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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.
| 50301. |
(a) Explain the following observations, (i) Aerated water bottles are kept under water during summer. (ii) Liquid ammonia bottle is cooled before opening the seal. |
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Answer» Solution :(i)In acrated water bottles, `CO_2`, gas is passed through the aqueous solution under pressure because the solubility of the gas in water is not very high in summer, the solubility of the eas in water is likely to decrease because of the rise in temperature Thus, in summer, more of gas will be present above the liquid face in the glass BOTTLE. In case, the pressure of the gas becomes too high, the glass will not be able to withstand the pressure and the bottle may explode To avond this the bottles are kept under water. As a result, the temperature is likely to decrease and the solubility of `CO_2` is likely to increase in aqueous solution resulting in decreased pressure (ii) Liquid ammonia bottle CONTAINS the gas under very high pressure. If the bottle is opened as such, then the sudden decrease in pressure will lead to a large increase in volume of the gas As a result, the gas will come out of the bottle all of a sudden with force. This will lead to the breakage of the bottle and also causes accident. However, if the bottle is COOLED under tap water for sometime, there will be a decrease in the volume of a gas to a large extent. If the seal is opened now, the gas will come out of the bottle at a SLOWER rate, reduces the CHANCES of accident. |
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| 50302. |
A examples of covalent hydride is ____________ |
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Answer» `CaH_2` |
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| 50303. |
(A): Even though diamond is covalent, it has a high melting point (R) : Diamond is a three dimensional gaint molecule. The C-C in it are very strong. |
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Answer» Both A and R are TRUE. R is the CORRECT EXPLANATION for A. |
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| 50304. |
Ketone secondary amines, ethers etc show metamerism |
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Answer» A and R are TRUE, R EXPLAINS A |
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| 50305. |
(A) : Equivalent weight of Cu in CuO is 63.6 and in Cu_(2)O is 31.8. (R) : Equivalent weight of an element =("Atomic weight of an element")/("Valency of the element") |
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Answer» Both A and R are true and R is the correct EXPLANATION |
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| 50306. |
(A): Entropy of a perfect crystalline substance at absolute zero is zero (R ): At absolute zero translation kinetic energy of a system is zero. |
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Answer» Both A and R are TRUE and R is THECORRECT EXPLANATION |
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| 50307. |
(A) Entropy decreases when a egg is boiled. (R) It is solidified due to denaturation of albumin. |
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Answer» Both (R) and (A) are TRUE and reason is the CORRECT explanation of assertion. |
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| 50308. |
(A): Enthalpy of graphite is lower than that of diamond. (R): Entropy of graphite is greater than that of diamond. |
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Answer» Both A and R are TRUE R is the correct EXPLANATION. |
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| 50309. |
(A): Emssion spectrum produced due to the transition of an electron from M shell to L shell (R) : The ratio of energy and frequency of a photon is 6.625xx10^(-27) "erg-sec" |
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Answer» Both A and R are true and R is the CORRECT explanation of A |
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| 50310. |
Assertion A: Empiricial formula of glucose or that of acetic acid is CH_(2)O. Reason(R):If percentage composition of elements is same, then empirical formula is same. The correct answer is |
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Answer» Both A and R are TRUE and R is the CORRECT EXPLANATION |
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| 50311. |
An element M has an atomic mass 19 and atomic number 9. Its ion is represented by |
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Answer» `M^(+)` |
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| 50312. |
