Which one of the following complexes
Question: Which one of the following complexes is violet in colour?$\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4}$$\left[\mathrm{Fe}(\mathrm{SCN})_{6}\right]^{4}$$\mathrm{Fe}_{4}\left[\mathrm{Fe}\left(\mathrm{CN}_{6}\right)\right]_{3} \cdot \mathrm{H}_{2} \mathrm{O}$$\left[\mathrm{Fe}(\mathrm{CN})_{5} \mathrm{NOS}\right]^{4}$Correct Option: 4, Solution: (1) $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4} \rightarrow$ Pale yellow solution (2) $\left[\mathrm{Fe}(\mathrm{SCN})_{6}\right]^{4} \righta...
Read More →Given below are two statements :
Question: Given below are two statements : Statement-I : Retardation factor $\left(\mathrm{R}_{\mathrm{f}}\right)$ can be measured in meter/centimeter. Statement-II : $R_{f}$ value of a compound remains constant in all solvents. Choose the most appropriate answer from the options given below: Statement-I is true but statement-II is falseBoth statement-I and statement-II are trueBoth statement-I and statement-II are falseStatement-I is false but statement-II is trueCorrect Option: , 3 Solution: $...
Read More →In chromotography technique, the purification of compound is independent of :
Question: In chromotography technique, the purification of compound is independent of :Mobility or flow of solvent systemSolubility of the compoundLength of the column or TLC PlatePhysical state of the pure compoundCorrect Option: , 4 Solution: In chromotography technique, the purification of a compound is independent of the physical state of the pure compound....
Read More →The equivalents of ethylene diamine required to replace the neutral ligands from the coordination sphere of the trans-complex
Question: The equivalents of ethylene diamine required to replace the neutral ligands from the coordination sphere of the trans-complex of $\mathrm{CoCl}_{3} .4 \mathrm{NH}_{3}$ is_________ (Round off to the Nearest Integer). Solution: trans $-\mathrm{CoCl}_{3} .4 \mathrm{NH}_{3}$ or trans- $\left[\mathrm{CO}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{C} \ell$ As we know that ethylene diamine is a bidentate ligand and ammonia is a mono dentate ligand. It means overall two eth...
Read More →The pressure exerted by a non-reactive gaseous mixture
Question: The pressure exerted by a non-reactive gaseous mixture of $6.4 \mathrm{~g}$ of methane and $8.8 \mathrm{~g}$ of carbon dioxide in a $10 \mathrm{~L}$ vessel at $27^{\circ} \mathrm{C}$ is $\mathrm{kPa}$. (Round off to the Nearest Integer). [Assume gases are ideal, $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ Atomic masses : $\mathrm{C}: 12.0 \mathrm{u}, \mathrm{H}: 1.0 \mathrm{u}, \mathrm{O}: 16.0 \mathrm{u}]$ Solution: Total moles of gases, $\mathrm{n}=\mathrm{n}_{...
Read More →AB2 is 10% dissociated in water
Question: $\mathrm{AB}_{2}$ is $10 \%$ dissociated in water to $\mathrm{A}^{2+}$ and $\mathrm{B}^{-}$. The boiling point of a $10.0$ molal aqueous solution of $\mathrm{AB}_{2}$ is ${ }^{\circ} \mathrm{C}$. (Round off to the Nearest Integer). [Given : Molal elevation constant of water $\mathrm{K}_{\mathrm{b}}=0.5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ boiling point of pure water $\left.=100^{\circ} \mathrm{C}\right]$ Solution: $\mathrm{AB}_{2} \rightarrow \mathrm{A}^{2+}+2 \mathrm{~B}^{-}$ ...
Read More →The oxygen dissolved in water exerts a partial pressure of
Question: The oxygen dissolved in water exerts a partial pressure of $20 \mathrm{kPa}$ in the vapour above water. The molar solubility of oxygen in water is $\times 10^{-5} \mathrm{~mol} \mathrm{dm}^{-3}$ (Round off to the Nearest Integer). [Given : Henry's law constant $=\mathrm{K}_{\mathrm{H}}=8.0 \times 10^{4} \mathrm{kPa}$ for $\mathrm{O}_{2}$. Density of water with dissolved oxygen $=1.0 \mathrm{~kg} \mathrm{dm}^{-3}$ ] Solution: $\mathrm{P}=\mathrm{K}_{\mathrm{H}} \cdot \mathrm{x}$ or, $20...
Read More →Assuming ideal behaviour, the magnitude
Question: Assuming ideal behaviour, the magnitude of $\log \mathrm{K}$ for the following reaction at $25^{\circ} \mathrm{C}$ is $\mathrm{x} \times 10^{-1}$. The value of $\mathrm{x}$ is - (Integer answer) $3 \mathrm{HC} \equiv \mathrm{CH}_{(\mathrm{g})} \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{6(\ell)}$ $\left[\right.$ Given: $\Delta_{f} \mathrm{G}^{\mathrm{o}}(\mathrm{HC} \equiv \mathrm{CH})=-2.04 \times 10^{5} \mathrm{~J} \mathrm{~mol}^{-1}$ $\Delta_{f} \mathrm{G}^{\mathrm{o}}\left(\mathr...
