Saved Bookmarks
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 901. |
Iron (II) oxide has a cubic structure and each unit cell has side 5 Å. If the density of the oxide is `4 g cm^(-3)`, calculate the number of `Fe^(2+)` and `O^(2-)` ions present in each unit cell (Molar mass of FeO=`72 g mol^(-1)`, `N_A=6.02xx10^23 "mol"^(-1)`) |
| Answer» Correct Answer - Z=4, i.e., `4 Fe^(2+)` and `4 O^(2-)` ion | |
| 902. |
Experimentally it was found that a metal oxide has formula `M_0.98 O`. Metal M is present as `M^(2+)` and `M^(3+)` in its oxide. Fraction of the metal which exists as `M^(3+)` would beA. `5.08%`B. `7.01%`C. `4.08%`D. `6.05%` |
|
Answer» Correct Answer - C The formula `M_0.98 O` shows that if there were 100 O-atoms present as `O^(2-)` ions, then there 98 M atoms present as `M^(2+)` and `M^(3+)` . Suppose `M^(3+) =x` , then `M^(2+)` =98-x . As the compound as a whole is neutral, Total charge on `M^(3+)` and `M^(2+)` =Total charge on 100 `O^(2-)` ions x (+3) + (98-x) (+2)=100 x 2 or 3x + 196 - 2x =200 or x=4 `therefore` % of `M^(3+)=4/98xx100`=4.08 % |
|
| 903. |
A metal (atomic mass 50) has a body centred cubic crystal structure. The density of the metal is `5.96 "g cm"^(-3)`. Find the volume of the unit cell (`N_O=6023xx10^23 "atoms mol"^(-1)`) |
|
Answer» Correct Answer - `2.786xx10^(-23) cm^3` Mass of the unit cell = Mass of one atom x No. of atoms present in the unit cell `=(50/(6.023xx10^23)g)xx2` [For BCC, Z=2] Volume of the unit cell=`"Mass"/"Density"=(50xx2)/(6.023xx10^23)gxx1/(5.96 g cm^(-3))=2.786xx10^(-23) cm^(3)` |
|
| 904. |
Analysis shows that a metal oxide has the empirical formula of `M_0.96 O_1.00` Calculate the percentage of `M^(2+)` and `M^(3+)` ions in this crystal. |
| Answer» Correct Answer - `M^(2+)=91.7%, M^(3+)=8.3%` | |
| 905. |
`Al^(3+)` ions replace `Na^+` ions at the edge centres of NaCl lattice. The number of vacancies in one mole NaCl is found to be ` x xx 10^23` . The value of x approximately is |
| Answer» Correct Answer - 3 | |
| 906. |
Classify the following into ionic, molecular , covalent and metallic crystals. Bronze, Dry ice, Nitre and Diamond |
| Answer» Bronze=Metallic, Dry ice=Covalent, Nitre=Ionic, Diamond=covalent | |
| 907. |
The volume of `2.8 g` of carbon monoxide at `27^(@) C` and `0.821 atm` pressure `(R = 0.821 "atm" K^(-1) "mol"^(-1))`A. `0.3 L`B. `1.5 L`C. `3 L`D. `30 L` |
|
Answer» Correct Answer - C According to the ideal gas equation , we have `pV = nRT` or `V = (nRT)/(p)` `n_(CO_(2)) = (m_(CO_(2)))/("Molar mass"_(CO_(2))) = (2.8 g)/(28 g "mol"^(-1)) = 0.1 "mol"` `:. V = ((0.1 "mol")(0.0821 (L "atm")/(K "mol")) (300 K))/((0.821 "atm"))` `= 32` |
|
| 908. |
The volume of 2.8 g of carbon monoxide at `27^(@)`C and 0.821 atm pressure isA. 30 LB. 3 LC. 0.3 LD. 1.5 L |
| Answer» (b) pV=nRT or V=`(nRT)/(p)`=`(2.8xx0.0821xx300)/(28xx0.821)`=3 L | |
| 909. |
The volume of 2.89 g of carbon monoxide at `27^(@)C` and 0.821 atm pressure isA. 2.5 LB. 4 LC. 3.5 LD. 3 L |