(A) Duma's method is more applicable to nitrogen containing organic compounds than Kjeldahl's method. (R) Kjeldahl's method does not give satisfactory results for compounds in which nitrogen is linked to oxygen. |
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Answer» If both ASSERTION and REASON are correct and reason is the correct EXPLANATION of the assertion |
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| 50313. |
A drug which is structurally related to adrenaline is : |
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Answer» salbutamol |
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| 50314. |
A drug marijuna owes its activity to tetrahydrocarbinol, which contains 70% as many C atoms as oxygen atoms. The number of mole of compound in a gm of it is 0.00318. the molecualr formular will be : |
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Answer» `C_(20)H_(30)O_(2)` |
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| 50315. |
(a) Draw the shapes of the following orbitals. (i)3d_(xy)(ii)d_(x^(2)) What is the total number of orbitals associated with the principal quantum number n=3 ? (c) Using s,p,d,f notations , describe the orbitalwith thefollowing quantum numbers :- |
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Answer» |
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| 50316. |
A doctor by mistake administres a dilute Ba(NO_(3)) solution to a patient for radiographic investigations. Which of the following should be the best to prevent the absorption of soluble barium and subsequent barium poisoning |
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Answer» `NaCl` |
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| 50317. |
A discharge tube containing nitrogen gas at 25^(@)C in evacuated till the pressure is 2xx10^(-2) mm. If the volume of discharge tube is 2 litre. Calculate the number of nitrogen molecules still present in the tube (R=0*0821 L "atm mol"^(-1)K^(-1)) |
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Answer» Solution :STEP : 1 CALCULATION of number of moles of nitrogen. Here `P=2xx10^(-2) mm =2xx10^(-2) xx(1)/(760) atm` `V=2L, T=25^(@)C=298*15K` According to ideal gas equation. `PV=nRT:. N(PV)/(RT)=2xx10^(-2)(1)/(760)xx(2)/(0*821)xx(1)/(298*15)` `:.n=2*15xx10^(-6)` Step: 2 Calculation of number of molecules. 1 MOL of nitrogen contains `=6*02xx10^(23)` molecules. `2*15xx10^(-6)` mol of nitrogen CONTAIN `=6*02xx10^(23)xx2*15xx10^(-6)` `1*29xx10^(18)` molecules |
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| 50318. |
A dispersion of AgCl in water is |
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Answer» HYDROPHILIC colloid |
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| 50319. |
(a) Discuss the concept of hybridization. What are its different types in a carbon atom. (b) What is the type of hybridiz.ation of carbon atoms marked with star. (a) underset(*)CH_(2) = CH - CH - ""* overset(O)overset(||)(C) - O - H (b) CH_(3) - ""^(*)CH_(2) - OH (c) CH_(3) - CH_(2) - overset(O) overset(||)(C*) - H (d)""^(*)CH_(3) - CH = CH - CH_(3) (e) ""^(*)CH_(3) - overset(*)(C) equiv CH |
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Answer» Solution :Hybridization : The process of intermixing of the orbitals of slightly different energy or of same energy to produce entirely new orbitals of equivalent energy, identical shapes and symmetrically disposed in plane. New orbitals formed are called hybrid orbitals. Only the orbitals of an isolated single atom can undergo hybridization. The hybrid orbitals generated are equal in number to that of the atomic orbitals which mix together . Hybrid orbitals do not make `pi` pi-bonds. If there are `pi` bonds, equal number of atomic orbitals must be left unhybridised for `pi`-bonding. Like atomic orbitals, hybri,d orbitals cannot have more than two electrons of opposite spins. Types of hybridization in carbon atoms : (a) (i) Diagonal or sp-hybridization : All compounds of carbon containing C `equiv` C triple bond like ethyne `(C_(2) H_(2))`. (ii) Trigonal or `sp^(2)`-hybridization : All compounds of carbon containing C = C (double bond) like ethene`(C_(2) H_(4))` . (iii) Tetrahedral or `sp^(3)`-hybridization : All compounds of carbon containing C -C single bonds only like ethane`(C_(2) H_(6))` . (i) ` underset(sp^(2) (3s))(*CH_(2)) = CH -underset(sp^(2) (3s))overset(O)overset(||)(C) -O - H ` (ii) `CH_(3) underset(sp^(3) (4 SIGMA))(*CH_(2)) OH "(iii)" CH_(3) - CH_(2) - underset((3 sigma) )underset(sp^(2))overset(O)overset(||)(C^(*)) - H ` (iv) ` underset((4 