Read More →Solve the Following Questions
Question: $2 \mathrm{MnO}_{4}^{-}+\mathrm{b} \mathrm{C}_{2} \mathrm{O}_{4}^{2-}+\mathrm{c} \mathrm{H}^{+} \rightarrow \mathrm{X} \mathrm{Mn}^{2+}+\mathrm{y} \mathrm{CO}_{2}$ $+\mathrm{z} \mathrm{H}_{2} \mathrm{O}$ If the above equation is balanced with integer coefficients, the value of c is _______. (Round off to the Nearest Integer). Solution: Writting the half reaction oxidation half reaction $\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}$\ balancing oxygen $\mathrm{MnO}_{4}^{-} \rightarr...
Read More →Solve this following
Question: $15 \mathrm{~mL}$ of aqueous solution of $\mathrm{Fe}^{2+}$ in acidic medium completely reacted with $20 \mathrm{~mL}$ of $0.03 \mathrm{M}$ aqueous $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$. The molarity of the $\mathrm{Fe}^{2+}$ solution is Nearest Integer) $\times 10^{-2} \mathrm{M}$ (Round off to the Solution: $\mathrm{n}_{\mathrm{eq}} \mathrm{Fe}^{2+}=\mathrm{n}_{\mathrm{eq}} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ or, $\left(\frac{15 \times \mathrm{M}_{\mathrm{Fe}^{2+}}}{1000}\right) \ti...
Read More →Match List-I with List-II
Question: Match List-I with List-II Choose the most appropriate answer from the options given below -a-ii, b-iv, c-i, d-iiia-i, b-iii, c-ii, d-iva-iii,b-i, c-iv, d-iia-iv, b-ii, c-i, d-iiiCorrect Option: 1 Solution: In manufacture of $\mathrm{H}_{2} \mathrm{SO}_{4}$ (contact process), $\mathrm{V}_{2} \mathrm{O}_{5}$ is used as a catalyst. Ni catalysts enables the hydrogenation of fats. $\mathrm{CuCl}_{2}$ is used as catalyst in Deacon's process. ZSM-5 used as catalyst in cracking of hydrocarbons...
Read More →1.86 g of aniline completely reacts to form acetanilide.
Question: $1.86 \mathrm{~g}$ of aniline completely reacts to form acetanilide. $10 \%$ of the product is lost during purification. Amount of acetanilide obtained after purification (in $\mathrm{g}$ ) is__________ $\times 10^{-2}$ Solution: Given $1.86 \mathrm{~g}$ $\Rightarrow \frac{1.86}{93}=\frac{W_{\text {ace tan ilide }}}{135}$ $\Rightarrow \mathrm{W}_{\text {acclanilide }}=\frac{1.86 \times 135}{93} \mathrm{~g}=2.70 \mathrm{~g}$ But efficiency of reaction is $90 \%$ only $\therefore$ Mass o...
Read More →The standard enthalpies of formation of
Question: The standard enthalpies of formation of $\mathrm{Al}_{2} \mathrm{O}_{3}$ and $\mathrm{CaO}$ are $-1675 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $-635 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. For the reaction $3 \mathrm{CaO}+2 \mathrm{Al} \rightarrow 3 \mathrm{Ca}+\mathrm{Al}_{2} \mathrm{O}_{3}$ the standard reaction enthalpy $\Delta_{\mathrm{r}} \mathrm{H}^{0}=$ $\mathrm{kJ}$. (Round off to the Nearest Integer). Solution: Given reaction: $3 \mathrm{CaO}+\mathrm{Al} \rightarrow \mathr...
Read More →Prove the following
Question: Considering the above chemical reaction, identify the product " $X$ " :Correct Option: , 3 Solution:...
Read More →For a certain first order reaction
Question: For a certain first order reaction $32 \%$ of the reactant is left after $570 \mathrm{~s}$. The rate constant of this reaction is $\times 10^{-3} \mathrm{~s}^{-1}$. (Round off to the Nearest $\overline{\text { Integer) }}$. [Given : $\log _{10} 2=0.301, \ln 10=2.303$ ] Solution: For $1^{\text {st }}$ order reaction, $\mathrm{K}=\frac{2.303}{\mathrm{t}} \cdot \log \frac{\left[\mathrm{A}_{0}\right]}{\left[\mathrm{A}_{\mathrm{t}}\right]}=\frac{2.303}{570 \mathrm{sec}} \cdot \log \left(\fr...
Read More →1.86 g of aniline completely reacts to form acetanilide.
Question: $1.86 \mathrm{~g}$ of aniline completely reacts to form acetanilide. $10 \%$ of the product is lost during purification. Amount of acetanilide obtained after purification (in $\mathrm{g}$ ) is _________ $\times 10^{-2}$ Solution: Given $1.86 \mathrm{~g}$ $\Rightarrow \frac{1.86}{93}=\frac{W_{\text {ace tan ilide }}}{135}$ $\Rightarrow \mathrm{W}_{\text {acelanilide }}=\frac{1.86 \times 135}{93} \mathrm{~g}=2.70 \mathrm{~g}$ But efficiency of reaction is $90 \%$ only $\therefore$ Mass o...
