Some Basic Concepts of Chemistry JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Some Basic Concepts of Chemistry, free to read — no sign-in needed. The full chapter has 21 questions; sign in to attempt the remaining 16 in the exam simulator.

1. Q1JEE Advanced Adv 2024 (Paper 2)
   To form a complete monolayer of acetic acid on $1 \mathrm{~g}$ of charcoal, $100 \mathrm{~mL}$ of $0.5 \mathrm{M}$ acetic acid was used. Some of the acetic acid remained unadsorbed. To neutralize the unadsorbed acetic acid, 40 $\mathrm{mL}$ of $1 \mathrm{M} \mathrm{NaOH}$ solution was required. If each molecule of acetic acid occupies $\mathbf{P} \times 10^{-23} \mathrm{~m}^2$ surface area on charcoal, the value of $\mathbf{P}$ is _______ [Use given data: Surface area of charcoal $=1.5 \times 10^2 \mathrm{~m}^2 \mathrm{~g}^{-1} ;$ Avogadro's number $\left(\mathrm{N}_{\mathrm{A}}\right)=6.0 \times 10^{23}$ $\left.\mathrm{mol}^{-1}\right]$
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   Answer: 2500

   $\begin{aligned} & \text { Millimole of acid taken }=100 \times 0.5=50 \\ & \text { Millimole of } \mathrm{NaOH} \text { used }=40 \times 1=40 \\ & \text { Millimole of acid adsorbed }=50-40=10 \\ & \text { Molecules of acid adsorbed }=10 \times 10^{-3} \times 6 \times 10^{23}=6 \times 10^{21} \\ & \text { Surface area occupied per molecule }=\frac{1.5 \times 10^2}{6 \times 10^{21}}=0.25 \times 10^{-19}=2500 \times 10^{-23}\end{aligned}$
2. Q2JEE Advanced Adv 2022 (Paper 2)
   To check the principle of multiple proportions, a series of pure binary compounds $\left({P}_{m}{Q}_{n}\right)$ were analysed and their composition is tabulated below. The correct option(s) is(are) Compound Weight $$ of $P$ Weight $$ of $Q$ $1$ $50$ $50$ $2$ $44.4$ $55.6$ $3$ $40$ $60$
   1. A.If empirical formula of compound $3$ is ${P}_{3}{Q}_{4}$, then the empirical formula of compound $2$ is ${P}_{3}{Q}_{5}$.
   2. B.If empirical formula of compound $3$ is ${P}_{3}{Q}_{2}$ and atomic weight of element $P$ is $20$ , then the atomic weight of $Q$ is $45.$
   3. C.If empirical formula of compound $2$ is $PQ$, then the empirical formula of the compound $1$ is ${P}_{5}{Q}_{4}$.
   4. D.If atomic weight of $P$ and $Q$ are $70$ and $35$, respectively, then the empirical formula of compound $1$ is ${P}_{2}Q$.
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   Answer: B,C

   (A) If empirical formula of $3$ is ${P}_{3}{Q}_{4}$, then molecular formula is ${\left({P}_{3}{Q}_{4}\right)}_{n}$ $\Rightarrow \dfrac{3{M}_{P}}{3{M}_{P}+4{M}_{Q}}=\dfrac{40}{100}=\dfrac{2}{5}$ $15{M}_{p}=6{M}_{p}+8{M}_{Q}$ $9{M}_{P}=8{M}_{Q}$ $∴$ For ${P}_{3}{Q}_{5},$ of $P=\left(\dfrac{3{M}_{P}}{3{M}_{P}+5{M}_{Q}}\right)\times 100=\dfrac{8{M}_{Q}}{23{M}_{Q}}\times 100=34.78$ (B) If empirical formula of compound $3$ is ${P}_{3}{Q}_{2}$, we have $\dfrac{3{M}_{P}}{3{M}_{P}+2{M}_{Q}}=\dfrac{2}{5}$ $15{M}_{P}=6{M}_{P}+4{M}_{Q}$ $9Mp=4{M}_{Q}$ If ${M}_{p}=20$ ${M}_{Q}=\dfrac{180}{4}=45$ (C) If empirical formula of $2$ is $PQ$ $∴\dfrac{{M}_{P}}{{M}_{P}+{M}_{Q}}=\dfrac{4}{9}=\dfrac{44.44}{100}$ $9Mp=5Mp+4MQ$ $5{M}_{P}=4{M}_{Q}$ If empirical formula is assumed as ${P}_{5}{Q}_{4}$, $$ of $P=\left(\dfrac{5{M}_{P}}{5{M}_{P}+4{M}_{Q}}\right)\times 100=50$ Hence ${P}_{5}{Q}_{4}$ is the empirical formula of compound (D) If empirical formula of $I$ is ${P}_{2}Q$, we have $$ $4{M}_{p}=2{M}_{p}+{M}_{Q}$ $2{M}_{p}={M}_{Q}$ Hence, atomic weight of $P$ and $Q$ cannot be $70$ and $35$ respectively.
3. Q3JEE Advanced Adv 2022 (Paper 1)
   The treatment of an aqueous solution of $3.74g$ of $Cu{\left({NO}_{3}\right)}_{2}$ with excess $KI$ results in a brown solution along with the formation of a precipitate. Passing ${H}_{2}S$ through this brown solution gives another precipitate $X$. The amount of $X$ (in $g$) is____[Given: Atomic mass of $H=1,N=14,O=16,S=32,K=39,Cu=63,I=127$]
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   Answer: 0.32