|
Answer» Correct Answer - D `n=(m)/(M)=(2.8)/(28)=0.1` `V=(nRT)/(P)=(0.1xx0.0821xx300)/(0.821)=3L` |
|
| 910. |
The stop cock connecting two bulbs of volume 5 litre and 10 litre containing an ideal gas at 9 atm and 6 atm respectively, is opened. What is the final pressure in the two bulbs |
|
Answer» `P_(1) V_(1) + P_(2) V_(2) = P_(R) (V_(1) + V_(2))` `9 xx 5 + 6 xx 10 = P_(R) (15)` `P_(R) = 7 atm` where, `P_(R) =` resultant pressure after mixing. |
|
| 911. |
A 1000 mL sample of a gas at `-73^(@)C` and 2 atmosphere is heated to `123^(@)C` and the pressure is reduced to 0.5 atmosphere. What will be the find volume ? |
|
Answer» `{:("Inital conditions","Final conditions",),(P_(1) = 2 atm,P_(2) = 0.5 atm,),(V_(1) = 1000 mL,V_(2) = ?,),(T_(1) = - 73^(@)C = (-73 + 273),T_(2) = 123^(@)C = (123 + 273),),(= 200 K,= 400 K,):}` We know that, `(P_(1) V_(1))/(T_(1)) = (P_(2) V_(2))/(T_(2))` So, `(2 xx 1000)/(200) = (0.5 xx V_(2))/(400)` or `V_(2) = (2 xx 1000 xx 400)/(200 xx 0.5) = 8000 mL` |
|
| 912. |
When the temperature is increased surface tension of water .A. increasesB. decreasesC. remain constantD. shows irregular behaviour. |
|
Answer» Correct Answer - B The increase in temperature decreases the intermolecular forces and also the surface tension. |
|
| 913. |
Which one of the following gases has the highest critical temperature ?A. NitrogenB. AmmoniaC. Water vapourD. Carbon dioxide |
|
Answer» Correct Answer - C The most easily liquefiable gas/vapour has the highest critical temperature. |
|
| 914. |
Surface tension of water is 73 dyne `cm^(-1)` at `20^(@)C`. If surface area is increased by `0.10" m"^(2)`, work done will beA. 73 ergsB. 730 ergsC. 7300 ergsD. 73000 ergs |
|
Answer» Correct Answer - D Work done=Surface tension xxIncrease in surface area `=73" dyne "cm^(-1)xx0.1" m"^(2)` `=73" dynwe "cm^(-1)xx0.1xx10^(4)"ergs"` `=7.3xx10^(4)" dyne " cm=7.3xx10^(4)" ergs"` |
|
| 915. |
Assuming that dry air contains 79% `N_(2)` and 21% `O_(2)` by volume, calculate the density of dry air at if it has a relative humidity of 40%. The vapour pressure of water at `25^(@)C` is 23.76 mm. |
|
Answer» Calculation of density of dry air. Density of air means the mass of air per litre. first let us convert 1 L `(1000 cm^(3))` of air at `25^(@)C` 1 atm pressure to the volume at S.T.P. `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` `:. " " (1 atmxx1000" cm"^(3))/(298" K")=(1 atm xxV_(2))/(273 K) " or " V_(2)=916.1" cm"^(3)` As air contains 79%`N_(2)` and 21% `O_(2)`, `:. " "` Volume of `N_(2)` at S.T.P. in the air `=(79)/(100)xx916.1cm^(3)=723.72" cm"^(3)` Volume of `O_(2)` at S.T.P. in the air `=(21)/(100)xx916.1" cm"^(3)=193.38" cm"^(3)` Mass of `723.72" cm"^(3)` of `N_(2)` at S.T.P. `=(28)/(22400xx723.72=0.905" g"` Mass of `193.38" cm"^(3)` of `O_(2)` at S.T.P. `=(32)/(22400)xx923.38=0.276" g"` `:. ` Total mass of 1 L of air at `25^(@)C` and 1 atm `=0.905+0.276" g"=1.181" g"` Hence, density