52671 + Interview Questions in Chemistry in Class 11

1.	Calculate the volume of air, contained 21% of oxygen by volume required for the complete combustion of 10 L of ethylene under similar conditions
Answer» Solution :Combustion of ethylene is given by the equation `C_(2)H_(4)+3O_(2) to 2CO_(2)+2H_(2)O` ONE mole of `C_(2)H_(4)="3 moles of "O_2` one volume of `C_(2)H_(4)="3 volumes of "O_(2)` 10mL of `C_(2)H_(4)=?` The volume of `O_(2)` required for the combustion of 10 L of `C_(2)H_(4)=10 XX 3=30L` But AIR contains only 21% `O_(2)` by volume 21 L of `O_(2)` =100 of air 30 L of `O_(2)=?` The volume of air required for the combustion `=(30)/(21) xx 100=142.86L`

Discussion

2.	Calculate the volume of 20 g of hydrogen gas at NTP.
Answer» Solution :Moles of hydrogen gas = `20/2=10` . (Rule 1) VOLUME of the gas at NTP = no. of moles `XX 22.4` `=10xx 22.4 = 224` LITRES

Discussion

3.	Calculate the volume of 10 volume H_(2)O solution that will react with 200 mLof 2N KMnO_(4)in acidic medium.
Answer» Solution :Normality of 10 VOLUME `H_(2)O_(2)=("Volume strength")/(5.6)=(10)/(5.6) N` APPLYING normality equation, `N_(1)V_(1)(H_(2)O_(2))=N_(2)V_(2)(KMnO_(4)` `(10)/(5.6)xxV_(1)=2xx200` or `V_(1)=(2xx200xx5.6)/(10)=224` mL.

Discussion

4.	Calculate the volume of 14.3m NH3, solution needed to prepare 1L of 0.1M solution.
Answer» ANSWER :`-6.77 ML `

Discussion

5.	Calculate the volume of 0.1 M acetic acid solution to be mixed with 50cm^(3) of 0.2M sodium acetate solution, in order to prepare a standard buffer of pH 4.94 (pK_(a) of acetic acid = 4.74).
Answer» SOLUTION :`pH=pK_(a)+"log"(["salt"])/(["acid"])` `4.94=4.74+"log"(["salt"])/(["acid"])` `"log"(["salt"])/(["acid"])=4.94-4.74=0.20` `(["salt"])/(["acid"])="ANTI"log0.20=1.58` That is, the ratio of concentrations of salt : acid is 1.58 : 1 Let V be VOLUME of 0.1 M acetic acid to be mixed `therefore(["salt"])/(["acid"])=((50xx2)/(1000))/((Vxx0.1)/(1000))=(1.58)/(1)` `therefore` Volume of acetic acid, `V=(50xx0.2xx1)/(0.1xx1.58)=90.4cm^(3)`.

Discussion

6.	Calculate the volume of 0.1 M NaOH required to compute neutralization 300 mL HCl having 2.25 pH.
Answer» SOLUTION :16.9 ML

Discussion

7.	Calculate the volume of 0.05 M KMnO_(4) solution required to oxidise completely 2.70 of oxalic acid (H_(2)C_(2)O_(4)) in acidic medium
Answer» Solution :Balanced EQUATION for the REDOX REACTION is `2KMnO_(4)+5(COOH)_(2)+3H_(2)SO_(4)rarrK_(2)SO_(4)+2MnSO_(4)+10CO_(2)+8H_(2)O` No of moles of oxalic acid `=2.70//90 =0.03` MOLE From the balanced equation 5 moles of `(COOH)_(2)=2 "moles of" KmnO_(4)` `therefore 0.03 "mole of" (COOH)_(2)=2//5xx0.03=0.012 "mole of" KMnO_(4)` Now 0.05 mole of `KmnO_(4)` is present in solution =`(1000xx0.012)/(0.05)=240` mL

Discussion

8.	Calculate the volume occupied by 8.8g " of "CO_2 at 31.1^(@)C and 1 bar pressure. R=0.083 L " bar "K^(-1)mol^(-1).
Answer» SOLUTION :`5.05 L`

Discussion

9.	Calculate the volume occupied by 8.8 g of CO_(2) at 31.1^(@)C and 1 bar pressure (R=0.083 "bar" LK^(-1) "mol"^(-1))
Answer» SOLUTION :No. of MOLES `=("mass")/("moleculear mass") =(88)/(44)=02, T=311+273` `PV=nRT or V=(nRT)/(P)=(02xx083xx3041)/(1)=50481L` `:. V=50481 L`

