Looking for classes? Ksquare Career Institute, Bengaluru →
Chemistry
Start Practice Clear paper filter
-
If a gas expands at constant temperature, it indicates that :The average translational kinetic energy of one molecule of an ideal gas is $KE = \dfrac{3}{2}\dfrac{R}{N_A}T = \dfrac{3}{2}kT$
Here $\dfrac{R}{N_A} = k$ is the Boltzmann constant.
So $E_t \propto T$: the kinetic energy depends only on the absolute temperature.
Thus, at constant temperature, the kinetic energy of the molecules remains the same even though the gas expands.
During isothermal expansion the volume increases, so the pressure decreases (Boyle's law, $PV$ = constant); it does not increase.
The number of molecules of a fixed mass of gas does not change on expansion. -
What volume of oxygen gas ($O_2$) measured at $0^\circ C$ and 1 atm, is needed to burn completely one litre of propane gas ($C_3H_8$) measured under the same conditions ?Balanced equation for the combustion of propane: $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$
Volume ratio of the gases: 1 : 5 : 3 : 4
At the same temperature and pressure, equal volumes of gases contain equal numbers of moles (Avogadro's law), so mole ratios are also volume ratios.
1 volume of propane requires 5 volumes of $O_2$.
So 1 litre of propane requires 5 litres of $O_2$ to burn completely. -
In DNA, the complementary bases are -DNA contains two types of nitrogenous bases.
Purines: Adenine (A) and Guanine (G).
Pyrimidines: Cytosine (C) and Thymine (T).
In the double helix a purine always pairs with a specific pyrimidine on the opposite strand.
Adenine pairs with thymine (A : T) by two hydrogen bonds.
Guanine pairs with cytosine (G : C) by three hydrogen bonds.
Uracil is present in RNA in place of thymine; it is not a base of DNA.
So the complementary bases in DNA are adenine and thymine, guanine and cytosine. -
The sequence of ionic mobility in aqueous solutions is -Smaller the size of the cation, higher is its charge density and higher will be its hydration.
Extent of hydration: $Na^+ \gt K^+ \gt Rb^+ \gt Cs^+$
A more heavily hydrated ion carries a larger shell of water molecules, so its effective (hydrated) size increases.
Hydrated radius: $Na^+(aq) \gt K^+(aq) \gt Rb^+(aq) \gt Cs^+(aq)$
A bigger hydrated ion moves more slowly, so its mobility in aqueous solution decreases.
Hence the ionic mobility follows the sequence: $Cs^+ \gt Rb^+ \gt K^+ \gt Na^+$ -
Which one of the following is most reactive towards electrophilic attack ?Reactivity towards an electrophile depends on the electron density of the benzene ring: higher the electron density, faster the attack.
The electron density of the phenol ring is the highest among chlorobenzene, benzyl alcohol and nitrobenzene.
In phenol, the OH group shows both the $+M$ (resonance) effect and the $-I$ (inductive) effect, but the $+M$ effect of OH dominates over its $-I$ effect.
The lone pair on oxygen is delocalised into the ring, increasing the electron density, especially at the ortho and para positions. So OH is a strongly activating group.
In chlorobenzene the $-I$ effect of Cl dominates over its $+M$ effect, so the ring is weakly deactivated.
In benzyl alcohol the $CH_2OH$ group has no lone pair directly on the ring atom and is slightly electron withdrawing, so there is no strong activation.
In nitrobenzene the $NO_2$ group has strong $-M$ and $-I$ effects, so the ring is strongly deactivated.
Hence phenol is the most reactive towards electrophilic attack. -
Volume occupied by one molecule of water (density = 1 g cm$^{-3}$) is :Weight of $6.023 \times 10^{23}$ molecules (1 mole) of water = 18 g
Volume occupied by $6.023 \times 10^{23}$ molecules of water (density = 1 g cm$^{-3}$) $= \dfrac{\text{mass}}{\text{density}} = \dfrac{18\ g}{1\ g\ cm^{-3}}$
$= 18$ cm$^3$ or 18 mL
So the volume occupied by one molecule of water $= \dfrac{18}{6.023 \times 10^{23}}$ cm$^3$
$= 2.988 \times 10^{-23}$ cm$^3$
$\approx 3.0 \times 10^{-23}$ cm$^3$ -
Which one of the following statements is not true ?Natural rubber is a linear 1,4-addition polymer of isoprene (2-methylbuta-1,3-diene).
