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51551.

13.8 g of N_2O_4 was placed in a 1 L reaction vessel at 400 K and allowed to attain equilibrium, N_2O_(4(g)) hArr 2NO_(2(g)) The total pressure at equilibrium was found to be 9.15 bar. Calculate K_c, K_p and partial pressure at equilibrium.

Answer»

Solution :Molecular mass of `N_2O_4`=2(N)+4(O)
=2(14)+4(16) = 92 g `"mol"^(-1)`
Mole of `N_2O_4 = n = "WEIGHT"/"Molecular mass"= "13.8 g"/"92 g mol"^(-1)`
= 0.15
Temperature T=400 K
Gas constant = R= 0.083 bar L `mol^(-1) K^(-1)`
pV=nRT where , p=Partial pressure `N_2O_4`
`therefore p=(nRT)/V`
`therefore p=((0.1 "mol")(0.083 "bar L mol"^(-1) K^(-1))(400 K))/(1 L)`
= 4.98 bar partial pressure of `N_2O_4`
`{:("Equilibrium reaction :",N_2O_(4(g)) hArr, 2NO_(2(g))),("INITIAL pressure :", "4.98 bar","0 bar"),("Pressure change :","-x bar" , "+2X bar"),("Partial pressure at equilibrium :","(4.98-x)bar","2x bar"):}`
`therefore` Total pressure at equilibrium = Addition of partial pressure
`p_"total"=p_(N_2O_4)+p_(NO_2)`
`therefore` 9.15=(4.98-x)+2x
`therefore` 9.15=4.98 +x
`therefore` x=(9.15-4.98)=4.17 bar
So, partial pressure at equilibrium
`p_(NO_2)`= 2x=2(4.17)=8.34 bar
`p_(N_2O_4)`=(4.98-x)=(4.98-4.17)=0.81 bar
The expression of `K_p`,
`K_p=(p_(NO_2))^2/(p_(N_2O_4))="(8.34 bar)"^2/"0.81 bar"`=85.87 bar
The calculation of `K_c` from `K_p`:
`K_p=K_c(RT)^(Deltan)` So, `K_c=K_p/(RT)^(Deltan_(g))`
where , `K_p`=85.87 bar, R=0.083 L bar `mol^(-1) K^(-1)`
T=400 K , `Deltan_((g))`=(2-1)=+1
`therefore K_c=(85.87)/(400xx0.083)^1=2.5864 approx` 2.586 M
Discussion
51552.

13.8 g ethyl alcohol is dissolved in 7.1 g water give the ratio of mole of ethyl alcohol and water.

Answer»

`3:4`
`1:2`
`1:4`
1:1`

ANSWER :A::C::D
Discussion
51553.

13.8 g element consist 4.6xx10^(22) atoms which the atomic mass of element ?

Answer»

290
180
34.4
10.4

Solution :`4.6xx10^(22)` atoms `= 13.8g`
`6.022xx10^(23)` atoms `=(?)`
Atomic mass `=(6.022xx10^(23)xx13.8)/(4.6xx10^(22))`
`= 180.66 ~~180 G "MOLE"^(-1)`
Discussion
51554.

1.375 g of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper that remained was 1.098 g In another experiment, 1.179 g of copper was dissolved in nitric acid and the resulting copper nitrate converted into cupric oxide by ignition. The weight of cupric oxide formed was 1.476 g. Show that these result illustrate the law of constant composition.

Answer»

Solution : In the first EXPERIMENT:
Mass of CuO taken = 1.375 g
Mass of Cu obtained = 1.098 g
`therefore` Mass of oxygen that combined with
`Cu = 1.375 - 1.098 = 0.277g`
Thus, the percentage of oxygen in the given sample of
`CuO = 0.277/(1.375)XX 100 = 20.14 %`
In the serial experiment:
Since, 1.178 g of Cu yielded 1.467 g of CuO, the mass of oxygen added would be:
`=1.476 - 1.178 = 0.298 g`
`therefore` Percentage of oxygen in this SIMPLE
`=(0.298 xx 100)/1.476 = 20.19%`
Since, the percentage of oxygen is almost the same (within the limit of experimental errors) in the two cases, the data are in accordance to the LAW of constant proportion and prove it.
Discussion
51555.