Discussion

10.	Calculate the volume occupied by 8.8 g of CO_(2) at 31.1^(@)C and 1 bar pressure (R=0.083 bar L K^(-1) mol^(-1))
Answer» <P> SOLUTION :PV=NRT `""or"" V=(nRT)/(P)=(w)/(M)=(RT)/(P)""(`:'` n=(w)/(M))` `:. "" V=(8.8 G)/(44 g mol^(-1))xx(0.083 bar" LK"^(-1)mol^(-1)xx(273+31.1K))/(1bar)=5.05" L"`

Discussion

11.	Calculate the volume occupied by 8.8 g of CO_(2) at 31.1^(@)C and 1 bar pressure. R = 0.083 bar L K^(-1)mol^(-1).
Answer» SOLUTION :ACCORDING to ideal gas equations, PV = nRT `therefore V=(nRT)/(p)` but mole `= n=("Weight"_((g)))/("Molar mass (M)")` `therefore V=(gRT)/(Mp)` where, g = 8.8 gram R = 0.083 bar L `K^(-1)mol^(-1)` `T=31.1^(@)C=(31.1+273)K=304.1K` M = molar mass of `CO_(2)=(12+32)=44g mol^(-1)` p = 1 bar `V=((8.8g)(0.083" bar L K"^(-1) mol^(-1))(304.1 K))/((44g mol^(-1))("1 bar"))` `= 5.04806 L ~~ 5.048 L`

Discussion

12.	Calculate the volume occupied by 8.8 g of CO_(2) at 31.1^(@)C and 1 bar pressure. R= 0.083 bar L K^(-1)mol^(-1).
Answer» Solution :According to gas EQUATION, `PV = nRT = m/M xxRxxT` or `1X V = (8.8)/44xx0.083xx (273 + 31.1)` or`V = (8.8xx0.083xx304.1)/44 = 5.05 dm^3`

Discussion

13.	Calculate the volume occupied by 5*09 ethyne gasat 50^(@)C and 740 mm pressure.
Answer» SOLUTION :`PV=NRT, v=(nRT)/(P)` since 760mm Hg=1 atm `n=("MASS")/("mol""mass")=(5)/(26),T =273+50=323K` Volume, `v=(5)/(26)XX(8314xx323)/(98659)=000523 m^(3)` `:. V=523dm^(3)`

Discussion

14.	Calculate the volume occupied by 25g. of carbon dioxide at 303K and 0.974 atm. Pressure
Answer» ANSWER :13.8 LIT

Discussion

15.	Calculate the volume occupied by 16 grm O_(2)at 300 K and 8.31 MPa if (P_(c)V_( c))/(RT_( c))= 3/8 and (P_(t)V_(t))/T_(t)=2.21(Given : R = 8.314 Mpa/K-mol)
Answer» 125.31 ML 124.41 mL 248.62 mL NONE of these Solution :`(PV_m)/(RT) = 3/8 XX 2.21` `V_m o= 3/8 xx 2.21 xx (RT)/P = 248.625 ml`. `V_L = 16/32 xx 248.625 = 124.31 ml`.

Discussion

16.	Calculatethe volume at STP, if certain mass, of gas occupies 0.5 L at 30^(@)C and 700 torr.
Answer» SOLUTION :0.415 L

Discussion

17.	Calculate the volume at N.T.P occupied by (i) 14 g of nitrogen (ii) 1.5 gram moles of carbon dioxide (iii) 10^(21) molecules of oxygen.
Answer» Solution :(i) 28.0 g of NITROGEN at N.T.P occupy = 22.4 L 14.0 g of nitrogen at N.T.P occupy `= (14.0)/(28.0)xx22.4 = 11.2 L` (ii) 1.0 mole of `CO_(2)` at N.T.P occupy = 22.4 L 1.5 moles of `CO_(2)` at N.T.P occupy at N.T.P `= 22.4 xx 1.5 = 33.6 L` (III) `6.022 xx 10^(23)` molecules of oxygen at oxygen at N.T.P occupy `= 22400 CM^(3)` `10^(21)` molecules of oxygen at N.T.P occupy `= (10^(21))/(6.022xx10^(23))xx22400=37.2 cm^(3)`.

