NEET Chemistry Questions: Some Basic Concepts in Chemistry

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A gaseous compound is composed of 85.7% by mass carbon and 14.3% by mass hydrogen. It's density is 2.28 g/litre at 300 K and 1.0 atm pressure. Determine the molecular formula of the compound:





d=PMRT  M=dRTP=2.28×0.0821×3001=56.15 g/molE.F.=85.712:14.31=7.14 : 14.3=1 : 2; E.F. is CH2; M.F.=CH2nwhere n=56.1512+24 M.F. is C4H8

Calculate the % of free SO3 in oleum (a solution of SO3 in H2SO4) that is labelled 109% H2SO4.





Percentage above 100 represents the mass of H2O that reacts with dissolved SO3 in oleum to give H2SO4, i.e., 9 g H2O reacts with free SO3 to produce H2SO4

     H2O+SO3  H2SO4

     18 g H2O reacts with 80 g SO3

 9 g H2O will react with 40 g SO3

or  % of free SO3 = 40

Suppose two elements X and Y combine to form two compounds XY2 and X2Y3when 0.05 mole ofXY2 weighs 5 g while 3.011×1023 molecules of X2Y3 weighs 85 g. The atomic masses of x and y are respectively:





Mol. wt. of XY2=50.05=100Mol. wt. of X2Y3=853.011×1023×NA=170Let molar mass of X and Y are a and b respectively        a+2b=100          2a+3b=170;       a=40;       b=30

40 milligram diatomic volatile substance X2 is converted to vapour that displaced 4.92 mL of air at 1 atm and 300 K. Atomic weight of element X is nearly:





4.921000×1 = 40×10-3M×0.0821×300;M200; Atomic mass of X=100

For the reaction; 2FeNO33 + 3Na2CO3  Fe2CO33+6NaNO3

Initially if 2.5 mole of FeNO32 and 3.6 mole of Na2CO3 is taken. If 6.3 mole of NaNO3 is obtained then % yield of given reaction is:





                                                                2FeNO33 + 3Na2CO3  Fe2CO33+6NaNO3mole                                                               2.5                  3.6mole/stoichiometric coefficient                1.25                 1.2Limiting reagent is Na2CO3 so moles of NaNO3 should be formed = 3.6×2=7.2                                                                                                % yield = 6.37.3×100=87.5

0.8 mole of a mixture of CO and CO2 requires exactly 40 gram of NaOH in solution for complete conversion of all the CO2 into Na2CO3. How many moles more of NaOH would it require for conversion into Na2CO3, if mixture (0.8 mole) is completely oxidised to CO2?





CO2+2NaOH  Na2CO3+H2OnNaOH=1;      CO2 present in mixture=0.5 and Co present=0.3 moleWhen more CO2 produced=0.3, more NaOH requried=0.3×2=0.6 mole

The impure 6 g of NaCl is dissolved in water and then treated with excess of silver nitrate solution. The weight of precipitate of silver chloride is found to be 14 g. The % purity of NaCl solution would be:





The reaction that takes place isNaCl+AgNO3  AgCl+NaNO3 143.5 g of AgCl is produced from 58.5 g NaCl 14 g of AgCl will produce from             58.5×14143.5=5.70gThis is the amount of NaCl in common slat;% purity=5.706×100=95%

25.4 of I2 and 14.2 g  of Cl2 are made to react completely to yield a mixture of ICI and ICI3. Calculate mole of ICI and ICI3 formed.





                                                      I2          +           2Cl2        ICI        +     ICI3Mole before reaction               25.4254                       14.271                    0                       0                                                    =0.1                   =0.2                0                      0Mole after reaction                       0                         0                     0.1                  0.1                                                                                                         Note: Mole ratio of reactionMole of ICI formed = 0.1Mole of ICI3 formed = 0.1

The vapour density of a mixture containing NO2 and N2O4  is 38.3. Calculate the mole of NO2 in 100 g mixture.





Given vapour density of mixture = 38.3 Molar mass of mixture = 2×38.3 = 76.6Let a g of NO2 be present in 100 g mixture,    g of N2O4 in mixture = 100-aTherefore,      a46+100-a92=10076.6                      a = 20.10 g Mole of NO2 in mixture = 20.10/46                                              = 0.437

Copper forms two oxides. For the same amount of copper, twice as much oxygen was used to form first oxide than to form second one. What is the ratio of the valencies of copper in first and second oxides?





Let valencies of Cu in two oxides be x and y, then

           I oxides is Cu2Ox

           II oxides is Cu2Oy

In I oxide:

            Equivalent of Cu = Equivalent of oxygen

                             wA/x=a8                     ....i

where w, A, x and a are mass of Cu, at. mass of Cu, valency of Cu and mass of oxygen

In II oxide:

                              wA/y=a8                     ....ii

By Eqs. (i) and (ii)

                             xy=21

 Valency of Cu in I and II oxides are in the ratio 2:1

5 mL of a gaseous hydrocarbon was exposed to 30 mL of O2. The resultant gas, on cooling is found to measure 25 mL of which 10 mL is absorbed by NaOH and the remainder by pyrogallol. Determine molecular formula of hydrocarbon. All measurements are made at constant room temperature.





Suppose formula of hydrocarbon is CaHb

      CaHb+a+b/4O2  aCO2 + b/2H2Ol

Volume taken initially

      5 mL           30 mL                      -                         -

Volume after reaction

      0           30-5a+b/4              5a                         -

 Given that, O2 left is absorbed by pyrogallol

Also volume absorbed by NaOH = 10 mL = Volume of CO2 formed

           5a=10        a=2

Total volume of gases left after reaction = 25 mL

or Volume of O2 left + Volume of CO2 formed = 25 mL

   Volume of O2 left = 25-10 =15

   30-5 [a+(b/4)]=15    b=4

Thus, hydrocarbon is C2H4.

A sample of CaCO3 and MgCO3 weighed 2.21 g is ignited to constant mass of 1.152 g. What is the composition of mixture? Also calculate the volume of CO2 evolved at 0°C and 76 cm of pressure.

 





Let the mixture contains a g CaCO3 and b g MgCO3.                                   a+b=2.21                               .....iOn heating    CaCO3  CaO+CO2                        MgCO3  MgO+CO2 Molar of CaO=mole of CaCO3=a/100 Mass of CaO=a/100×56 gSimilarly, mass of MgO=b/84×40 gTherefore,    56a100+40b84=1.152                                 .....ii                                                                           i.e., mass of residue leftSolving Eqs. i and ii     a=1.19 g;  b=1.02 gAlso, mole of CO2 formed  = mole of CaCO3+mole of MgCO3                                               = 1.19100+1.0284=0.0241 Volume of CO2 at NTP = 0.0241×22400                                              = 539.8 mL

The vapour density of a volume chloride of a metal is 95 and the specific heat of the metal is 0.13 cal/g. The equivalent mass of the metal will be:





Molar mass of metal chloride=95×2=190 g mol-1Atomic mass of metal=6.40.13=49.23 g mol-1Let the metal chloride be MClnThen 49.23+n×35.5=190                         n=3.94;Thus exact atomic massa of metal can be derived by            a+4×35.5 = 190                          a = 48                          E = 484                                  = 12 g eq-1

Which mixture is lighter than humid air?





Humid air N2+O2+H2Ov is lighter than dry air N2+O2 as its average molar mass is less. Presence of He in air N2+O2 gives lower molar mass than humid air. He has lower molar mass than H2Ov.

O3 is absorbed by:





O3 reacts with turpentine oil or oil of cinnamon.