.5M CH_3CO OH & then molarity of acetic acid reduces to 0.49 M. The no. of milli moles of acetic acid adsorbed is _____
Answer» SOLUTION :`M = (WT)/(GMt) cdot 1000/V , 0.5 (wt)/(60)cdot (1000)/(100)` ,after `0.49 = (wt)/(60) cdot 1000/100` w.t = 3 , w.t. 2.94 , wt REDUCED = 0.06 , no. of mole = `(0.06)/(60) = 0.001,` Milimoles = 1

Discussion

26.	One gram of a water insoluble substance of density 0.8 g cm^(-3) is despersed in 1 L of water forming a colloidal solution having 10^(13) particles of spherical shape per mm^(3). Calculate the radius of the particles.
Answer» Solution :Particles present in `1 mm ^(3)` of the solution `=10^(13)` `therefore` PATICLES present in `1 CM^(3)` of the solution `=10^(13)xx10^(3)=10^(16)` Particles present in `1 L (1000cm^(3))` of the solution `=10^(16)xx10^(3)=10^(19)` Mass of the substance dispersed =1 g DENSITY of the substance `=0.8 g cm^(-3)` `therefore` Volume of the dispersed substance `=(1.0)/(8.0)=1.25cm^(3)` `therefore` Volume of 1 particle `=(1.25)/(10^(19))=1.25xx10^(19)cm^(3)` But volume `=4/3pi r^(3)=1.25xx10^(19)cm^(3)` or `r^(3)=3/4xx7/22xx1.25xx10^(-19)=29.8xx10^(-21)cm^(3)` `r=(29.8)^(1//3)xx10^(-7)cm=3.1xx10^(-7)cm=310xx10^(-9)m=310 nm`

Discussion

27.	One gram of activated carbon has a surface area of 1000cm^2 considering complete coverage as well as monomolecular adsorption. How much ammonia at latm and 273K would be adsorbed on the surface of 44/7gr. carbon. If radius of a ammonia molecule is 10^(-8)cm .........
Answer» Solution :Total surface area of carbon `44/7 xx 10^(7) cm^(2) , r = 10^(-8) cm` surface area of `NH_3 = pi r^2 = 22/7 xx 10^(-16) cm^2` No. of `NH_3` molecules adsorbed ` = ((44)/(7) xx 10^(7))/(22/7 xx 10^(-16)) = 2 xx 10^(23)` vol. of `NH_3` adsorbed at STP `= (2 xx 10^(23))/(6 xx 10^(23)) xx 22.4 = 7.46 ` lit

Discussion

28.	One gram of a mixture of potassium and sodium chlorides on treatment with excess of silver nitrate gave 2 g AgCl. What was the composition of the two salts in the original mixture?
Answer» ANSWER :`NaCl=14%, KCl=86%`

Discussion

29.	One gram of a mixture of Ma_(2)CO_(3) and NaHCO_(3) consumes y gram equivalent of HCl for complete neutralization. One gram of the mixture is strongly heated, then cooled and the residue treated with HCl. The gram equivalent of HCl now required for complete neutralization will be
Answer» 2 y 3 y y y/2 Solution :`2NaHCO_(3)OVERSET(DELTA)rarrNa_(2)CO_(3)+H_(2)O+CO_(2)` `Na_(2)CO_(3)overset(Delta)rarr" No decomposition"` As 1 g eq of `NaHCO_(3)` on heating PRODUCES 1 g eq of `Na_(2)CO_(3)`, number of gram equivalents of the mixture remains the same. Hence, number of gram equivalents of HCL consumed will also remain the same.

Discussion

30.	One gram of a chloride was found to contain 0.835 g of chlorine. Its vapour density is 85. calculate its molecular formula.
Answer» Solution :MASS of metal chloride=1g Mass of chlorine=0.835g Mass of metal =(1-0.835)=0.165 G Equivalent mass of metal `=(0.165xx35.5)/(0.835)` VALENCY of the metal `=(2V.D.)/(E+35.5)` `=(2xx85)/(7.01+35.5)` =4 Formula of the chloride `=MCI_(4)` .