Discussion

18.	Calculatethe vlueof (n+l)for3d
Answer» SOLUTION :(n+1) = 3 + 2 =5

Discussion

19.	Calculate the velocity of electron in the first Bohr orbit of hydrogen atom. Given that Bohr radius = 0.529 Å, Planck's constant, h = 6.626 xx 10^(-34)Js, mass of electron = 9.11 xx 10^(-31) kg and 1J = 1kg m^(2) s^(-2). Also calculate the velocity of electron in third orbit of He^(+) ion
Answer» Solution :`mvr = (nh)/(2pi) or v = (nh)/(2pi mr) = ((1) (6.626 xx 10^(-34) Js))/(2 xx 3.14 xx (9.11 xx 10^(-31) KG) xx (0.529 xx 10^(-10)m))` But `1J = 1 kg m^(2) s^(-2)` Hence, `v = 2.189 xx 10^(6) ms^(-1)` For H-like particles, velocity of electron in nth ORBIT is given by `v_(n) = (Z)/(n) xx v_(0)` (`v_(0)`= velocity of electron in 1st orbit of H-atom) For `He^(+), Z = 2` `:. v_(3) = (2)/(3) xx 2.189 xx 10^(6) ms^(-1) = 1.459 xx 10^(6) ms^(-1)`

Discussion

20.	Calculate the velocity of an alpha-particle which begins to reverse its direction at a distance of 2xx10^(-14) m from a scattering gold nucleus (Z=79)
Answer» ANSWER :`2.346xx10^(7)m//sec`

Discussion

21.	Calculate the vapour pressure of the solution. The molefraction of the solute is 0.25. The vapour pressure of the pure solvent is 0.8atm.
Answer» <P> SOLUTION :`(p^(@)-p)/(p^(@))=x_(2)` `(0.8-p)/(0.8)=0.25` `p=0.6atm` `"Vapour PRESSURE of the solution = 0.6 ATM"`

Discussion

22.	Calculate the values of DeltaU and DeltaH for an ideal gas in terms of C_P and C_V.
Answer» SOLUTION :Calculation of `DELTAU` and `DeltaH`: For one mole of an IDEAL gas , we have `C_V=(dU)/(dT)` `dU=C_V. dT` For a finite change , we have `DeltaU=(U_2-U_1)=C_V(T_2-T_1)` and for n moles of an ideal gas we get `DeltaU=n C_V (T_2-T_1)`...(1) We KNOW, `DeltaH=Delta(U+PV)` `DeltaH=DeltaU+Delta(PV)` `DeltaH=DeltaU+DeltaRT`[`because` PV=RT] `DeltaH=DeltaU+RDeltaT` `DeltaH=C_V(T_2-T_1)+R(T_2-T_1)` `DeltaH=(C_V+R)(T_2-T_1)` `DeltaH=C_P(T_2-T_1) "" [because C_P -C_V=R]` For n moles of an ideal gas we get `DeltaH=n C_P(T_2-T_1)`....(2)

Discussion

23.	Calculate the value of z: where z = 3 xx ("Number of" sigma "bonds")/(DBE) xx (3^(@) "Carbon atoms")/("Number of" pi "bonds")
Answer» ANSWER :90

Discussion

24.	Calculate the value of log K_(p) for the reaction N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g) at 25^(@)C. The standardenthalpy of formation of NH_(3)(g)is-46kJ and standard entropiesof N_(2)(g), H_(2)(g) andNH_(3)(g) areare 191,130,193JK^(-1) mol^(-1) respectively . ( R = 8.3 JK^(-1) mol^(-1))
Answer» SOLUTION :`Delta_(f)G^(@)(NH_(3)) = DeltaH^(@) - TDeltaS^(@) = -46000- ( 298 xx192) J MOL^(-1) = - 103.2 KJ mol^(-1)` `Delta_(f)G^(@)(N_(2))= DeltaH^(@) - T DeltaS^(@) = 0 -298 xx191 J mol^(-1) = -56.9 kJ mol^(-1)` `Delta_(f)G^(@)( H_(2)) = DeltaH^(@) - T DeltaS^(@) =0 - 298 xx 130 J mol^(-1) = - 38.74kJ mol^(-1)` `DeltaG^(@)` for reaction `Delta _(r)G^(@) = 2 xx Delta_(f)G^(@) ( NH_(3))- Delta_(f)G^(@) ( N_(2))-3Delta_(f)(H_(2)) =- 206.4 +56.9 +116.2 = - 33.28 kJ mol^(-1)- DeltaG^(@) =2.303 RT log K` `33.28 = 2.303 xx (8.314 xx 10^(-3)) log K`or `log K =1.738 xx 10^(3)`