1.355 g of pyrolusite sample are added to 50 mL of 1 N oxalic acid solution containing sulphuric acid. After the reaction is completed, the contents are transferred to a measuring flask and the volume made up to 200 mL. 20 mL of this solution is titrated against KMnO_(4) solution whose strength is 2 g//L and 31.6 mL of KMnO_(4) solution are required. Calculate the percentage purity in the given sample of pyrolusite.

Answer»


SOLUTION :N//A
Discussion
51556.

1,3,5,7-Octatetraene contains (X)________ sigma-bonds and (Y)_______pi-bonds. 'X' and 'Y' are

Answer»

23.4
17,4
18,5
33,2

Solution :`CH_(2)=CH-CH=CH-CH=CH-CH=CH_(2)` NOL. Of `sigma`-BONDS =17, No. of `pi`-bonds=4
Discussion
51557.

1.325 g of anhydrous sodium carbonate are dissolved in water and the solution made up to 250 mL. On titration, 25 mL of this solution neutralise 20 mL of a solution of sulphuric acid. How much water should be added to 450 mL of this acid solution to make it exactly N/12 ?

Answer»


ANSWER :225 ML
Discussion
51558.

1.325 g of anhydrous sodium carbonate are dissolved in water and the solution made up to 250 mL. On titration 25 mL of this solution neutralise 20 mL of a solution of suphuric acid. How much water should be added to 450 mL of this acid solution to make it exactly N//12 ?

Answer»

Solution :EQ. mass of `Na_(2)CO_(3)=("Mol.mass")/(2)=(106)/(2)=53`
250 mL of the SODIUM carbonate solution contains = 1.325 g
1000 mL of the sodium carbonate solution contains
`=(1.325g)/(250)xx1000=5.300` g
NORMALITY of `Na_(2)CO_(3)` solution` =("Strength "(g//L))/("Eq. mass")`
`=(5.30)/(53)=(1)/(10)N`
Applying `UNDERSET((Na_(2)CO_(3)))(N_(1)V_(1))-=underset((H_(2)SO_(4)))(N_(2)V_(2))`
`(1)/(10)xx25=N_(2)xx20`
`N_(2)=(25)/(10xx20)=(1)/(8)`
Applying`underset(("Before dilution"))(N_(B)V_(B))-=underset(("After dilution"))(N_(A)V_(A))`
`=(1)/(8)xx450=(1)/(12)xxV_(A)`
`V_(A)=(450xx12)/(8)=675 mL`
Water to be added for dilution =(675-450)=225 mL
Discussion
51559.

1,3-pentadiene and 1,4-pentadiene are compared with respect to their intrinsic stability and reaction with HI. The correct statement is :

Answer»

1,3-pentadiene is more STABLE and more REACTIVE than 1,4-pentadiene
1,3-pentadiene is LESS stable and less reactive than 1,4-pentadiene
1,3-pentadiene is morestable butless reactive than 1,4-pentadiene
1,3-pentadiene is lessstable butmore reactive than 1,4-pentadiene

Answer :A
Discussion
51560.

1,3-Dibromopropane reacts with metallic zinc to form

Answer»

Propene
Propane
Cyclopropane
Hexane

Solution :
Discussion
51561.

1,3-butadiene reacts wwith ethylene to form

Answer»

Benzene
Cyclohexane
Cyclohexene
2,3-Dimethyl BUTANE

ANSWER :C
Discussion
51562.

12ml of H_(2)SO_(4) is completely neutralised with 15ml of 0.1N NaOH.What is the strength of acid in gL^(-1)?