Discussion

31.	One gram of a mixture of KCl and KI dissolved in water and precipitated with silver nitrate 1.618g of silver halides are obtained calculate the percentage composition of mixture.
Answer» ANSWER :KCL = `39.6%` and KI= `60.4%`

Discussion

32.	One gram molecule of a gas at NTP occupies 22.4 L. This fact was derived from
Answer» AVOGADRO's law Charles law Dalton's law Boyle's law Answer :A

Discussion

33.	One gram mole of a gas at NTP occupies 22.4 litres. This fact was derived from
Answer» law of gaseous volumes Avogadro's HYPOTHESIS Berzelius hypothesis DALTON's atomic theory Answer :B

Discussion

34.	One gram mole of a gas at N.T.P. occupies 22.4 litres. This fact was derived from :
Answer» LAW of GASEOUS volumes Avogadro's HYPOTHESIS BERZELIUS hypothesis Dalton's ATOMIC theory Solution :Avogadro.s hypothesis

Discussion

35.	one gram mole of a gas at N.T.P. occupies 22.4 L. This fact was derived from
Answer» Law of GASEOUS volumes Avogadro's HYPOTHESIS Berzerlius hypothesis Dalton's hypothese Solution :According to Avogadro's hypothesis GRAM MOLAR volume is 22.4 L at N.T.P.

Discussion

36.	One gram is more than
Answer» 0.1 MOL of `CO_2` mass of `6.02 xx 10^22`molecules of water mass of 2.24 L of HYDROGEN GAS at N.T.P. 0.1 gram ATOM of carbon Solution :mass of 2.24 L of hydrogen gas at N.T.P.

Discussion

37.	One gram equivalent of a substance is liberated at an electrode by :
Answer» `6.22xx10^(23)` ELECTRONS 96500 C 1 amp of current for ONE second 1 amp current for 96500 C Solution :(a,B,d) are CORRECT.

Discussion

38.	One gram atom of which of the following compound will require maximum amount of electricity for its decomposition ?
Answer» `AgNO_(3)` `MgSO_(4)` `AlCl_(3)` `CuCl_(2)` Solution :`AlCl_(3)` contain `Al^(+3)` cation. `therefore Al^(3+)+3e^(-) to Al`

Discussion

39.	One gram atom of zinc can reduce
Answer» 1 MOL CONC.`HNO_(3)` 2 mol conc.`HNO_(3)` 3 mol conc.`HNO_(3)` 4 mol conc.`HNO_(3)` Answer :2

Discussion

40.	One gm of impure sodium carbonate is dissolved in water and the solution is made up to 250ml. To 50ml of this solution, 50ml of 0.1N – HCl is added and the mixture after shaking well required 10ml of 0.16N NaOH solution for complete titration. Calculate the % purity of the sample.
Answer» ANSWER :0.901

Discussion

41.	One gram - atom of Ca burnt in excess of oxygen wasdissolved in waterto makeup a litre solution. Calculate the normality alkaline solution .
Answer» Solution :First Method `Ca (O_(2))/(" BURNT") CaO (" disolved in water ") to Ca(OH)_(2)` m.e of Ca = m.e of CaO= m.eof 1000 mL of `Ca(OH)_(2)` solution ` :.` m.e of 1000 mL of `Ca(OH)_(2)`solution = m.e of Ca ` = eq xx 1000` = (gram - atom ` xx` VALENCY of Ca) `xx` 1000 ` = 1 xx 2 xx 1000` = 2000 ` :. `normalityof `Ca(OH)_(2)` solution = `(m.e)/(" vol in mL")` ` (2000)/(1000 ) = 2 N `

Discussion

42.	One gram atom of aluminium can reduce how many moles of chromic oxide?
Answer» ANSWER :0.5

Discussion

43.	One gas bleaches the colour of flowers by reduction and another gas by oxidation . The gases respectively are
Answer» `SO_(2)` and `Cl_(2)` CO and `Cl_(2)` `NH_(3)` and `SO_(2)` `H_(2)S ` and `Br_(2)` ANSWER :A

Discussion

44.	One gas bleaches the colour of the flowers by reduction while the other by oxidation. The gases are:
Answer» `CO, CI_2` `H_2S, Br_2` `SO_2,CI_2` `NH_3,SO_3` ANSWER :C

Discussion

45.	One gas bleaches the colour of flowers by reduction while the other by oxidation :
Answer» Solution :`SO_(2)`bleaches by REDUCTION while chlorine bleaches colour of flowers by OXIDATION.