sigma))underset( sp^(3)) (*CH_(3) ) - overset(sp^(3)) overset((4 sigma))(CH) = CH - CH_(3) ` `CH_(3) - underset((2sigma))underset(sp)(*C) equiv CH` Direction (Q. Nos. 65-68) : Con1prehension given below is followed by s01nc n1ultiple choice questions. Each question has one correct option. Choose the correct option. Molecular orbitals are fonned by the overlap of atomic orbitals. Two atomic orbitals combine to form two molecular orbitals called bonding molecular orbital (BMO) and anti bonding molecular orbital (ABMO). Energy of anti bonding orbital is raised above the parent atomic orbitals that have combined and the energy of the bonding orbital ls lowered than the parent atomic orbitals. Energies of various molecular orbitals for ELEMENTS hydrogen to nitrogen increase in the order : `sigma 1s lt sigma^(**) 1s lt sigma 2s lt sigma^(**) 2s lt (pi 2p_(x)) = pi 2p_(y)) lt sigma 2p_(z) lt (pi^(**) 2p_(x) = pi^(**) 2p_(x)` and for oxygen and fluorine order of energy of molecular orbitals is given below : `sigma 1s lt sigma^(**) 1s lt sigma 2s lt sigma^(**) 2s sigma 2p_(z) lt (pi 2p_(x)) = pi 2p_(y)) lt (pi^(**) 2p_(x) = pi^(**) 2p_(x)` Different atomic orbitals of one atom combine with those atomic orbitals of the SECOND atom which have com arable energies and proper orjentation. Further, if the overlapping Is head on, the molecular orbital Is called .Sigma., `(sigma)` and if the overlap is lateral, the molecular orbital is called .pi. `(pi)` . The molecular orbitals are filled with electrons according to the same rules as followed for filling of atomic orbitals. However, the order for filling is not the same for all molecules or their Jons. Bond order is one of the most important parameters to compare the strength of bonds. |
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| 50320. |
A discharge tube of 2 litre capacity containing hydrogen gas was evacuated till the pressure inside is 1 xx 10^(-5) atm. If the tube is maintained at a temperature of 27^@C, calculate the number of hydrogen molecules still present in the tube. |
| Answer» SOLUTION :`4.89 XX 10^(17)` MOLECULES | |
| 50321. |
A dilute solution of KCl was placed between two Pt electrodes 10cm apart across which a potential difference of 10 volt was applied. Which is /are correct statement (Given: molar conductivity of K^+at infinite dilution is 96.5 Scm^2mol^(-1)) |
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Answer» Ionic MOBILITY of `K^+` is `10^(-3)CM^2sec^(-1)"VOLT"^(-1)` |
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| 50322. |
A dibasic organic acid gave the following results: C = 34.62%, H = 3.84%, 0.1075g of this acid consumes 20mL of 0.1N NaOH for complete neutralisation. Find out the molecular formula of the acid. |
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Answer» Solution :Calculation of empirical formula: `{:("Element","Percentage","At.mass","Relative number of atoms","Simplest ratio of atoms"),("Carbon",34.62,12,(34.62)/(12)=2.88,(2.88)/(2.88)=1xx3=3),("Hydrogen",3.84,1,(3.84)/(1)=3.84,(3.84)/(2.88)=1.33xx3 =4),("Oxygen",overset(61.54)("by difference"),16,(61.54)/(16)=3.84,(3.84)/(2.88)=1.33 xx 3 =4):}` Empirical formula of the acid `= C_(3)H_(4)O_(4)` Empirical formula mass `= (3 xx 12) +(4 xx1) +(4 xx 16) = 104` Calculation of molecular mass: `20mL 0.1N NaOH -= 0.1075g` acid `20 xx 0.1 ML N NaOH -= 0.1075g` acid So, `1000 mL 1N NaOH -= (0.1075)/(20 xx 0.1) xx 1000g` acid `-= 53.75g` acid Eq. mass of the acid `= 53.75` MOL. mass of the acid = Eq. mass `xx` Basicity `= 53.75 xx 2 = 107.50` `n = ("Mol. mass")/("Emp. mass") = (107.50)/(104.0) ~~1` Molecular formula `= C_(3)H_(4)O_(4)` |
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| 50323. |
A diabasic acid containing C,H and O was found to contain C=26.7% and H=2.2%. The vapour density of its dimethyl ester was found to bc 73. The molecular formula of the acid is |
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Answer» `CH_(2) O_(2)` |
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| 50324. |
A diatomic molecule has a dipole moment 1.2D. If the internuclear distance is 1A^@, what is the fraction of charge exists on each atom? |
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Answer» |
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| 50325. |
(A): Diamond is harder than silica (R): The Si-O-Si bonds in silica are weaker than C-C bonds in diamond |
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Answer» A and R are true, R explains A |
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| 50326. |
Diamond is covalent. Yet it has high melting point Why? |