Answer»


ANSWER :6.125
Discussion
51563.

12L of SO_2 gas diffused through a porous membrane in 20 minutes. Under same conditions 36L of X gas diffused in 30 minutes. The molar mass of the gas 'X' in g mol^(-1) is :

Answer»

64
32
16
8

Answer :C
Discussion
51564.

1.2g of organic compound on kjeldahlization librates ammonia which consumes 30 cm^(3) of 1N HCI. The percentage of nitrogen in the organic compound is:

Answer»

30
35
46.67
20.28

Answer :B
Discussion
51565.

1.2g of Mg atoms in vapour phase absorb 50.0 kJ of energy to convert Mg atoms into Mg ions. The energy absorbed is needed for the following changes : Mg(g) rarr Mg^(+) (g)+ e^(-) , DeltaH =750 kJ mol^(-1) Mg^(+)(g) rarrMg^(2+) (g) + e^(-) , Delta H = 1450 kJ mol^(-1) Calculate the percentage of Mg^(+) and Mg^(2+) ions in the final mixture.

Answer»

Solution :No. of moles of Mg `=( "Mass")/( "ATOMIC mass") = (1.2)/( 24) = 0.05`
Suppose final mixture contains 'a' moles of `Mg^(+)` ions and 'b' moles of `Mg^(2+)` ions. Then
`a+b=0.05`….(i)
`Mararr Mg^(+) ` ABSORB `750 kJ mol^(-1)` and `Mg rarr Mg^(2+)`absorb `750 + 1450 = 2200 kJ mol^(-1)`
Hence, `a xx 750 + b xx 2200 = 50` ( Given )
or ` 15 a + 44b = 1`....(ii)
or ` 15 a + 44 (0.05 -a) = 1`
or ` 29a = 1.2 ` or `a= 0.04`
`:.b = 0.05 - 0.04 = 0.01`
`%`mole of `Mg^(+). ( 0.04)/( 0.05) xx 100 = 80%, % ` mole of `Mg^(2+) = 20%`
Discussion
51566.

12g of alkaline earth metal gives 14.8g of nitiride. Atomic weight of metal is

Answer»

12
20
40
14.8

Answer :C
Discussion
51567.

1.2g of Mg (At. mass 24) will produce MgO equal to

Answer»

0.05mol
40g
40mg
4g

Answer :B
Discussion
51568.

1.2g of a monoprotic acid HA,is titrated with 0.222 M NaOH solution. The pH of the analyte solution is tiration curve is shown in the diagram. The different indicators used in this titratation and their pK_(a) values are as follows : Methyl Red (pK_(a) = 5.8) , Neutral Red (pK_(a) = 7.2), Cresol Purple (pK_(a) = 8.6) Choose the correct statement from the following

Answer»

In the titration using METHYL red a premature end pointis OBTAINED and the CALCULATED molar mass of HA is found to be greater than the actual value.
In the titration using methyl red a premature end pointis obtained and the calculated molar mass of HA is found to be SMALLER than the actual value.
In the titration using cresol purple end POINT is crossed and the calculated molar mass of HA is found to be smaller than the actual value.
In the titration using cresol purple the correct choice of indicator for this titration is neutral red.

Solution :N//A
Discussion
51569.

1.2g of a monoprotic acid HA,is titrated with 0.222 M NaOH solution. The pH of the analyte solution is tiration curve is shown in the diagram. The different indicators used in this titratation and their pK_(a) values are as follows : Methyl Red (pK_(a) = 5.8) , Neutral Red (pK_(a) = 7.2), Cresol Purple (pK_(a) = 8.6) Which of the following terms most appropriately decribe the beheviour of the analyte solution between points P and Q

Answer»

ACIDIC
Basic
Neutral
Buffer

Solution :`:'` meq of acid `GT` meq of base
`:. [OH^(+)] = (N_(B)V_(b) - N_(a)V_(a))/(V_(a) + V_(b))`
Discussion
51570.