of dry air `=1.181" g "L^(-1)` Calculation of density of moist air % Relative humidity`=("Partial pressure of "H_(2)O "vapours")/("Vapour pressure of "H_(2)O "at the same temp")xx100` `:. " "` Partial pressure of `H_(2)O` vapours`=(40xx23.76)/(100) =9.5" mm"` As total pressure =1 atm=760 mm, `:. " "` Partial pressure of `N_(2)=(79)/(100)xx750.5" mm "=592.9" mm "` and Partial pressure of `O_(2)=750.5-529.9=157.6" mm"` Applying gas equation PV=nRT, i.e., `n=(PV)/(RT)` No. of moles of `H_(2)O` in the air `(n_(H_(2)O))=((9.5//760)"atm "xx1L))/(0.0821" L atm "K^(-1)mol^(-1)xx298" K ")=5.1xx10^(-4)` No. of moles of `N_(2)` in the air `(n_(N_(2)))=((592.9//760)xx1)/(0.0821xx298)=3.19xx10^(-2)` No. of moles of `O_(2)` in the air `(n_(O_(2))=((157.6//760)xx1)/(0.0821xx298)=8.48xx10^(-2)` Total mass of 1 L of air `=5.1xx10^(-4)xx18+3.19xx10^(-2)xx28+8.48xx10^(-3)xx32` `=0.00918+0.8932+0.02714=1.174" g"` Hence, density of moist air `=1.174" g "L^(-1)` Note. Density of moist dry air can also be calculated by this method. |
|
| 916. |
Air contains `79% N_(2)` and `21% O_(2)` by volume. If the barometric pressure is `750 mm Hg`. The partial pressure of oxygen isA. `157.7mmHg`B. `175.5mmHg`C. `315.0mmHg`D. none |
|
Answer» Correct Answer - A Calculate `M_(eff)` |
|
| 917. |
A cylinder of 20.0 L capacity contains 160 g oxygen gas at `25^(@)C`. What mass of oxygen must be released to reduce the pressure of the cylinder to 1.2 atm ? |
|
Answer» Number of moles of oxygen gas present initially in the cylinder`=(160g)/(32"g mol"^(-1))=5" moles"` To calculate the number of moles now present, we have P=1.2 atm, T=298 K, V=20.0L Applying the relation, PV=nRT, we have `n=(PV)/(RT)=(1.2" atm "xx20.0" L")/(0.0821" L atm " K^(-1)mol^(-1)xx298" K")=0.98" mol"` `:.` Number of moles of `O_(2)` required to be released =5-0.98=4.02 mol or Mass of `O_(2)` required to be released `=4.02xx32" g"=128.64" g"` |
|
| 918. |
A `34.0 dm^(3)` cylinder contains `212 g` of oxygen gas at `21^(@)C`. What mass of oxygen must be released to reduce the pressure in the cylinder to `1.24` bar ? |
|
Answer» Step I. Calculation of no. of moles of `O_(2)` left in cylinder. `P = 1.24` bar , `V = 34 dm^(3)` `T = (21 + 273) = 294 K` , `R = 0.083 dm^(3) "bar" K^(-1) mol^(-1)` According to ideal gas equation, `PV = nRT` `n = (PV)/(RT) = ((1.24 "bar") xx (34 dm^(3)))/((0.083 dm^(3) "bar" K^(-1) mol^(-1)) xx (294 K)) = 1.727` mol Step II. Calculation of mass of hydrogen released Mass of `O_(2)` left in the cylinder `= n xx M = (1.727 "mol") xx (32 g "mol"^(-1)) = 55.26 g` Mass of `O_(2)` initially present `= 212 g` `:.` Mass of `O_(2)` released `= (212 - 55.26) = 156.74 g` |
|
| 919. |
`1 L` of a gaseous mixture is effused in `5 min 11s`, while `1 L` of oxygen takes `10 min`. The gaseous mixture contains methane and hydrogen. Calculate (`a`) The density of gaseous mixture. (`b`) The percentage by volume of each gas in mixture. |
|