It is cis-1,4-polyisoprene and has only the cis-configuration at every double bond.
Because of the cis arrangement the chains cannot pack closely; they are held by weak van der Waals forces and take up a coiled structure, which gives rubber its elasticity.
The all-trans polymer of isoprene is gutta-percha, which is hard and non-elastic.
So the statement that natural rubber has the trans-configuration at every double bond is not true.
The other statements are true: Buna-S is a copolymer of buta-1,3-diene and styrene; natural rubber is a 1,4-polymer of isoprene; and in vulcanization the sulphur cross-links (bridges) formed between different chains make the rubber harder and stronger. -
For the gas phase reaction,
$PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$
Which of the following conditions are correct ?$PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$
The dissociation of $PCl_5$ involves the breaking of two P-Cl bonds, which needs energy, so the reaction is endothermic.
$\Delta H = \Delta E + \Delta n_g RT$, where $\Delta n_g$ = change in the number of moles of gaseous products and reactants.
Here $\Delta n_g = 2 - 1 = +1$, i.e. positive, so $\Delta H$ = +ve.
$\Delta S = S_{products} - S_{reactants}$
One mole of gas gives two moles of gas, so the randomness increases and $\Delta S$ = +ve.
Hence $\Delta H \gt 0$ and $\Delta S \gt 0$. -
The measurement of the electron position is associated with an uncertainty in momentum, which is equal to $1 \times 10^{-18}$ g cm s$^{-1}$. The uncertainty in electron velocity is :
(Mass of an electron is $9 \times 10^{-28}$ g)Momentum $p = mv$, so uncertainty in momentum $\Delta p = m\,\Delta v$
Uncertainty in momentum $(m\,\Delta v) = 1 \times 10^{-18}$ g cm s$^{-1}$
Uncertainty in velocity: $\Delta v = \dfrac{\Delta p}{m}$
$\Delta v = \dfrac{1 \times 10^{-18}}{9 \times 10^{-28}}$
$\Delta v = 1.1 \times 10^{9}$ cm s$^{-1}$
$\approx 1 \times 10^{9}$ cm s$^{-1}$ -
The angular shape of ozone molecule ($O_3$) consists of -In ozone the central oxygen atom is $sp^2$ hybridised and the molecule is angular (bent) with a bond angle of about $117^\circ$.
Each resonance structure is written as $O=O^+-O^-$: the central oxygen forms one double bond and one single bond.
The single bond is one sigma bond.
The double bond consists of one sigma bond and one pi bond.
The central oxygen also carries one lone pair.
Total: 2 sigma bonds and 1 pi bond.
The two resonance structures make both O-O bonds equal in length (about 128 pm), but the count of bonds remains 2 sigma and 1 pi. -
Percentage of free space in a body centered cubic unit cell is -In a body centred cubic (bcc) unit cell the atoms touch along the body diagonal: $\sqrt{3}\,a = 4r \Rightarrow a = \dfrac{4r}{\sqrt{3}}$
Number of atoms per bcc unit cell $= 8 \times \dfrac{1}{8} + 1 = 2$
Volume occupied by the atoms $= 2 \times \dfrac{4}{3}\pi r^3 = \dfrac{8}{3}\pi r^3$
Volume of the unit cell $= a^3 = \left(\dfrac{4r}{\sqrt{3}}\right)^3 = \dfrac{64r^3}{3\sqrt{3}}$
Packing fraction $= \dfrac{\frac{8}{3}\pi r^3}{\frac{64 r^3}{3\sqrt{3}}} = \dfrac{\sqrt{3}\,\pi}{8} = 0.68$
So 68 percent of the unit cell is occupied by atoms.
Percentage of free space $= 100 - 68 = 32$ percent. -
The value of equilibrium constant of the reaction,
$HI(g) \rightleftharpoons \dfrac{1}{2}H_2(g) + \dfrac{1}{2}I_2(g)$ is 8.0.