Discussion

25.	Calculate the value of log K_(p) for the N_(2)(g) + 3H_(2)(g) iff 2NH_(3)(g) at 25^(@)C. The standard enthalpy of formation of nh_(3) is -46 kJ and standard entropies of N_(2)(g), H_(2)(g) and NH_(3)(g) are 191, 130 and 192 J K^(-1) "mol"^(-1) respectively.
Answer» Solution :`N_(2)(g) + 3H_(2)(g) iff 2NH_(3)(g)` `DeltaH^(@) = 2 Delta_(f)H^(@)(NH_(3)) - [Delta_(f)H^(@)(N_(2)) + 3Delta_(f)H^(@)(H_(2))]` `= 2 xx -46 = -92 KJ` Now , `DeltaS^(@) = 2S^(@)(NH_(3))-[S^(@)(N_(2))+ 3S^(@)(H_(2))]` ` = 2 xx 192 - [191 + 3 xx 130]` ` = -197 JK^(-1) or = -0.197 kJ K^(-1)` `DeltaG^(@) = DeltaH^(@) -TDeltaS^(@)` ` = -92 -298 xx (-0.197)` ` =-33.294 kJ = - 33.294 xx 10^(3)J` Now , `DeltaG^(@)= -2.3030 RT "log K_(p)` or `log K_(p) = -(DeltaG)/(2.3030RT)` `= - (-33.294 xx 10^(3))/(2.303 xx 8.314 xx 298) = 5.835`.

Discussion

26.	Calculate the value of DeltaU and DeltaH on heating 128.0 g of oxygen from 0^@C to 100^@C. CV and C_p on an average are 21 and 29 J "mol"^(-1) K^(-1)(The difference is 8J "mol"^(-1) K^(-1)which is approximately equal to R )
Answer» Solution :We KNOW `DeltaU=n C_V(T_2-T_1)` `DELTAH=n C_P (T_2-T_1)` Here `n=128/32` 4 moles , `T_2=100^@C`=373 K , `T_1=0^@`C =273 K `DeltaU=nC_V(T_2-T_1)` `DeltaU` = 4 x 21 x (373-273 ) `DeltaU` =8400 J `DeltaU` =8.4 kJ `DeltaH= nC_P (T_2-T_1)` `DeltaH`=4 x 29 x (373-273) `DeltaH` =11600 J `DeltaH`=11.6 kJ

Discussion

27.	Calculate the value of Delta G for the system of water in atmosphere having 100 K temperature.
Answer» Solution :`X^(@) C + 273 = 100K` `therefore x= -173^(@) C` `therefore DELTA G = -ve`

Discussion

28.	Calculate the value of (DeltaH - DeltaE)in calories for the combustion of CH_4 at 27^@C.
Answer» Solution :`CH _(4 (g)) + 2O _(2 (g)) to CO _(2 (g)) + 2H _(2) O _((L))` CHANGE in the number of moles, `DELTAN n = n _(P) - n _(R) =1 -3 =-2, Delta H = Delta E + Delta n RT` `Delta H - Delta E =- 2 XX 2 xx 300 =-1200 ` calories.

Discussion

29.	Calculate the value of Avogadro's number from the following data: Density of NaCl=2.165 g cm^(-3). Distance between Na^+ and Cl^- in NaCl=281 pm
Answer» Solution :A unit cell of NaCl CONTAINS 4 NaCl units, THEREFORE , Z=4 , M=23+35.5=`58.5 g MOL^(-1), RHO =2.165 g cm^(-3)` As distance between `Na^+` and `Cl^-` =281 pm `therefore` Edge of the unit cell =`2xx281`=562 pm (Edge is the distance from `Na^+` to `Na^+` ion or `Cl^-` to `Cl^-` ion. Substituting these values in the expression , `rho=(ZxxM)/(a^3xxN_0)` `2.165 g cm^(-3)=(4xx58.5 g mol^(-1))/((562xx10^(-10) cm)^3 xxN_0)` or `N_0=6.09xx10^23 mol^(-1)`