1.2g of a monoprotic acid HA,is titrated with 0.222 M NaOH solution. The pH of the analyte solution is tiration curve is shown in the diagram. The different indicators used in this titratation and their pK_(a) values are as follows : Methyl Red (pK_(a) = 5.8) , Neutral Red (pK_(a) = 7.2), Cresol Purple (pK_(a) = 8.6) What is the pH of analyte at the equivalence point

Answer»

`3.50`
`7.00`
`8.40`
`10`

SOLUTION :N//A
Discussion
51571.

1.2g of a monoprotic acid HA,is titrated with 0.222 M NaOH solution. The pH of the analyte solution is tiration curve is shown in the diagram. The different indicators used in this titratation and their pK_(a) values are as follows : Methyl Red (pK_(a) = 5.8) , Neutral Red (pK_(a) = 7.2), Cresol Purple (pK_(a) = 8.6) What is the molar mass of HA

Answer»

`180`
`222`
`282`
`390`

Solution :`N_(A)V_(A) = N_(B)V_(B)`
Discussion
51572.

1.2g of a monoprotic acid HA,is titrated with 0.222 M NaOH solution. The pH of the analyte solution is tiration curve is shown in the diagram. The different indicators used in this titratation and their pK_(a) values are as follows : Methyl Red (pK_(a) = 5.8) , Neutral Red (pK_(a) = 7.2), Cresol Purple (pK_(a) = 8.6) How many mL of NaOH is required to bring about the titration to its equivalence point

Answer»

`4.00`
`9.00`
`19.00`
`21.00`

Solution :From the GRAPH the equivalence point LIES between PH of `8` to `9`
`:.` vol. NaOH `= 19.0 ML`
Discussion
51573.

12.8 gm of an organilc compound containing C_(1)H_(1)O and undergoes combustion to produce 25.56 gm CO_(2) and 10.46 gm of H_(2)O. What is the empirical formula of compound.

Answer»

`CH_(2)O_(2)`
`CH_(2)O`
`C_(2)H_(4)O`
`CHO`

Solution :
Discussion
51574.

127 mL of a gas at 136°C and 758 mm pressure weigh 0.4524 g. If 1 mL of hydrogen at S.T.P. weighs 0.00009 g, calculate the vapour density and molecular mass of the gas.

Answer»


ANSWER :59.45, 118.90
Discussion
51575.

1.26 g of a dibasic acid were dissovled in water and made up to 200 mL. 20 mL of this solution were completely neutralised by 10 mL of N//5 caustic soda solution. Calculate the equivalent mass and molecular mass of the acid.

Answer»


SOLUTION :N//A
Discussion
51576.

1.26 g of a dibasic acid were dissolved in water and the solution made up to 200 mL. 20 mL of this solution were completely neutralised by 10 mL of N/5 NaOH solution. Calculate the equivalent mass and molecular mass of the acid.

Answer»


ANSWER :63, 126
Discussion
51577.

12.53 cm^(3)of 0.51 M SeO_(2) reacts exactly with 25.5 cm^(3) of 0.1 M CrSO_(4) which is oxidised Cr(SO_(4))_(3) To what oxidation state is the selenium converted during the reaction ?

Answer»

Solution :Let O.N of se in the new compound =X

Now 12.53 `cm^(3)` of 0.051 M `SeO_(2)=12.53xx0.51=0.63` millimoles of `SeO_(2)`
and 25.5 `cm^(3)` of 0.1 M `CrSO_(4)=25.5xx0.1=2.55` millimole of `CrSO_(2)`
But according to balancedredox equation (4-x) moles of `CrSO_(4)` REDUCE 1 mole of `SeO_(2)`
`therefore2.55` millimoles of `CrSO_(4)` it reduce `SeO_(2)=(2.55)/(4-x)` millimoles
But `SeO_(2)` actually REDUCED= 0.64 millimoles
Equating these two values we have`(2.55)/(4-x)=0.64 or x=0`
Discussion
51578.