Answer» `{:(O_(2),"Mixture"),(d_(1)=16, d_(2)=?),(t_(1)=10xx60=600 s, t_(2)=5xx60+11=311 s):}` Volume is same, so `(t_(2))/(t_(1))=sqrt((d_(2))/(d_(1)))` `(311)/(600)=sqrt((d_(2))/(16))` `implies ((311)^(2))/((600)^(2))=(d_(2))/(16)` `implies d_(2)=4.3` `d_(H_(2))=1` and `d_(CH_(4))=8`. So `d_(mix)=(d_(CH_(4))xxx+(100-x)d_(H_(2)))/(100)` `4.3xx100=8x+(100-x)xx1` `430=8x+100-x` `330=7x` `x=47.14%` `%CH_(4)=47.14`, `%H_(2)=52.86` |
|
| 920. |
How many `Cl^(-)` ions are there around `Na^(+)` ion in a NaCl crystal?A. 3B. 4C. 6D. 8 |
| Answer» (c ) In NaCl, both `Na^(+)` and `Cl^(-)` ions, have coordination number 6 | |
| 921. |
Assertion. In a unit cell of NaCl , all `Cl^-` ions touch `Na^+` ion as all well they touch each other . Reason. Radius ratio `r_+//r_-` in NaCl is 0.414.A. If both assertion and reason are true, and reason is the true explanation of the assertion.B. If both assertion and reason are true, but reason is not the true explanation of the assertion.C. If assertion is true , but reason is falseD. If both assertion and reason are false. |
|
Answer» Correct Answer - D Correct A . In a unit cell of NaCl, all `Cl^-` ions touch `Na^+` ion but do not touch each other. Correct R . Actual radius ratio `r_(Na^+)//r_(Cl^-)` is 0.525 and not 0.414 |
|
| 922. |
Assertion. Size of the cation is larger in tetrahedral hole than in an octahedral hole. Reason. The cations occupy more space than anions in crystal packing.A. If both assertion and reason are true, and reason is the true explanation of the assertion.B. If both assertion and reason are true, but reason is not the true explanation of the assertion.C. If assertion is true , but reason is falseD. If both assertion and reason are false. |
|
Answer» Correct Answer - D Correct A.Octahedral hole is larger in size than tetrahedral hole. Correct R. Cations are generally smaller. They occupy the voids and hence occupy less space. |
|
| 923. |
Which of the following are not the units of gas constant (R) ?A. dynes `K^(-1) mol^(-1)`B. ergs `"degree"^(-1) mol^(-1)`C. `"cm"^(3) K^(-1) mol^(-1)`D. `Pa dm^(3) K^(-1) mol^(-1)` |
|
Answer» Correct Answer - A::C are not the units of gas constant R. |
|
| 924. |
Units of pressure are :A. NewtonB. PascalC. BarD. Torr. |
|
Answer» Correct Answer - B::C::D are all units of pressure. |
|
| 925. |
Which of the following are incorrect?A. At altitude, the pressure of air is lowB. At altitude, the density of air is lowC. At altitude, cooking takes place slowerD. At altitude, water will boil at temperature greater than `100^(@)C`. |
|
Answer» Correct Answer - C::D Cooking will be fast, b.p. of water will be less than `100^(@)C`. |
|
| 926. |
A gas at a pressure of `5.0 atm` is heated from `0^(@)C` to `546^(@)C` and simultaneously compressed to one-third of its original volume. Hence final presseure isA. `10.0 atm`B. `30.0 atm`C. `45.0 atm`D. `5.0 atm` |
|