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Answer» A and R are true, R EXPLAINS A |
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| 50327. |
A. Diamagnetic C_(2) molecule have been detected in vapour phase in which doubel bond consists of both pi bonds R: In C_(2) molecule foru electrons are presentin two I molecular orbitals |
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Answer» Both A and R are TRUE and R is the corret EXPLANATION of LA |
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| 50328. |
(a) Describe Fajan's rule |
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Answer» Solution :(a) The ability of a cation to polarise an anion is called its polarising ability and the tendency of an anion to get polarised is called its polsrisability. The EXTENT of polarisation in an ionic compound is given by the Fajans rules Fajans Rules : (1) To show greater covalent character, both the cation and anion should have high charge on them. Higher the positive charge on the cation, grater will be the attraction on the electron cloud of the anion. Similarly higher the magnitude of nagative charge on the anion, greater is its polarisability . Hence, the increase in charge on cation or in anion increases the covalent character Let us CONSIDER three jonic compounds ALUMINUM chloride, magnesium chloride and sodium chloride. Since the charge of the cation increase in the order `Na^(+) lt Mg^(2+) lt Al^(3+)`, the covalent character also follows the same order `NaCl lt MgCl_(2) lt AlCl_(3)`. (ii) The smaller cation and larger anion show greater covalent character due to the grater extent of polarisation. Lithium chloride is more covalent than sodium chloride e size of `Li^(+)` is smaller than `Na^(+)` and hence the polarising power of `Li^(+)` is more. Lithium iodide is more covalent than lithium chloride as the size of `I^(-)` is larger than the `Cl^(-)`. Hence `I^(-)` will be more polarised than `Cl^(-)` by the cation , `Li^(+)`. (iii) Cations having `ns^(2) np^(6) nd^(10)` configuration show greater polarising power than the cation with `ns^(2) np^(6)` configuration. Hence, they show greater covalent character. `CuCl` is more covalent than `NaCl`. Compared to `Na^(+) (1.13overset(" "@)(A))`. `CU^(+)(0.6 overset(" "@)(A))` is smalland have `3s^(3) 3p^(6) 3d^(10)` conguration Electronic conguration of `Cu^(+)` [Ar] `3s^(2), 3p^(6), 3d^(10)` Electronic Conguration of `Na^(+)` [He] `2S^(2), p^(6)` |
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| 50329. |
(a) Describe chemical properties of air, water, dihydrogen and halogen. (b) Explain reduction potential of alkali metal elements. (c) Which type of solutions are formed on reaction of alkali metals with liquid ammonia ? |
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Answer» Solution : The alkali metals are highly reactive due to their large size and low ionization enthalpy. The reactivity of thcsc mctals increases down the group. Reactivity towards air : The alkali metals tarnish in dry air due to the formation of their oxides which in turn react with moisture to form hydroxides. They burn vigorously in oxygen forming oxides. Lithium forms monoxide, sodium forms peroxide, the other metals form superoxides. The superoxide `(O_(2)^(-))` ion is stable only in the PRESENCE of large cations such as K, Rb, Cs. `4Li + O_(2) to 2Li_(2)O` (oxide) `2Na + O_(2) to Na_(2)O_(2)` (peroxide) `M + O_(2) to MO_(2)` (superoxide) (M = K, Rb, Cs) In all these oxides the oxidation state of the alkali METAL is +1. Lithium shows exceptional behaviour in reacting directly with nitrogen of air to form the nitride, `Li_(3)N` as well. Because of their high reactivity towards air and water, alkali metals are normally kept in kerosene oil. Reactivity towards water: The alkali metals react with water to form hydroxide and dihydrogen. `2M + 2H_(2)O to 2M^(+) + 2OH^(-) + H_(2)`(M = an alkali metal) It may be noted that although lithium has most negative `E^("ϴ")`value, its reaction with water is less vigorous than that of sodium which has the least negative `E^("ϴ")`value among the alkali metals. This behavior of lithium is attributed to its small size and very high hydration energy. Other metals of the group react EXPLOSIVELY with water. They also react with PROTON donors such as alcohol, gaseous ammonia and alkynes. Reactivity towards dihydrogen : The alkali metals react with dihydrogen at about 673K (lithium at 1073K) to form hydrides. All the alkali metal hydrides are ionic solids with high melting points. `2M + H_(2) to 2M^(+)H^(-)` Reactivity towards halogens : The alkali metals readily react