1.250 g of a sample of octane (C_(8)H_(18)) is burnt in excess of oxygen in a bomb calorimeter. The temperature of the calorimeter rises from 298 K to 304.73 K. If heat capacity of calorimeter is 8.93 kJ/K. Calculate the heat transferred. Also calculate Delta U and Delta H of the reaction at 298 K. The reaction involved is C_(8)H_(18(l)) + (25)/(2)O_(2(g)) rarr 8CO_(2(g)) + 9H_(2)O_((l)).

Answer»

Solution :Let q be the QUANTITY of heat transferred to the calorimeter.
Here, `Delta T = 304.73 - 298.08 = 6.73 K`
Heat absorbed by the calorimeter `= C_(v) xx Delta T`
`= 8.93(KJ K^(-1)) xx 6.73 (K) = 60.1 kJ`
Molar mass of octane `(C_(8)H_(18))`
Heat produced by 1.250 g of `C_(8)H_(18) = 60.1 kJ`
Heat produced by 1 mol `= 114g` of `C_(8)H_(18)`
`= (60.1 xx 114)/(1.250) = 5481.1 kJ`
Thus, `Delta U = -5481.1 kJ mol^(-1)`
`Delta n_((g))` of the reaction `= 8 - (25)/(2) = -4.5`
`Delta H = Delta U + Delta nRT`
`= -5481 + (-4.5) xx 8.314 xx 10^(-3) xx 298`
`= -5492.2 kJ mol^(-1)`
Discussion
51579.

12.5 mL of a solution containing 6 g of a dibasic acid in one litre was found to be neutralized by 10 mL of a decinormal solution of NaOH. The molecular mass of the acid is :

Answer»

110
75
120
150

Solution :`N_(1)V_(1) ("ACID")=N_(2)V_(2)("base")`
`N_(1)xx12.5=(1)/(10)xx10`
`N_(1)=(1)/(12.5)`
STRENGTH `N=xx Ew`
`6 = (1)/(12.5)xx Ew`
Ew=75
MOLAR mass = Equivalent mass `xx` Basicity
`=75xx2=150`
Discussion
51580.

1.249g of a sample fo pure BaCO_(3) and impure CaCO_(3) containingsome Cao was treatedwith dil.HCI and it BaCrO_(4) was precipitaate was dissolvbed in dilute H_(2)SO_(4) and with KI solution libetarediodine which required exactly 20mL of 0.05N Na_(2)S_(2) O_(3). Calculate percentage of CaO in the sample.

Answer»


ANSWER :`14.09%`
Discussion
51581.

1.22g of a gas mesured over water at 15^(@)C and undr a pressure of 1.02 bar of mercury occupied 0.9 dm^(3). Calculate the volume of the volume of the dry gas at N.T.P. Vapour pressure of water at 15^(@)C is 0.018 bar.

Answer»

Solution :Pressure of dry gas = Pressure of moist gas - AQUEOUS tension
`= 1.02 - 0.018 = 1.002` bar
From the available data: `V_(1) = 0.9 dm^(3) , V_(2) = ?`
`P_(1) = 1.002` bar , `P_(2) = 1.013` bar
`T_(1) = 15+273 = 288 K , T_(2) = 273 K`
According the Gas equation , `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` or `V_(2) = (P_(1)V_(1)T_(2))/(T_(1)P_(2))`
SUBSTITUTING the values, `V_(2) = ((1.022 "bar")xx (0.9 dm^(3)) xx (273 K))/((288 K) xx (1.013 "bar" )) = 0.844 dm^(3)`
Discussion
51582.