   Number of moles of $Cu{\left({NO}_{3}\right)}_{2}=\dfrac{3.74}{187}=0.02$ $2Cu{\left({NO}_{3}\right)}_{2}+4KI\rightarrow {Cu}_{2}{I}_{2}↓+{I}_{2}+4{KNO}_{3}$ Number of moles of ${Cu}_{2}{l}_{2}$ precipitated $=0.01$ ${I}_{2}\left(brownsolution\right)+{H}_{2}S\rightarrow S↓+2HI$ Number of moles of $S$ precipitated $=0.01$ Mass of $S$ precipitates $=\left(0.01\times 32\right)g=0.32g$
4. Q4JEE Advanced Adv 2020 (Paper 1)
   Aluminium reacts with sulfuric acid to form aluminium sulfate and hydrogen. What is the volume of hydrogen gas in liters $(L)$ produced at $300K$ and $1.0$ atm pressure, when $5.4g$ of aluminium and $50.0mL$ of $5.0M$ sulfuric acid are combined for the reaction? (Use molar mass of aluminium as $27.0g{mol}^{-1}$, $R=0.082atmL{mol}^{-1}{K}^{-1}$)
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   Answer: 6.15

   $2AI+3{H}_{2}{SO}_{4}⟶{AI}_{2}{\left({SO}_{4}\right)}_{3}+3{H}_{2}$ $\dfrac{5.7}{27}$ $\dfrac{50\times 5}{1000}$ $=0.2mole$ $=0.25 mole_{(Limiting reagent)}$ $∴$ $0.25 mole{\text{H}}_{2}$ is formed $PV=nRT$ $V=\dfrac{nRT}{P}=\dfrac{0.25\times 0.082\times 300}{1atm}$ $=6.15$ litre
5. Q5JEE Advanced Adv 2019 (Paper 2)
   The mole fraction of urea in an aqueous urea solution containing $900g$ of water is $0.05$ . If the density of the solution is $1.2g{cm}^{-3}$ , the molarity of urea solution is _______ (Given data: Molar masses of urea and water are $60g{mol}^{-1}$ and $18g{mol}^{-1}$ , respectively)
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   Answer: 2.98

   Mole fraction $\Rightarrow 0.05$ Number of moles of water $\Rightarrow \dfrac{900}{18}=50$ $∴\dfrac{{n}_{urea}}{{n}_{urea}+50}=0.05$ ${n}_{urea}=2.63$ Mass of urea $=\left({W}_{urea}\right)=2.63\times 60\Rightarrow 157.8 g$ Total mass of solution $\Rightarrow 157.8+900\Rightarrow 1057.8 g$ Volume of solution $\Rightarrow \dfrac{1057.8}{1.2}ml\Rightarrow 881.5 ml$ Molarity $\Rightarrow \dfrac{no. of moles of solutions}{volume of solution in L}\Rightarrow \dfrac{2.63}{0.8815}\Rightarrow 2.98 M$