Discussion

30.	Calculate the value of Avogadro's number from the following data : Density of NaCl = 2.165 g cm^(-3), Distance betweenNa^(+) and Cl^(-)in NaCl = 281 pm.
Answer» Solution :A unit cell of NaCl contains 4 NaCl units, therefore, Z = 4 M = 23 + 35.5 = 58.5g ` "MOL"^(-1)` , p = 2.165 g` cm^(-3)` As distance between ` Na^(+) and Cl= 281` pm` therefore ` Edge of the unit cell ` = 2 xx 281 = 562 ` pm ( Edge is the distance from ` Na^(+) "to" Na^(+) "ION or " Cl^(-) "to" Cl^(-)` .see Fig .1.33 g , page 1/26) substituting these values in the EXPRESSION , ` p = ( Z xx M)/ (a^(3) xx N_(0)) ` `2.165 "gcm" ^(-3)= ( 4 xx 58.5"g mol"-1)/( 562 xx 10^(-10)"cm") ^(3) xx N_(0) or N_(0) = 6.09 xx 10^(23)mol ^(-1)`

Discussion

31.	Calculate the value of Avogadro's number from the following data : Density of KF =2.48 g cm^(-3). Distance between K^+ and F^- in KF =269 pm. (Atomic masses : K=39 and F=19 amu)
Answer» ANSWER :`6.0xx10^23`

Discussion

32.	Calculate the value of A. A= (E_1,2)/(2E_(2,1))Where E_(n,z) = Energy of electron in n^(th) orbit , Z = atomic number of hydrogen like specie.
Answer» SOLUTION :`A= (E_(1,2))/(2E_(2,1)) = (-13.6 xx 2^2 xx 2^2)/( 2XX 1^2 xx (-13.6)xx 1^2)= 16/(2) = 8`

Discussion

33.	Calculate the uncertainty in the velocity of an electron, if the uncertainity in position is 100 pm.
Answer» Solution :Given `Deltaxx="100 PM "=100xx10^(-12)m=10^(-10)m` `"mass of electron, m"=9.1xx10^(-31)kg` `Deltav=?""pi=3.1416` `Deltax xx Deltav=(h)/(4pim)` `Deltav=(h)/(4pim. Deltax)=(6.625xx10^(-34)"kgm"^(2)s^(-1))/(4xx3.1416xx9.1xx10^(-31)kgxx10^(-10))` `=5.8xx10^(5)MS^(-1)` `"The uncertainty in VELOCITY"=5.8xx10^(5)ms^(-1)`

Discussion

34.	Calculate the uncertainty in velocity (Delta v) of a cricket ball (mass = 0.15 kg) if the uncertainty in its position (Delta x) is of the order of 1Å (i.e., 10^(-10)m) or 100 pm
Answer» Solution :`3.5 XX 10^(-24) ms^(-1)`

Discussion

35.	Calculate the uncertainty in the velocity of a wagon of mass 3000kg whose position is known to an accuracy of +- 10 pm (Planck's constant = 6.63 xx 10^(-34) Js)
Answer» Solution :Here, we are GIVEN: `m = 3000 kg, DELTA x = 10` pm `= 10 xx 10^(-12)m = 10^(-11)m` `:.` By uncertainty principle, `Delta v = (H)/(4pi xx m xx Delta x) = (6.63 xx 10^(-34) kg m^(2) s^(-1))/(4 xx (22)/(7) xx 3000 kg xx 10^(-11)m)` `= 1.76 xx 10^(-27) ms^(-1)`

Discussion

36.	Calculate the uncertainty in the position of an electron,if the uncertainty in its velocity is 5.7xx10^(5)ms^(-1)
Answer» SOLUTION :Uncertainty in velocity`/_\v=5.7xx10^(5)MS^(-1)` MASS of electron=m`=9.110^(-31)kg` uncertainty in POSITION=`/_\X=?` `/_\x.m./_\v=h/(4pi)``/_\x=(h)/(m./_\x4pi)=(6.626xx10^(-34))/(9.1xx10^(-34)xx5.7xx10^(5)xx4pi)` `/_\x=(6.626xx10^(-34)/(9.1xx10^(-31)xx5.7xx10^(5)xx4xx3.14))` `=0.010xx106^(-8)=1xx10^(-10)m.` Uncertainty in position-`1xx10^(-10)m`.