12.25 g of KClO_3 and 5.85 g of NaCl ar heated together. The residue obtained at the end of burning is dissolved in water to prepare a 500 mL solution. To the solution obtained, excess of AgNO_3 is added. Find the moles of white precipitate formed. Also find the molarity of the solution after filtering out the precipitate with respect to NaNO_3 and KNO_3 (molecular mass of KClO_3 is 122.5 and that of NaCl is 58.50)

Answer»

Solution :`12.25 g KClO_3-=0.1 MOL`
`5.85 g NaCl-=0.1 mol`
`2KClO_3overset(Delta)to2KCl+3O_2`
`NaCloverset(Delta)toNaCl` (does not DECOMPOSE)
`{:(KCloverset(AgNO_3)toAgCl+KNO_3),(NaClunderset(AgNO_3)toAgCl+NaNO_2):}`
`implies` MOLES of `AGCL=(0.1)+(0.1)=0.2`
Moles of `KNO_3=0.1`
`M=0.2` M
SIMILARLY, `M_(NaNO_3)=0.2M`
Discussion
51583.

1.224g of a metallic carbonate on being strongly heated liberates 1309.28 mL of CO_(2) at 27^(@)C and 750 mm pressure of Hg. Find the equivalent mass of the metal carbonate.

Answer»

SOLUTION :`("MASS of the metallic HYDROXIDE")/("mass of the metallic oxide")`
`=("equivalent mass of the metallic hydroxide")/("equivalent mass of the metallic oxide")`
or, `(1.872)/(1.224) =(E+17)/(E+8)""[therefore E=8""E_(OH^(-))=17,E_(O)=8]`
Discussion
51584.

1.216 g of an organic compound was Kjeldahlised and the ammonia evolved was absorbed in 100 mL of 1-N H_(2)SO_(4). The remaining acid solution was made upto 500 mL by dilution with water. 20 mL of this diluted solution required 32 mL of N//10 NaOH solution for complete neutralisation. calculate the percentage of nitrogen in the compound.

Answer»

Solution :Step I. Calculation of the normality of diluted `H_(2)SO_(4)` solution.
Volume of `NaOH` solution = 32 mL
Normality of `NaOH` solution `= N//10`
Volume of the `H_(2)SO_(4)` solution `= 20` mL
Normality of `H_(2)SO_(4)` solution can be calculated by applying normality equation
`ubrace(N_(1)V_(1))_("Acid")= ubrace(N_(2)V_(2))_("Base")`
`Nxx20=1/10xx32, N=32/(10xx20)`
Step II. Calculation of the volume of unused `H_(2)SO_(4)`
`ubrace(N_(1)V_(1))_("Conc. acid")= ubrace(N_(2)V_(2))_("Diluted acid")`
`1xxV=(32xx500)/(10xx20)=80 mL`.
Step III. Calculation of the volume of used `H_(2)SO_(4)`
Total volume of `H_(2)SO_(4)` yaken = 100 mL
Volume of `H_(2)SO_(4)` unused = 80 mL
Volume of `H_(2)SO_(4)` used `100-80=20` mL.
Step IV. Calculation of the percentage of nitrogen.
Percentage of`(1.4xx"Normality of acid used"xx"Volume of acid used")/("Mass of the compound")`
`=(1.4xx1xx20)/(1.216)=23.03 %`
Discussion
51585.

1.216 g of an organic compound was Kjeldahlised and the ammonia evolved was absorbed in 100 mL 1 N H_(2)SO_(4). The remaining acid solution was made upto 500 mL by addition of water. 20 mL of this dilute solution required 32mL of N/10 caustic soda solution for complete neutralisation. Calculate the percentage of nitrogen in the organic compound.

Answer»

Solution :20 mL of diluted unreacted `H_(2)SO_(4) -= 32 mL` of N/10 NAOH solution
`:.` 500 mL of diluted unreacted `H_(2)SO_(4) -= (32)/(20) xx 500 mL` of `(N)/(10) NaOH -= (32)/(20) xx (500)/(10)mL` 1N NaOH
But 80 mL 1 N `NaOH -= 80 mL` of `1 N H_(2)SO_(4) :.`Acid left unused = 80 mL 1 `N H_(2)SO_(4)`
or Acid used for NEUTRALIZATION of `NH_(3) = (100 - 80) = 20 mL 1 N H_(2)SO_(4)`
`:. % N = (14 xx "NORMALITY of the acid" xx "Vol. of acid used")/("Mass of the substance taken") = (1.4 xx 1 xx 20)/(1.216) = 23.026`
Discussion
51586.