Discussion

46.	One Faraday of electricity will liberate half gram atom of the metal from a solution of _________.
Answer» `AuCl_(2)` `CUCL` `BaCl_(2)` NaCl Answer :C

Discussion

47.	One faraday of electricity is passed through molten Al_(2)O_(3), aqueous solution of CuSO_(4) and molten NaCl taken in three different electrolytic cells connected in series. The mole ratio of Al, Cu and Na deposited at the respective cathodes is
Answer» `2:3:6` `6:2:3` `6:3:2` `1:2:3` SOLUTION :`Ag^(3+)+3e^(-)TOAL` `Cu^(2+)+2e^(-)toCu` `Na^(+)+e^(-)toNa` thus 1F will be deposit `(1)/(3)` mole of `AL,(1)/(2)` mole of Cu and 1 mole of Na. Hence, MOLAR ratio `=(1)/(3):(1)/(2):1=2:3:6`

Discussion

48.	One Faraday of electricity will liberate 1 gram atom of the metal from the solution of
Answer» `CuCl_2` `CuSO_4` `AgNO_3` `AuCl_3` ANSWER :C

Discussion

49.	One Faraday of electricity is passed through molten Al_(2)O_(3), aqueous solution of CuSO_(4) and molten NaCl taken in three different electrolytic cells connected in series . The mole ratio of Al , Cu and Na deposited at the respective cathode is :
Answer» `2: 3: 6` `6 : 2 : 3` `6 : 3 : 2 ` `1 : 2 : 3` SOLUTION :(a) `Al^(3+)+underset(1F)underset(3F)3e^(-) to nderset(1//3"mol")underset(1"mol")(Al)` `Cu^(2+)+underset(1F)underset(2F)2E^(-) to underset(1//2mol)underset(1mol)(Cu)` `Na^(+)+underset(1F)e^(-) to underset(1mol)(Na)` Mole RATIO of Al, Cu and Nadeposited at the respective CATHODES is : `1//3 : 1//2 : 1//1 "or"2:3:6`

Discussion

50.	One Faraday of electricity is passed through molten Al_(2)O_(3), aqueous solution of CuSO_(4)and molten NaCl taken in three different electrolytic cells connected in series. The mole ratio of AI, Cu and Na deposited at the respective cathode is
Answer» `2:3:6` `6:2:3` `6:3:2` `1:2:3` Solution :The charge carried by 1 mole of electrons is ONE faraday. Thus for a reaction `M^(n+) +nc ^(-) to M` `nF=1` mole of M `{:(Al^(3+)+3e^(-) to Al),(3F """1mole"),(IF"""1/3mole"):} ` `{:(Cu^(2+) +2C^(-) to Cu),(2F """1mole"),(IF """1mole"):}` `{:(Na^(+)+E ^(-) to Na),(IF """mole"):}` The mole ration of Al, Cu and Na deposited at eh respectivecat ODE is `1/3. 1/2: 1 sigma 2:3:6`

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

One litre of a gas at S.T.P. weighs 1.16 g. It can possibly be

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One litre CO_2 is passed over hot coke . The volume becomes 1.4 litre . The percent composition of products is :

One litre aqueous solution of sucrose (molar mass = "342 g mol"^(-1)) weighing 1015 g is found to record an osmotic pressure of 4.82 atm at 292 K. What is the molality of the sucrose solution? (R = 0.0821 atm "mol"^(-1)"K"^(-1)).

One industrial processs for the synthesis of 1-butanol begins with ethanol. Show how this synthesis might be carried out.

One hundred millilitres of 0.8 M copper sulphate is electrolysed for 30 minutes by passing a current of 5 amp. Calculate the amount of copper sulphate in grams in the solutions.