The equilibrium constant of the reaction,
$H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ will be -$HI(g) \rightleftharpoons \dfrac{1}{2}H_2(g) + \dfrac{1}{2}I_2(g)$
$K = \dfrac{[H_2]^{1/2}[I_2]^{1/2}}{[HI]} = 8$
The required reaction is $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$, with $K' = \dfrac{[HI]^2}{[H_2][I_2]}$
This is the first reaction reversed (so the constant becomes $\dfrac{1}{K}$) and multiplied by 2 (so the constant is squared).
$K' = \left(\dfrac{1}{K}\right)^2 = \left(\dfrac{1}{8}\right)^2$
$K' = \dfrac{1}{64}$ -
Which one of the following is an amine hormone ?On the basis of chemical nature, hormones are classified as steroid hormones, peptide (protein) hormones and amine hormones.
Thyroxine and adrenaline are amine hormones. Thyroxine is an iodine containing derivative of the amino acid tyrosine and is secreted by the thyroid gland.
Amine hormones are water soluble hormones having amino groups.
Insulin is a peptide (protein) hormone made of 51 amino acids.
Progesterone is a steroid hormone.
Oxypurin (oxypurine) is a purine derivative, not an amine hormone.
So the amine hormone among these is thyroxine. -
Kohlrausch's Law states that at :Kohlrausch's law of independent migration of ions: at infinite dilution, when dissociation is complete, each ion makes a definite contribution to the equivalent (molar) conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte.
$\Lambda_m^\infty = \lambda_+^\infty + \lambda_-^\infty$
Here $\lambda_+^\infty$ and $\lambda_-^\infty$ are the ionic conductances at infinite dilution for the cation and the anion respectively.
The law holds only at infinite dilution, because only then are the ions far apart and free from inter-ionic attractions; so the statement with 'finite dilution' is wrong.
The contribution of an ion is independent of the other ion, so the statement 'depending on the nature of the other ion' is wrong.
The contribution is to the equivalent (or molar) conductance, not simply to the conductance.
The law is used to find $\Lambda^\infty$ of weak electrolytes such as acetic acid. -
Green chemistry means such reactions which -Green chemistry is the design of chemical products and processes that reduce or eliminate the use and production of hazardous substances.
It aims at preventing pollution at the source rather than treating the waste afterwards.
Examples: using $H_2O_2$ in place of chlorine for bleaching paper, using liquefied $CO_2$ with a suitable detergent in place of tetrachloroethene for dry cleaning of clothes, and carrying out reactions in water.
The word 'green' here does not refer to plants or to colour, and green chemistry is not the study of ozone depletion.
So green chemistry means such reactions which reduce the use and production of hazardous chemicals. -
With which one of the following elements silicon should be doped so as to give p-type of semiconductor ?Silicon belongs to group 14 and has four valence electrons, each forming a covalent bond with a neighbouring silicon atom.
If silicon is doped with any element of group 13 (B, Al, Ga etc.) of the periodic table, a p-type semiconductor is obtained.
A group 13 atom such as boron has only three valence electrons, so it can form only three bonds; the fourth bond lacks an electron and a positive hole is created.
These holes move through the crystal and conduct electricity, hence the name p-type (positive).
Arsenic (group 15) has one extra electron and gives an n-type semiconductor; selenium (group 16) also supplies extra electrons.
Germanium is a group 14 element like silicon and creates neither holes nor extra electrons.
So silicon should be doped with boron. -
An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the compound would be :Percentage of oxygen $= 100 - (38.71 + 9.67) = 51.62$
Atomic mass of C = 12, H = 1 and O = 16.
C: mole ratio $= \dfrac{38.71}{12} = 3.22$; simple ratio $= \dfrac{3.22}{3.22} = 1$
H: mole ratio $= \dfrac{9.67}{1} = 9.67$; simple ratio $= \dfrac{9.67}{3.22} = 3$
O: mole ratio $= \dfrac{51.62}{16} = 3.22$; simple ratio $= \dfrac{3.22}{3.22} = 1$
C : H : O = 1 : 3 : 1
Thus the empirical formula of the compound is $CH_3O$. -
In which of the following co-ordination entities the magnitude of $\Delta_0$ (CFSE in octahedral field) will be maximum ?