1.2046xx10^(24) molecular of HCl dissolve in 1 dm^(3) solution. The concentration of solution is.....

Answer»

6 N
2 N
4 N
8 N

Answer :B
Discussion
51587.

120 mm of pressure is developed at equilibrium when PCl_(5) at 100 mm is subjected to decomposition . Then the percentage of dissociation of PCl_(5) is

Answer»

0.1
0.2
0.4
0.5

Answer :B
Discussion
51588.

120 g Mg was burnt in air to give a mixture of MgO and Mg_(3)N_(2). The mixture is now dissolved in HCl to form MgCl_(2) and NH_(4)Cl, if 107 gram NH_(4)Cl is produced. Then the moles of MgCl_(2) formed is: Mg + (1/2)O_(2) rightarrow MgO...........................(i) 3Mg + N_(2) rightarrow Mg_(3)N_(2) .............................(ii) MgO + 2HCl rightarrow MgCl_(2) + H_(2)O ..........................(iii) Mg_(3)N_(2) + 8HCl rightarrow 2NH_(4)Cl + 3MgCl_(2).......................(iv)

Answer»

3 MOLES
6 moles
5 moles
10 moles

Answer :C
Discussion
51589.

12% (w/v) aqueous caustic soda has a density of 1.2g cc^(-1). Calculate the mole fraction of the components.

Answer»


ANSWER :0.0476 and 0.9524
Discussion
51590.

1.2 moles of SO_(3)are allowed to dissociate in a 2 litre vessel the reaction is 2SO_(3(g)) hArr2SO_(2(g)) +O_(2(g))and the concent ration of oxygen at equilibrium is 0.1 mole per litre. The total number of moles at equilibrium will be

Answer»

2
1.4
0.8
1.6

Solution :
Given `(O_(2))=(x)/(2)=0.1 IMPLIES x=0.2`
TOTAL no. of moles at EQUILIBRIUM` = 1.2-2x+2x+x=1.2+x=1.4`
Discussion
51591.

12 ml of a gaseous mixture of hydrocarbons C_(n)H_(2n) and C_(n)H_(2n+2) (n is same for both ) required exactly 57mL of oxygen for combustion . If the volume of carbondioxide obtained under similar conditions is 36mL, What is the volume ratio of hydrocarbons in the given mixture.

Answer»


ANSWER : 1:1
Discussion
51592.

12 mL of 0.2 N sulphuric acid is neutralised with 15 mL of sodium hydroxide solution on titration. Calculate the normality of sodium hydroxide solution.

Answer»


SOLUTION :N//A
Discussion
51593.

12 ml of saturatedsolution .001gr of CaCO_(3) were required in the titration of 100 ml of water to produce a good lether its degree of hardness is

Answer»

100 ppm
120 ppm
60 ppm
80 ppm

SOLUTION :In 1 ml0.001 g of `CaCO_3` is RESENT
12 ML = ` 12 xx 0.001 g = 0.012g`
hardness = `0.012/100 xx 10^6 = 120` ppm
Discussion
51594.

12 L H_2O_2solution contain 680 gm H_2O_2 then % W/V and volume of the solution is respectively ........ and ……….

Answer»

SOLUTION :5.66,18.7
Discussion
51595.

12 grams of a mixture of sand and calcium carbonate on strong heating produced 7.6 grams of residue. How many grams of sand is present in the mixture?

Answer»


Solution :`CaCO_(3)rarrCaO+CO_(2)`
`100g-56gimplies10g-5.6g`
`THEREFORE` weight of sand = `12-10=2g`
weight of RESIDUE = `5.6+2=7.6g`
Discussion
51596.