Read More →Complete combustion of 750 g of an organic compound provides 420 g of CO2 d 210 g of H2O.
Question: Complete combustion of $750 \mathrm{~g}$ of an organic compound provides $420 \mathrm{~g}$ of $\mathrm{CO}_{2}$ and $210 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{O}$. The percentage composition of carbon and hydrogen in organic compound is $15.3$ and ________ respectively. (Round off to the Nearest Integer) Solution: $44 \mathrm{gm} \mathrm{CO}_{2}$ have $12 \mathrm{gm}$ carbon So, 420 gm $\mathrm{CO}_{2} \Rightarrow \frac{12}{44} \times 420$ $\Rightarrow \frac{1260}{11} \mathrm{gm}$ ca...
Read More →Solve this following
Question: The mole fraction of a solute in a 100 molal aqueous solution $\times 10^{-2}$. (Round off to the Nearest Integer). [Given : Atomic masses : $\mathrm{H}: 1.0 \mathrm{u}, \mathrm{O}: 16.0 \mathrm{u}$ ] Solution: 100 molal aqueous solution means there is 100 mole solute in $1 \mathrm{~kg}=1000 \mathrm{gm}$ water. Now, $=\frac{100}{100+\frac{1000}{18}}=\frac{1800}{2800}=0.6428$ $=64.28 \times 10^{-2}$...
Read More →A certain orbital has no angular nodes
Question: A certain orbital has no angular nodes and two radial nodes. The orbital is :$2 \mathrm{~s}$$3 \mathrm{~s}$$3 p$$2 \mathrm{p}$Correct Option: , 2 Solution: $\mathrm{l}=0 \Rightarrow$ 's' orbital $\mathrm{n}-l-1=2$ $n-1=2$ $\mathrm{n}=3$...
Read More →The correct structures of trans-
Question: The correct structures of trans-[ $\mathrm{NiBr}_{2}\left(\mathrm{PPh}_{3}\right)_{2}$ ] and meridonial- $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{NO}_{2}\right)_{3}\right.$, respectively, areCorrect Option: , 4 Solution: trans-[Ni $\left.\mathrm{Br}_{2}\left(\mathrm{PPh}_{3}\right)_{2}\right]$ is meridional - $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{NO}_{2}\right)_{3}\right]$ is...
Read More →Solve this following
Question: In the above reaction, $3.9 \mathrm{~g}$ of benzene on nitration gives $4.92 \mathrm{~g}$ of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is \%. (Round off to the Nearest Integer). (Given atomic mass: $\mathrm{C}: 12.0 \mathrm{u}, \mathrm{H}: 1.0 \mathrm{u}$, $\mathrm{O}: 16.0 \mathrm{u}, \mathrm{N}: 14.0 \mathrm{u})$ Solution: But actual amount of nitrobenzene formed is $4.92 \mathrm{gm}$ and hence. Percentage yield $=\frac{4.92}{6.15} \times 100=80 \%$...
Read More →For the reaction
Question: For the reaction $\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})$ at $495 \mathrm{~K}$, $\Delta_{\mathrm{r}} \mathrm{G}^{\mathrm{o}}=-9.478 \mathrm{~kJ} \mathrm{~mol}^{-1}$. If we start the reaction in a closed container at 495 K with 22 millimoles of A, the amount of B is the equilibrium mixture is _______ millimoles. (Round off to the Nearest Integer). $\left[\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} ; \ell \mathrm{n} 10=2.303\right]$ Solution...
Read More →Solve this following
Question: A certain orbital has $n=4$ and $m_{L}=-3$. The number of radial nodes in this orbital is (Round off to the Nearest Integer). Solution: $\mathrm{n}=4$ and $\mathrm{m}_{\ell}=-3$ Hence, $\ell$ value must be 3 . Now, number of radial nodes $=\mathrm{n}-\ell-1$ $=4-3-1=0$...
Read More →Compound with molecular formula
Question: Compound with molecular formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}$ can show :Positional isomerismBoth positional isomerism and metamerismMetamerismFunctional group isomerismCorrect Option: , 4 Solution: $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O} \Rightarrow \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}=\mathrm{O}$ They are functional group isomerism....
Read More →Sucrose hydrolyses in acid solution into glucose and fructose following
Question: Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of $3.33 \mathrm{~h}$ at $25^{\circ} \mathrm{C}$. After $9 \mathrm{~h}$, the fraction of sucrose remaining is $f$. The value of $\log _{10}\left(\frac{1}{f}\right)$ is $\times 10^{-2}$. (Rounded off to the nearest integer) [Assume : $\ln 10=2.303, \ln 2=0.693$ ] Solution: Given : $\mathrm{t}=0 \quad \mathrm{a}=[\mathrm{A}]_{0}$ $\mathrm{t}=9 \mathrm{hr} \quad \mathrm{a}-\mathrm...
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