Answer» Correct Answer - C `P_(1)= 5.0 atm, T_(1)= 0^(@)C= 273, V_(1)=V` `P_(2)=? T_(2)= 546^(@)C= 819 K, V_(2)= (1)/(3)V` `:. From (P_(1)V_(1))/(P_(2)V_(2))=(T_(1))/(T_(2))` `implies (5xxV)/(P_(2)xx(1)/(3)V)= (273)/(819)` `implies (5xx3)/(P_(2))=(1)/(3)` `P_(2)= 45 atm` |
|
| 927. |
A gas at a pressure of 5.0 atm is heated from `0^(@)C` to `546^(@)C` and simultaneously compressed to one-third of its original volume. Hence, final pressure is:A. 15.0 atmB. 30.0 atmC. 45.0 atmD. `(5)/(9) atm` |
|
Answer» Correct Answer - C `(p_(1)V_(1))/T_(1)=(p_(2)V_(2))/(T_(2))rArr(5xxV_(1))/(273)=(p_(2)xxV_(1)/(3))/(819)` `p_(2)=45 " atm"` |
|
| 928. |
The fraction of the total volume occupied by the atoms present in a simple cube isA. `pi/4`B. `pi/6`C. `pi/sqrt(3sqrt2)`D. `pi/sqrt(4sqrt2)` |
|
Answer» Correct Answer - B In a simple cube, no. of atoms/unit cell =`8xx1/8=1` Volume of the atom =`4/3pir^3` Volume of the cube=`a^3=(2r)^3=8r^3 (because a=2r)` `therefore` Fraction occupied=`(4/3 pir^3)/(8r^3)=pi/6` |
|
| 929. |
The fraction of total volume occupied by atoms in a simple cube isA. `(pi)/(6)`B. `(pi)/(3sqrt2)`C. `(pi)/(4sqrt2)`D. `(pi)/(4)` |
|
Answer» Correct Answer - A For simple cube, `"Radius (r)"=(a)/(2)" [a = edge length]"` `"Volume of the atom"=(4)/(3)pi((a)/(2))^(3)` `therefore" Packing fraction"=((4)/(4)pi((a)/(2))^(3))/(a^(3))=(pi)/(6)` |
|
| 930. |
Assertion: Gases like `N_(2), O_(2)` behave as ideal gases at high temperature and low pressure. Reason: Molecular interaction diminihes at high temperature and low pressure.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
| Answer» Correct Answer - A | |
| 931. |
The ratio of rates of diffusion of `CO_(2) and SO_(2)` at the same pressure and temperatue is:A. `4:sqrt(11)`B. `11:4`C. `1:4`D. `1:6` |
|
Answer» Correct Answer - A `(r_(1))/(r_(2))=sqrt((M_(2))/(M_(1)))=sqrt((64)/(44))=(4)/(sqrt(11))` |
|
| 932. |
When helium gas (at room temperature) undergoes the Joule-Thomson expansion , heating of the gas is observed becauseA. helium is an ideal gasB. the inversion temperature of helium gas is very highC. helium is a noble gasD. the inversion temperature of helium gas is very low |
|
Answer» Correct Answer - D At inversion temperature , there is neither heating nor cooling. Below the inversion temperature, there is heating effect. Thus , when a compressed gas is allowed to expand thriugh a porous plug (joule - Thomson expansion) at a temperature above its inversion temperature, there is a rise in temperature. |
|
| 933. |
In a crystalline solid,atoms of X from fcc packing and the atoms of Y occupy all octahedral voids.If the atoms along one body diagnal are removed then the simplest formula of the crystalline solid will be:A. XYB. `X_(4)Y_(3)`C. `X_(5)Y_(4)`D. `X_(4)Y_(5)` |
|
Answer» Correct Answer - C Number of atoms of X in fcc packing (at corners and face centres of cubic unit cell) `=8xx(1)/(8)+6xx(1)/(2)=4` Number of atoms of Y at octahedral voids =4 Along one body diagonal there are two X atoms atoms and one Y atom. Number of effective atoms of X after removal `=4-2xx(1)/(8)=(15)/(4)` Number of effective atoms of Y after removal `=4-1=3` `X:Y=(15)/(4):3` `=5:4` `therefore` SImplest formula `=X_(5)Y_(4)` |