1.2 g of organic compound on Kjeldahilization liberates ammonia which consumes 30 cm^(3) of 1 N HCl. The percentange of nitrogen in the organic compound is

Answer»

30
35
`46.67`
`20.8`

Solution :`% N = (1.4 xx 30 xx 1 xx 1)/(1.2) = 35`
Discussion
51597.

12 g of Mg will react completely with an acid to give

Answer»

1 MOLE of O2
1/2 mole of H2
1 mole of H2
2 mole of H2

Solution :`underset(24g)(Mg) + underset(1 "mole")(H_(2)SO_(4)) to MgSO_(4) + H_(2)` (1 mole)
Discussion
51598.

12 g of an element combines with 32 g of oxygen. What is the equivalent weight of the element if the equivalent weight of oxygen is 8?

Answer»

Solution :SUPPOSE the element is M.
No. of EQ. of `M =` no. of eq. of O
`("WT. of M")/(E_M) = ("wt. of oxygen")/( E_O)`
`(12)/( E_M)= ( 32)/( 8) `
`E_M= ( 12 XX 8)/( 32) = 3`.
Discussion
51599.

1.2 g mixture of Na_(2)CO_(3) and K_(2)CO_(3) was dissolved in water to form 100 cm^(3) of the solution. 20 cm^(3) of this solution required 40 cm6(3) of 0.1 N HCl solution for neutralisation. Calculate the percentage composition of the mixture.

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Solution :Step I. Calculation of TOTAL no of gram equivalents in the mixture
Let the mass of `Na_(2)CO_(3)` in the mixture = x g
`:.` Mass of `K_(2)CO_(3)` in the mixture `= (1.2 - x)g`
Equivalent mass of `Na_(2)CO_(3)=("MOLECULAR mass")/(2)=(106)/(2)=53 `
Equivalent mass of `K_(2)CO_(3)=("Molecular mass")/(2)=(138)/(3)=69`
No. of gram equivalents of `Na_(2)CO_(3)=(x)/(53)`
No. of gram equivalent of `K_(2)CO_(3)=((1.2 - x))/(69)`
Total no. of gram equivalent of `Na_(2)CO_(3)` and `K_(2)CO_(3) = (x)/(53) +((1.2 -x))/(69)`
Step II. Calculation of the no. of gram equivalents by titration
No. of gram equivalents in `40 cm^(3)` of 0.1 HCl `= (0.1)/(1000) xx 40 = 4 xx 10^(-3)`
`20 cm^(3)` of the mixture solution NEUTRALISE `HCl = 4 xx 10^(-3) g eq`.
`100 cm^(3)` of the mixture solution neutralise `HCl = (4 xx 10^(-3))/(20) xx 100 = 2 xx 10^(-2) g aq.`
Since acids and bases react in proportions of their gram equivalents
`:.` No. of gram equivalents of `Na_(2)CO_(3) and K_(2)CO_(3)=2 xx 10^(-2)`
Step III. Calculation of percentage composition of the mixture
Equating eqns. (i) and (ii), `(x)/(53)+((1.2-x))/(69)=2xx10^(-2)=0.02`
`69x+63.6-53x=0.02xx53xx69=73.14`
`16x=73.14-63.6=98.54`
`:. x =(9.54)/(16)=0.596g`
% of `Na_(2)CO_(3)` in the mixture `= (0.596)/(1.2) xx 100 = 49.67`
% of `K_(2)CO_(3)` in the mixture `= 100-49.67 = 50.33`.
Discussion
51600.

1.2 g of a sample of CaOCl_(2) were suspended in water made up to 100 mL. 25 mL of this solution was treated with KI and the I_(2) liberated corresponded to 10 mL of N//25 hypo. Calculate thepercentage of Cl_(2) available in CaOCl_(2).

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SOLUTION :N//A
Discussion