vigorously with halogens to form ionic halides, `M^(+)X^(-)`. However, lithium halides are somewhat covalent. It is because of the high polarisation capability of lithium ion (The distortion of electron cloud of the anion by the cation is called polarisation). The `Li^(+)` ion is very small in size and has high tendency to distort electron cloud around the negative halide ion. Since anion with large size can be easily distorted, among halides, lithium iodide is the most covalent in NATURE. Reducing nature : The alkali metals are strong reducing agents, lithium being the most and sodium the least powerful. The standard electrode potential `(E^("ϴ"))`which measures the reducing power represents the overall change : `M_((s)) + M_((g))` (sublimation enthalpy) ` M_((g)) + M_((g))^(+) + e^(-)` (ionization enthalpy) `M_((g))^(+) + H_(2)O toM_((aq))^(+)` (hydration enthalpy) With the small size of its ion, lithium has the highest hydration enthalpy which accounts for its high negative`E^("ϴ")`value and its high reducing power. |
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| 50330. |
A definite amout of gaseous hydrocarbon was burnt with just sufficient of O_(2).The volume of all reactantswas 600ml,after the explosion the volume of the products [CO_(2)(g) and H_(2)O(g) was found to be 700ml under the similar conditions the molecular formula of the compound is . |
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Answer» `C_(3)H_(8)` |
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| 50331. |
(a) DefinePhotoelectric effect ? Mention its one practical application in daily life. (b) Electrons are emitted with zero velocity from a mental surface when it is exposed to radiation of wavelength 6800 Å. Calculatethresholds freqency (V_o) and work function (W_o) of the mental. |
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Answer» |
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| 50332. |
(a) Define ionization energy. (b) Prove that ionization energy is a periodic property. |
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Answer» Solution :(a) The energy required to remove the most loosely held electron from an isolated gaseous atom is called as ionization energy. (b) Variation in a PERIOD: On moving across a period from left to right, the ionization enthalpy value increases. This is due to the following REASONS: • Increase of NUCLEAR charge in a period • Decrease of atomic size in a period Because of these reasons, the valence ELECTRONS are held more tightly by the nucleus, thus ionization enthalpy increases. Hence, ionization energy is a periodic property. (ii) Variation in a group: As we move from top to bottom along a group, the ionization enthalpy decreases. This is due to the following reasons: • A gradual increase in atomic size • Increase of screening effect on the outermost electrons due to the increase of NUMBER of inner electrons. Hence, ionization enthalpy is a periodic property. |
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| 50333. |
(a) Define hydrogen bond and it types. |
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Answer» SOLUTION :(a) (1) Hydrogen bond : When hydrogen atom (H) is covalently bonded to highly electronegative atom (F or O or N), the bond is polarized in such a way that the hudrogen atom is able to from a weak bond (electrosatic attraction) between the hydrogen atom of a molecule and the electronegative atom a second molecule. The bond thus formed is CALLED a hydrogen bonds. (ii) Intermolecular Hydrogen : Intermolecular hydrogen bonds occur between two separate molecules. They can occur between any number of like or unlike molecules as long as hydrogen donors and acceptors are present an in position in which they can interact. Eg: Water, HF, etc., (iii) Intramolecular Hydrogen : This type of bond is formed between hydrogen atom and N. O or F atom of the same molecule. This type of hydrogen bonding is commonly called chelation and is more frequently FOUND in organic compounds. Eg: o-nitro phenol, salicylic acid . etc. |
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| 50334. |
(a) Define eutrophication and pneumoconiosis. (b) Write differences between photochemical smog and classical smog. |
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Answer» SOLUTION :Eutrophication: When the growth of algae increases in the surface of water, dissolved oxygen in water is GREATLY reduced. This phenomenon is known as eutrophication. DUE to this growth of fishes gets inhibited. PNEUMOCONIOSIS: It is a disease which irritates lungs. It causes scarring or FIBROSIS of the lungs. 