(Atomic number of Co = 27)In all four complexes cobalt is in the +3 oxidation state ($3d^6$) in an octahedral field, so the metal ion and the geometry are the same.
The CFSE in an octahedral field then depends only upon the nature of the ligands: stronger the ligand, larger will be the value of $\Delta_{oct}$.
Spectrochemical series (increasing field strength): $I^- \lt Br^- \lt Cl^- \lt F^- \lt OH^- \lt C_2O_4^{2-} \lt H_2O \lt NH_3 \lt en \lt CN^- \lt CO$
Among the given ligands the order of field strength is: $C_2O_4^{2-} \lt H_2O \lt NH_3 \lt CN^-$
$CN^-$ is the strongest field ligand here, so it causes the largest splitting.
Hence $\Delta_0$ is maximum for $[Co(CN)_6]^{3-}$. -
The alkali metals form salt-like hydrides by the direct synthesis at elevated temperature. The thermal stability of these hydrides decreases in which of the following orders ?Alkali metals form ionic (salt-like) hydrides $M^+H^-$ on heating with hydrogen: $2M + H_2 \rightarrow 2MH$
The size of the metal cation increases from $Li^+$ to $Cs^+$.
The small $H^-$ ion is held most strongly by the small $Li^+$ ion; as the cation becomes larger, the lattice energy of the hydride decreases.
So the M-H bond becomes weaker from LiH to CsH and the thermal stability of these hydrides decreases as follows:
LiH > NaH > KH > RbH > CsH
LiH is stable up to about 1000 K, while the others decompose at lower temperatures. -
Which of the following are not state functions ?
(I) $q + w$
(II) $q$
(III) $w$
(IV) $H - TS$State functions or state variables depend only on the state of the system and not on the path by which that state is reached.
Here '$w$' represents work done and '$q$' represents the amount of heat. Both depend on the path followed, so both of these are not state functions (they are path functions).
(I) $q + w = \Delta U$, the change in internal energy (first law of thermodynamics). Internal energy is a state function, so $q + w$ is a state function even though $q$ and $w$ separately are not.
(IV) $H - TS = G$, the Gibbs free energy. $H$, $T$ and $S$ are all state functions, so $G$ is also a state function.
Therefore the quantities that are not state functions are (II) $q$ and (III) $w$. -
In the hydrocarbon,
$\overset{6}{C}H_3-\overset{5}{C}H=\overset{4}{C}H-\overset{3}{C}H_2-\overset{2}{C}\equiv\overset{1}{C}H$
The state of hybridization of carbons 1, 3 and 5 are in the following sequence :The hybridisation of a carbon atom is decided by the number of sigma bonds it forms: 4 sigma bonds means $sp^3$, 3 sigma bonds (one double bond) means $sp^2$, and 2 sigma bonds (a triple bond) means $sp$.
Hybridisation of the carbons from C6 to C1: $sp^3$, $sp^2$, $sp^2$, $sp^3$, $sp$, $sp$
C1 is part of the triple bond ($C\equiv CH$), so it is $sp$ hybridised.
C3 is a $CH_2$ group with four single bonds, so it is $sp^3$ hybridised.
C5 is part of the double bond ($CH=CH$), so it is $sp^2$ hybridised.
The state of hybridization of carbons 1, 3 and 5 is $sp$, $sp^3$ and $sp^2$ respectively. -
A strong base can abstract an $\alpha$-hydrogen from:The hydrogen atoms on the carbon next to a carbonyl group ($\alpha$-hydrogens) are acidic.
The carbonyl group is strongly electron withdrawing, and the carbanion formed after the loss of the $\alpha$-hydrogen is stabilised by resonance with the $C=O$ group (enolate ion).
$R-CO-CH_3 + B^- \rightarrow R-CO-\overset{\ominus}{C}H_2 \leftrightarrow R-C(O^-)=CH_2 + BH$
So a strong base can abstract an $\alpha$-hydrogen from aldehydes and ketones to form a carbanion or the enolate ion. This is the first step of reactions like the aldol condensation.
In alkanes, alkenes and amines there is no such resonance stabilisation of the resulting carbanion, so their C-H hydrogens are not acidic enough to be removed in this way.
Hence the answer is ketone.