|
| 934. |
A crystalline solid is made of X, Y and Z elements. Atoms of X form fcc packing , atoms of Y occupy octahedral voids while atoms of Z occupy tetrahedral voids. What will be the simplest formula of solid if atoms along one body diagonal are removed ?A. `X_(5)Y_(4)Z_(8)`B. XYZC. `X_(8)Y_(4)Z_(5)`D. `X_(2)YZ` |
|
Answer» Correct Answer - A Number of atoms of X (at packing site, i.e., at corners and face-centres) `8xx(1)/(8)+6xx(1)/(2)=4` Number of atoms of Y=4 Number of atoms of Z=8 Along one body diagonal there will be two X atoms , one Y atoms and two Z atoms are found and are removed. Number of aotms of X will be `=4-(1)/(8)xx2=(15)/(4)` Number of atoms of Y will be 4-1=3 Number of atoms of Z will be =8-2=6 `X:Y:Z` `(15)/(4):3:6` `5:4:8` `therefore` Simplest formula will be `X_(5)Y_(4)Z_(8)`. |
|
| 935. |
The energy gap `(E_g)` between valence band and conduction band for diamond, silicon and germanium are in the orderA. `E_g "(diamond) gt " E_(g) "(silicon) gt " E_g "(germanium)"`B. `E_g "(diamond) lt " E_(g) "(silicon) lt " E_g "(germanium)"`C. `E_g "(diamond) = " E_(g) "(silicon) = " E_g "(germanium)"`D. `E_g "(diamond) gt " E_(g) "(germanium) gt " E_g "(silicon)"` |
|
Answer» Correct Answer - A (a) is correct option because diamond is an insulator and non-metallic character decreases down the group. |
|
| 936. |
To get an n-type semiconductor from silicon, it should be doped with a substances with valence _____A. 2B. 1C. 3D. 5 |
| Answer» Correct Answer - d | |
| 937. |
(a)Based on the nature of intermolecular forces , classify the following solids : Sodium sulphate, Hydrogen (b)What happen when AgCl is doped with `CdCl_2` (c )Why do ferrimagnetic substances show better magnetism than antiferromagnetic substances ? |
| Answer» (a)`Na_2SO_4`=Ionic solid, `H_2`=Non-polar molecular solid and (b)Electrical conductivity increases | |
| 938. |
Classify each of the following as being either a p-type or n-type semiconductor: (i) Ge doped with In (ii) B doped with Si. |
|
Answer» (i)Ge is Group 14 element and In is Group 13 element. Hence, an electron deficit hole created and therefore, it is p-type. (ii) B is Group 13 element and Si is Group 14 element, there will be a free electron. Hence, it is n-type |
|
| 939. |
The appearance of colour in solid alkali metal halides is generally due toA. F-centresB. Schottky defectC. Frenkel defectD. Interstitial positions |
|
Answer» Correct Answer - A F - centres are the sites where anions are missing and instead electrons are present and the appearance of colour in solid alkali metal halides is generally due to F- centres. |
|
| 940. |
The ratio of the rate of diffusion of helium and methane under indentical conditions of pressure and temperature will beA. 2B. 0.5C. 16D. 4 |