Discussion

37.	Calculate the uncertainty in the position of an electron if the uncertainty in its velocity is 5.7 xx 10^(5) m//sec (h = 6.6 xx 10^(-34) kg m^(2) s^(-1), mass of the electron = 9.1 xx 10^(-31) kg)
Answer» Solution :Here, we are given: `Delta v = 5.7 xx 10^(5) m s^(-1), m = 9.1 xx 10^(-31) kg, h = 6.6 xx 10^(-34) kg m^(2) s^(-1)` Substiuting these values in the euqation for uncertainty PRINCIPLE, i.e., `Delta x xx (m xx Delta v) = (h)/(4PI)`, we have, `Delta x = (h)/(4pi xx m xx Delta v) = (6.6 xx 10^(-34) kg m^(2) s^(-1))/(4 xx (22)/(7) xx 9.1 xx 10^(-31) kg xx 5.7 xx 10^(5) ms^(-1)) = 1.0 xx 10^(-10) m` i.e., Uncertainty in position `= +- 10^(-10)m`

Discussion

38.	Calculate the uncertainty in the velocity in the velocity of an electron if the uncertainty in its position is 1 Å or 100 pm (10^(-10) m)
Answer» SOLUTION :`5.77 xx10^(5) m s^(-1)`

Discussion

39.	Calculate the uncertainty in the position of a dust particle with mass equal to 1 mg if the uncertainty in its velocity is 5.5 xx 10^(-20) m s^(-1)
Answer» SOLUTION :`Delta x = (h)/(4PI m Delta V) = (6.6 xx 10^(-34) kg m^(2) s^(-1))/(4 xx 3.14 xx (1 xx 10^(-6) kh) xx (4.5 xx 10^(-20) ms^(-1))) = 9.55 xx 10^(-10) m`

Discussion

40.	Calculate the uncertainty in position of an electron, if Deltav = 0.1% and v = 2.2 xx 10^(6) ms^(-1)
Answer» Solution :MASS of an electron = `m= 9.1xx10^(-31) kg` ` Deltav = "Uncertainty in velocity" = (0.1)/(100)xx2.2xx10^(6) ms^(-1)` `Deltav = 0.22x10^(4) = 2.2 x 10^(3) ms^(-1)` `DeltaX.DeltaV.m=(H)/(4pi)` `DeltaX=(h)/(Deltav.mxx4pi)=(6.626xx10^(-34))/(2.2xx10^(3)xx9.1xx106(-31)xx4xx3.14)``Deltax=2.635xx10^(-8)` `"Uncertainty in position" = 2.635xx10^(-8)m`

Discussion

41.	Calculate the uncertain itin the velocity of a cricket ball of mass150 g, if uncertainity in its position is of the order of 1 Å.
Answer» ANSWER :`3.5 XX 10^(-24) ms^(1)`

Discussion

42.	Calculate the uncertainty in the momentum of an electron if it is confined to a linear region of length 1xx10^(-10)metre
Answer» `5.37xx10^(-27)" kg "ms^(-1)` `5.27xx10^(-27)" g "ms^(-1)` `5.37xx10^(-25)" g "ms^(-1)` `5.27xx10^(-25)" kg "ms^(-1)` ANSWER :D

Discussion

43.	Calculate the uncertaimty in position of an electron,if /_\v=0.1% and v=2.2 xx 10^(6) ms^(-1)
Answer» Solution :MASS of an electron`=m=9.1xx10^(-31)KG`. `/_\V`=UNCERTAINTY in velocity `=(0)/(100)xx2.2xx10^(6)ms-1` `/_\x./_\v.m=h/(4pi)` `/_\x=h/(/_\v.mxx4pi)` `=(6.626xx10^(-34))/(2.2xx10^(3)xx9.1xx10^(-31)xx4xx3.14)` `=(6.626xx10^(-6))/(251.45)` `=0.02635xx10^(-6)` `/_\x=2.65xx10^(-6)` Uncertainty in position`=2.635xx10^(-10)m`.