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| 50335. |
(a) Define atomic radius. (b) What are the difficulties in determining atomic radius? |
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Answer» Solution :(a) Atomic radius is the distance between the CENTRE of its nucleus and the outermost shell containing the electron. (b) Difficulties in determining atomic radius (i) The size of an atom is very small (`-1.2oversetoA` ie `1.2xx 10^(-10)`) (ii) The atom is not a rigid sphere, it is more like a spherical cotton ball rather than like a cricket ball. (iii) It is not possible to isolate an atom and measure its radius. (IV) The size of an atom depends upon the TYPE of atoms in its NEIGHBORHOOD and ALSO the nature of bonding between them. |
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| 50336. |
A define amount of impure sample of P_(4)O_(6) is treated with 20 mL of XM KMnO_(4) in acidic mediumto produceH_(3) PO_(4) and MnCl_(2). 20mL of same KMnO_(4) on treatementwith 0.2M FeSO_(4) requires exctly 10 mL of FeSO_(4) solution. What is amount of pure P_(4) O_(6)? If 1g sample is taken calculate % purity of P_(4)O_(6). |
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Answer» |
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| 50337. |
A cyllnder filled with a movable piston contains liquid water in equilibrium with water vapour at 25^(@)C. Which one of the following operations results in a decrease In the equilibrium vapour pressure ? |
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Answer» Moving PISTON downward a short distance |
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| 50338. |
A cylindrical balloon of 21 cm diameter is to be filled up with H_2 at NTP from a cylinder containing the gas at 20 atm " at "27^(@)C. The cylinder can hold 2.82 litre of water at NTP. Calculate the number of balloons that can be filled up. |
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Answer» Solution :Volume of 1 balloon which has to be filled `=4"/"3pi (21"/"2)^(3)` `=4851 mL=4.851` litre LET n balloons be filled, then volume of `H_(2)` OCCUPIED by balloons `=4.851xxn` Also, cylinder will not be empty and it will OCCUPY volume of `H_(2)= 2.82` litre. Total volume occupied by `H_(2)" at "NTP=4.851xxn +2.82` litre At STP `P_(2)=1 " atm Available "H_(2)` `V_(1)=4.851xxn+2.82""P_(2)=20 atm` `T_(1)=273K""T_(2)=300K` `P_(1)V_(1)"/"T_(2)=P_(2)V_(2)"/"T_(2)""V_(2)=2.82` litre or `1xx(4.851n+2.82"/"273)=20xx2.82"/"300"":. n=10` |
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| 50339. |
A cylinder of gas supplied by Bharat Petroleum is assumed to contain 14 kg of butane.If a normal family requires 20,000 kJ of energy per day for cooking butane gas in the cylinder last for....days(DeltaH_(c) of C_(4)H_(10) = -2658 kJ per mole) |
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Answer» 15 days 58 g of butane gives 2658 kJ of heat energy . 14 kg of butane will give heat energy `= (2658)/(58) xx 14 xx 10^(3)` `= 641.5862 xx 10^(3)` kJ DAILY energy requirement for cooking = 20, 000 kJ= `2 xx 10^(4) kJ` No. of days cylinder will last = `(641.5862 xx 10^(3)kJ) /(2 xx 10^(4) kJ"day^(-1))` = 32.08 days. |
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| 50340. |
A cylinder contains 6.023 xx 10^23 molecules of hydrogen and 5 xx 6.023 xx 10^22 molecules of oxygen. The partial pressure of oxygen is |
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Answer» 6/5 of the TOTAL PRESSURE |
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| 50341. |
A Cylinder contains 0.32 g of oxygen. The same volume of an unknown gas 'X' under similar conditions of temperature and pressure weighs 0.26 g. The gas 'X' is known the contain only carbon and hydrogen atoms in the ratio 1 : 1. Identify the gas 'X'. |