|
Answer» Correct Answer - A `(r_(He))/(r_(CH_(4)))=sqrt((M_(CH_(4)))/(M_(He)))=sqrt((16)/(4))=2`. |
|
| 941. |
Calculate the pressure exerted by `10^(23)` gas molecules each of mass `10^(-22) g` in a container of volume 1 litre the rms speed is `10^(5) cm s^(-1)` |
|
Answer» Using kinetic gas equation, `P = (1)/(3) (mnc^(2))/(V)` Given, `V = 1` litre `= 1000 mL = 1000 cm^(3), n = 10^(23), m = 10^(-22) g " and " c = 10^(5) cm s^(-1)` Substituting the values in the above equation, `P = (1)/(3) xx (10^(-22) xx 10^(23) xx 10^(10))/(1000) = 3.33 xx 10^(7) " dyne" cm^(-2)` |
|
| 942. |
What volume of air at N.T.P containing 21 % oxygen by volume is required to completely burn 1000 g of sulphur containing 4 % incombustible matter ? |
|
Answer» Correct Answer - `3360 L` Amount of incombustible matter in sulphur `= (1000 xx 4)/(100) = 40 g` Amount of pure matter `= 1000 - 40 = 960 g` Step I. Volume of oxygen required The chemical equation for the reaction is `underset(32g)(1//8S_(B))+underset(22.4L)(O_(2))rarrSO_(2)` `32g` of sulphur `(S_(8))` require `O_(2)` at `N.T.P.` `= 22.4 L` `960g` of sulphur `(S_(8))` require `O_(2)` at `N.T.P = ((22.4 L) xx (960g))/((32g)) = 972 L` Step II. Volume air required Volume of oxygen required `= 972 L` Volume of air required `= (672 L) xx (100)/(200) = 3360 L` |
|
| 943. |
A 10.0 litre container is filled with a gas to a pressure of 2.00 atm at `0^(@)C`. At what temperature will the pressure inside the container be 2.50 atm ? |
|
Answer» As volume of the container remains constant, applying pressure-temperature law, viz., , `(P_(1))/(T_(1))=(P_(2))/(T_(2))`, we get `(2 atm)/(273 K)=(2.50 atm)/(T_(2))" or " T_(2)=341 K=341-273^(@)C=68^(@)C` |
|
| 944. |
The critical temperature `(T_(c))` and critical pressure `(p_(c))` of `CO_(2)` are `30.98^(@)C` and 73 atm respectively. Can `CO_(2) (g)` be liquefied at `32^(@)C` and 80 atm pressure ? |
| Answer» No, the gas cannot be liquefied at `32^(@)C` because this temperature is above the critical temperature of `30.98^(@)C`. The effect of pressure is insignificant above the critical temperature. | |
| 945. |
An `LPG` cylinder weighs `14.8 kg` when empty. When full it weighs `29.0 kg` and the weight of the full cylinder reduces to `23.2 kg`. Find out the volume of the gas in cubic metres used up at the normal usage conditions and the final pressure inside the cylinder. Assume `LPG` to be `n`-butane with normal boiling point of `0^(@)C`. |
|
Answer» Weight of full cylinder `=29 kg` Weight after usage `=23.2 kg` Gas consumed `=29-23.2=5.8kg` or `5800 g` For butane `(C_(4)H_(10))`, `M=58` Normal pressure `=1 atm` Room temperature `=298 K` `PV=nRT` or `1xxV=(5800)/(58)xx0.0821xx298` or `V=2446.58 L~~2.447m^(3)` |
|
| 946. |
What volume of `O_(2)` at 2.00 atm pressure and `27^(@)C` is requried to burn 10.0 g of heptane `(C_(7)H_(16))` ? `C_(7)H_(16) + 11 O_(2) to 7 CO_(2) + 8 H_(2)O` |
|