One hundred grams of a food product was found to contain 6.0 g of nitrogen. If protein contains 16% nitrogen. What is the percentage of protein in food?

One hundred gram of a 5 m urea solutionare cooled to- 6^(@)C .Whatamountof urea will seprate out ? (K_(f) = 1.86 K m^(-1)) ?

One gram sample of ammonium nitrate is decomposed in a bomb calorimeter. The temperature of calorimeter increases by 6.12 K. The heat capacity of the system is 1.23 kJ//g//deg. What is the molar heat of decomposition of NH_(4)NO_(3)

One having high vapour pressure at temperature below its melting point:

One gram sample of NH_(4)NO_(3) is decomposed in a bomb calorimeter. The temperature of the calorimeter increases by 6.12 K the heat capacity of the system is 1.23 kJ/g/deg. What is the molar heat of decomposition for NH_(4)NO_(3)

One gram of silver gets distributed between "10 cm"^(3) of molten zinc and "100 cm"^(3) of molten lead of 800^(@)C. The percentage of silver in the zinc layer is approximately

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One gram of hydrogen is found to combine with 80 g of bromine one gram of calcium (valency = 2) combines with 4 g of bromine the equivalent weight of calcium is

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One gram of [Cr(H_(2)O)_(6)]Cl_(3) was passed through cationic exchanger and the HCl produced is diluted to 1 L. Find the molarity of acid.

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One gram of activated carbon has a surface area of 1000cm^2 considering complete coverage as well as monomolecular adsorption. How much ammonia at latm and 273K would be adsorbed on the surface of 44/7gr. carbon. If radius of a ammonia molecule is 10^(-8)cm .........

One gram of a mixture of potassium and sodium chlorides on treatment with excess of silver nitrate gave 2 g AgCl. What was the composition of the two salts in the original mixture?

One gram of a mixture of Ma_(2)CO_(3) and NaHCO_(3) consumes y gram equivalent of HCl for complete neutralization. One gram of the mixture is strongly heated, then cooled and the residue treated with HCl. The gram equivalent of HCl now required for complete neutralization will be

One gram of a chloride was found to contain 0.835 g of chlorine. Its vapour density is 85. calculate its molecular formula.

One gram of a mixture of KCl and KI dissolved in water and precipitated with silver nitrate 1.618g of silver halides are obtained calculate the percentage composition of mixture.

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One gram mole of a gas at NTP occupies 22.4 litres. This fact was derived from

One gram mole of a gas at N.T.P. occupies 22.4 litres. This fact was derived from :

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One gram atom of zinc can reduce

One gm of impure sodium carbonate is dissolved in water and the solution is made up to 250ml. To 50ml of this solution, 50ml of 0.1N – HCl is added and the mixture after shaking well required 10ml of 0.16N NaOH solution for complete titration. Calculate the % purity of the sample.

One gram - atom of Ca burnt in excess of oxygen wasdissolved in waterto makeup a litre solution. Calculate the normality alkaline solution .

One gram atom of aluminium can reduce how many moles of chromic oxide?

One gas bleaches the colour of flowers by reduction and another gas by oxidation . The gases respectively are

One gas bleaches the colour of the flowers by reduction while the other by oxidation. The gases are:

One gas bleaches the colour of flowers by reduction while the other by oxidation :

One Faraday of electricity will liberate half gram atom of the metal from a solution of _________.

One faraday of electricity is passed through molten Al_(2)O_(3), aqueous solution of CuSO_(4) and molten NaCl taken in three different electrolytic cells connected in series. The mole ratio of Al, Cu and Na deposited at the respective cathodes is

One Faraday of electricity will liberate 1 gram atom of the metal from the solution of

One Faraday of electricity is passed through molten Al_(2)O_(3), aqueous solution of CuSO_(4) and molten NaCl taken in three different electrolytic cells connected in series . The mole ratio of Al , Cu and Na deposited at the respective cathode is :

One Faraday of electricity is passed through molten Al_(2)O_(3), aqueous solution of CuSO_(4)and molten NaCl taken in three different electrolytic cells connected in series. The mole ratio of AI, Cu and Na deposited at the respective cathode is