Discussion

44.	Calculate the total pressure in a mixture of 8g of dioxygen and 4g of dihydrogen confined in a vessel of 1 dm^(3)" at "27^(@)C.
Answer» SOLUTION :`56.025` BAR

Discussion

45.	Calculate the total pressure in bar in a mixture of 8g of O_2 and 4g of H_2 confined in a vessel of 1dm^3 at 27^@C.
Answer» SOLUTION :NUMBER of MOLES of `O_2` = 8/32 = 0.25 Number of moles of `H_2` = 4/2 = 2 Total number of moles = 2.25 P = nRT/V = `(2.25xx0.083xx300)/1` = 56.025 bar

Discussion

46.	Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1dm^(3) at 27^(@)C. R =0.083 bar dm^(3)K^(-1)mol^(-1).
Answer» SOLUTION :Total NUMBER of moles present in the MIXTURE `=8/32 + 4/2 = 2.25` (Mol. mass of `O_2 = 32, " Mol. mass of " H_2 = 2`) According to the gas equation, `PV = NRT` or `P xx 1= 2.25 xx 0.083 xx 300` or `P= 2.25 xx 0.083 xx 300 = 56.025` bar

Discussion

47.	Calculate the total pressure in a mixture of 8 g of oxygen and 4 g of hydrogen confined in a vessel of 1 dm^(3) at 27^(@)C. R=0.083 bar dm^(3)K^(-1)mol^(-1).
Answer» Solution :MOLAR mass of `O_(2)`=32 g `mol^(-1)"":.8 g O_(2)=(8)/(32)mol=0.25 mol` Molar mass of `H_(2)=2 g mol^(-1)"" :.4 g O_(2)=(4)/(2)=2 mol` `:.` TOTAL number of moles (n)=2+0.25=2.25 V=1 `dm^(3)`, T=`27^(@)C=300 K,R=0.083 BAR dm^(3)K^(-1)mol^(-1)` PV=nRTorP`=(nRT)/(V)=((2.25 mol)(0.083 bar dm^(3)K^(-1)mol^(-1))(300 K))/(1 dm^(3))=56.025 bar`

Discussion

48.	Calculate the total pressure in a mixture of 8 g os dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm^(3) at 27^(@)C.(R = 0.083 bar dm^(3)K^(-1)mol^(-1))
Answer» Solution :Calculation of dioxygen mole `(n_(1))` :Molecular mass of dioxygen `= 2(16)=32 g mol^(-1)` Weight of mass = 8 g `therefore` Mole of dioxygen `= ("Mass")/("Molecular mass")` `=(8g)/(32 g mol^(-1))=0.25 mol = n_(1)` Calculation of dihydrogen mole `(n_(2))` :Molecular mass of dihydrogen `= 2.0 g mol^(-1)` Weight of mass = 4.0 g `therefore` Mole of dihydrogen `=("Mass")/("Molecular mass")` `= (4.0 g)/(2.0 g mol^(-1))=2.0 mol = n_(2)` Total mole `(n)=(n_(1)+n_(2))` `= 0.25 mol O_(2)+2.0 mol H_(2)` = 2.25 mol where, Volume of vessel `(V) = 1dm^(3)` Temperature `(T)=(27+273)K = 300K` R = 0.083 bar `dm^(3)K^(-1)mol^(-1)` pV = NRT and `p=(nRT)/(V)` `therefore p =((2.25 mol)(0.083" bar dm"^(3)K^(-1)mol^(-1))(300 K))/(1 dm^(3))` = 56.025 bar pressure.

Discussion

49.	Calculate the total pressure in a 10 L cylinder which contains 0.4 g of helium, cylinder. Assume idealbehaviour of gases.(R=0.082 L atm K6(-1)mol^(-1))
Answer» SOLUTION :Total no. of MOLES of all gases(N)`=(0.4)/(4)+(1.6)/(32)+(1.4)/(28)=0.2` VOLUME of the gasesour mixture (V)=10 L Temperature (T)=27+273=300 K, R=0.082 L atm `K^(-1)mol^(-1)` As gases have behaviour, PV=nRT or `P=(nRT)/(V)=(0.2xx0.082xx300)/(10)=0.492 atm` `"Partial pressure of helium "=(0.4//4)/(0.2)xx0.492=0.246 atm`

Discussion

50.	Calculate the total pressure in a 10 L cylinder which contains 0.4 g of helium, 1.6 g of oxygen and 1.4 g of nitrogen at 27^(@)C. Also calculate thepartial pressures of He gas in the cylinder. Assume Ideal behaviour for gases. R = 0.082 L atm k^(-1) mol^(-1)
Answer» <P> Solution :`p_("total") = 0.4926` ATM `p_(He) = 0.2463` atm `p_(O_(2)) = 0.1231` atm `p_(N_(2)) =0.123` atm

Discussion

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