Answer» Correct Answer - `13.71 L` Step I. Volume of `O_(2)` at N.T.P. The chemical equation for the reaction is : `{:(C_(7)H_(16),+,11O_(2),rarr7CO_(2)+8H_(2)O),(7xx12+16xx1,,11xx22.4,),(=100g,,-246.4L,):}` 100 g of `C_(7)H_(16)` require `O_(2) = 246.4 L , 10 g` of `C_(7)H_(16)` require `O_(2) = (246.4 xx 10)/(100) = 24.64 L` Volume of `O_(2)` under exprimental conditions `{:("N.T.P. conditions",,"Experimental conditions"),(V_(1)=24.64L,,V_(2)=?),(P_(1)=1.013"bar",,P_(2)=2 "bar"),(T_(1)=273 K,,T_(2)=27+273=300K):}` By applying 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)) = ((1.013 "bar") xx (24.64 L) xx (300 K))/((273 K) xx (2 "bar")) = 13.71 L` |
|
| 947. |
An LPG (liquefied petroleum gas) cylinder weighs 14.8 kg when empty. When full, it weighs 29.0 kg and shows a pressure of 2.5 atm. In the course of use at `27^(@)C`, the weight of the full cylinder reduced to 23.2 kg. Find out the volume of the cubic metres used up at the normal usage conditions, and the final pressure inside the cylinder. Assume LPG to be n-butane with normal boiling point of `0^(@)C` |
|
Answer» Decrease in the amount of PG`=29.0-23.2=5.8" kg"=(5800)/(58)" kg moles"=100" moles"` Volume of 100 moles at 1 atm `300" K "(nRT)/(P)=(100" moles"xx0.0821" L atmK^(-1)xx300 K)/(1 atm)` `=2463" L "=2463xx10^(-3)m^(3)=2.463m^(3)` Final pressure inside the cylinder. As the cylinder contains liquefied petroleum gas in equibrium with its vapours, therefore, so long as temperature remains constant and some LPG is present, pressure with remain constant. As the cylinder still contains LPG=23.2-14.8=8.4 kg, pressure inside the cylinder will be same, i.e., 2.5 atm. |
|
| 948. |
What is the volume at `N.T.P.` of a gas that occupies `43.0 mL` at `- 3^(@)C` and `0.98 "bar"` ? |
|
Answer» Correct Answer - `42.06 mL` `V_(1) = 43.0 mL , V_(2) = ?` `P_(1) = 0.98 "bar" , V_(2) = ?` `T_(1) = -3 + 273 = 270 K , T_(2) = 273 K` According to ideal 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)) = ((0.98 "bar" ) xx (43 mL) xx (273 K))/((270 K) xx (1.013 "bar")) = 42.06 mL` |
|
| 949. |
At `25^(@)C` and 760 mm of Hg pressure, a gas occupies 600 mL volume. What will be its pressure at a height where temperature is `10^(@)C` and volume of the gas is 640 mL ? |
|
Answer» `{:("Initial conditions","Finalconditions"),(P_(1)=760 mm,P_(2)=?),(V_(1)=600 mL,V_(2)=640 mL),(T_(1)=25^(@)C=298 K,T_(2)=10^(@)C=283 K):}` Applying gas equation (combined gas law), `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))=((760 mm)(600 mL))/(298 K)=(P_(2)(640 mL))/(283 K)` or `P_(2)=676.6" mm"` |
|
| 950. |
Calculate the temperature of 4.0 mol of a gas occupying d `dm^(3)` at 3.32 bar. (R=0.083 bar `dm^(3) K^(-1)mol^(-1))`. |
|
Answer» Correct Answer - 50 K Given, n = 4.0 mol V= 5 `dm^(3)` p = 3.32 bar R= 0.083 bar `dm^(3)K^(-1)mol^(-1)` The temperature (T) can be calculated using the ideal gas question as: pV = nRT `rArrT=(pV)/(nR)` `=(3.32xx5)/(4xx0.083)` = 50 K Hence, the required temperature is 50 K. |
|