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Answer» Volume of `O_(2)` corresponding to 0.1 mole = 2.24 L Volume of unknown gas 'X' = 2.24 L 2.24 L of unknown gas 'X' weigh = 0.26 g 22.4 L of unknown gas 'X' weigh `= ((0.26g))/((2.24L))xx(22.4L)=26g` Since the gas contains only carbon and HYDROGEN in the ratio 1 : 1 , its `C_(2)H_(2)` (Acetylene). |
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| 50342. |
A cyclotron cannot accelerate |
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Answer» protons |
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| 50343. |
(A) : Cycloalkanes can exhibit geometrical isomerism. (R ): In cycloalkanes, free rotation is not possible about the ring |
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Answer» A and R are TRUE, R explains A |
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| 50344. |
A cycloalkane having molecular mass 84 and four secondary carbon atoms will form four monochloro structure isomers on chlorination. Identify the structure of cycloalkane. |
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Answer» |
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| 50345. |
(a) CuSO_(4).5H_(2)O(s)hArrCuSO_(4).3H_(2)O(s)+2H_(2)O(g) K_(p)=4xx10^(-4)atm^(2) (b) Na_(2)SO_(4).10H_(2)O(s)hArrNa_(2)SO_(4).5H_(2)O(s)+5H_(2)O(g) K_(p)=2.43xx10^(-8) atm^(5) (c) Na_(2)S_(2)O_(3).5H_(2)O(s)hArrNa_(2)S_(2)O_(3).2H_(2)O(s)+3H_(2)O(g) K_(p)=6.4xx10^(-5) atm^(3) What is order of vapour pressure and relative huimidity respectively. |
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Answer» `cgtbgta` V.P. |
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| 50346. |
(a) CusO_(4) reacts with KI in acidic medium to liberate I_(2): 2CuSO_(4) + 4KI rarr Cu_(2)I_(2) + 2K_(2) SO_(4) + I_(2) (b) Merrcuric per isodate Hg_(5)(IO_(6))_(2) reacts with a mixture of KI and HCI follwingthe equaction: Hg_(5)(10_(6))_*(2) + 34KI + 24HCI rarr 5K_(2) HgI_(4) + 8I_(2) + 24KCI + 12H_(2)O (c) The liberted iodine s titrated against Na_(2)S_(2)O_(3) solution. One mL of whcih is equivalent to 0.0499g of CuSO_(4).5H_(2)O. 5H_(2)O. What volume in mL of Na_(2)S_(2)O_(3) solution will be required to react withI_(2) liberated from 0.7245g of Hg_(5) (10_(6))_(2)? M. wt, of Hg_(5) (10_(6))_(2) = 1448.5 and M.wt of CuSO_(4). 5H_(2)O = 249.5 |
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| 50347. |
A current of dry air was bubbled through a bulb containing 26.66g of an organic compound in 200g of water, then through a bulb at the same temperature, containing water and finally through a tube containing anhydrous calcium chloride. The loss of mass in bulb containing water was 0.087g and gain in mass of the calcium chloride tube was 2.036g. Calcium the molecular mass of the organic substance. |
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Answer» Solution :`(p_(0)-p_(s))/(p_(0))=("LOSS in mass of solvent bulb")/("Grain in mass of CaCl"_(2)"TUBE")` `=(0.087)/(2.036)` Let the molecular mass of the organic substance be `m`. ACCORDING to RAOULT's law, `(p_(0)-p_(s))/(p_(0))=(w//m)/((w)/(m)+(W)/(M))` `(0.087)/(2.036)=((26.66)/(m))/((26.66)/(m)+(200)/(18))=(26.66)/(26.66+(200)/(18)m)` `m=53.75`. |
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| 50348. |
A current of 3 amperes was passed through silver nitrate solution for 125 seconds. The amount of silver deposited at cathode was 0.42 g. Calculate the equivalent mass of silver. |
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| 50349. |
A cubical vessel has a side with ‘l’ cm length contained a gas at a pressure of ‘P’. When the side of the vessel is made l/2 cm, the pressure of the gas becomes |
| Answer» ANSWER :D | |
| 50350. |
A cubic unit cell has one atom on each corner and one atom on each body diagonal. The number of atoms in the unit cell is |
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Answer» `therefore "total atoms in the unit cell " = 5` |
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