States of Matter (Gases and Liquids)NEET Chemistry · Class 11 · NCERT Chapter 13

Solution

Using the ideal gas law, $PV=nRT$, and given equal masses, the volume of a gas is inversely proportional to its molar mass. The molar masses are $2\text{g/mol}$ for ${\text{H}}_2$, $32\text{g/mol}$ for ${\text{O}}_2$, and $16\text{g/mol}$ for methane. Therefore, the volume ratio is $\frac{1}{2}:\frac{1}{32}:\frac{1}{16}=16:1:2$, so option (c) is correct.

Solution

Real gases show ideal gas behaviour at high temperatures and low pressures.

Solution

n1 t1 n2 t2 = √ M2 M1 ⇒ n2 = 1 2 ,n1 = n′ ⇒ 2t2n′ t1x1 = √ M2 M1 = √ 2 32 = √ 1 16 = 1 4 Assuming t2 = t1 2n′t2 t1 = 1 4 ∴ n′ = 1 8

Solution

The gas with the highest van der Waals constant $a$ is most easily liquefied due to stronger intermolecular forces. Among the given values, ${\text{NH}}_3$ has the highest $a$ value of 4.17, so option (c) is correct.

Solution

The correction factor 'a' in the van der Waals equation accounts for the intermolecular forces of attraction between gas molecules. This correction is necessary to adjust for the deviation from ideal gas behavior, so option (d) is correct.

Solution

Tribromooctaoxide, ${\text{Br}}_3{\text{O}}_8$, has a structure where each bromine atom is bonded to two oxygen atoms, forming a chain. The correct structure is option (b), as it aligns with the molecular formula and bonding pattern described in NCERT XI chapter States of Matter.

Solution

For a body-centered cubic (bcc) structure, the relationship between the atomic radius $r$ and the cell edge $a$ is given by $r=\frac{\sqrt{3}}{4}a$. Substituting $a=288\text{pm}$, we get $r=\frac{\sqrt{3}}{4}\times 288\text{pm}=\frac{288\text{pm}}{4}\times \sqrt{3}$. Therefore, the correct option is (a).

Solution

Calculate the number of moles of each gas: $n_{{\text{N}}_2}=\frac{7}{28}=0.25\text{mol}$, $n_{\text{Ar}}=\frac{8}{40}=0.2\text{mol}$. The total number of moles is $0.25+0.2=0.45\text{mol}$. The partial pressure of ${\text{N}}_2$ is given by $P_{{\text{N}}_2}=\frac{n_{{\text{N}}_2}}{n_{\text{total}}}\times P_{\text{total}}=\frac{0.25}{0.45}\times 27=15\text{bar}$, so option (c) is correct.

Solution

CO${}_2$ (g) is indeed used as a refrigerant for ice-cream and frozen food, and ZSM-5 is used to convert alcohols into gasoline. However, the structure of C${}_{60}$ contains twelve five-carbon rings and twenty hexagonal rings, not twelve six-carbon rings and twenty five-carbon rings. CO is colorless and odorless, but this is not a relevant statement for the correct option. Therefore, option (b) is correct.

Solution

Boyle's law states that pressure is inversely proportional to volume at constant temperature, so the graph of pressure vs. volume is a hyperbola. The correct representation is a set of hyperbolic curves, one for each temperature, so option (a) is correct.

Solution

Use the ideal gas law $PV=nRT$. For ${\text{O}}_2$, $n=\frac{4\text{g}}{32\text{g/mol}}=0.125\text{mol}$; for ${\text{H}}_2$, $n=\frac{2\text{g}}{2\text{g/mol}}=1\text{mol}$. Total moles $n=0.125+1=1.125\text{mol}$. Substituting, $P\times 1\text{L}=1.125\text{mol}\times 0.082{\text{L atm mol}}^{-1}{\text{K}}^{-1}\times 273\text{K}\Rightarrow P=2.602\text{atm}$, so option (b) is correct.

Solution

• Dalton’s law of partial pressure states that the total pressure by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases. • pTotal = p1 + p2 + p3 • Also, pi = χip ; where pi and χi are the partial pressure and mole fraction of i th gas respectively and p is the total pressure. • pTotal = p1 + p2 + p3 = 1 2 3 RT RT RTn + n + nVVV = ( )1 2 3 RTn + n + n V - 29 - NEET (UG)-2022 (Code-Q1)

Solution

We know for ideal gas PV = nRT RTPn V= 64 0.0831 300P 32 10 ×=× P = 4.9 bar Pressure of O2 gas inside the flask = 4.9 bar

Solution

Helium is used to dilute oxygen in diving apparatus to reduce the risk of oxygen toxicity at high pressures. However, helium has low solubility in blood, not high solubility in ${\text{O}}_2$, so the reason R is false. Option (d) is correct.

Solution

Intermolecular forces include dipole-dipole forces, dipole-induced dipole forces, hydrogen bonding, and dispersion forces. Covalent bonding is an intramolecular force, not an intermolecular one, so option (d) is correct.

Solution

Boyle's Law states that pressure is inversely proportional to volume at constant temperature, so a graph of $P$ vs $\frac{1}{V}$ should be a straight line. The correct representation is a graph of $P$ vs $\frac{1}{V}$ with three lines labeled $T_1$, $T_2$, $T_3$ where $T_3>T_2>T_1$, so option (c) is correct.

Solution

Pumice stone is a solid sol, a type of colloid where solid particles are dispersed in a solid medium. NCERT XII chapter States of Matter classifies pumice stone as a solid sol, so option (d) is correct.

Solution

(1) 4 mol of He = 4 NA He atoms (2) 4 u of He = 4u 4u = 1 He atom (3) 4 g of Helium = 4g 4g mole = 1 mole = NA He atom (4) 2.2710982 of He at STP = 2.271 22.710982 mole = 0.1 mole = 0.1 NA He atom

Solution

Value of Henry’s law constant 1 Solubility of gas∝ Higher the value of KH at a given pressure, lower is the solubility of the gas in the liquid. KH value of gases (given) : A > C > B ∴ Order of solubility of gases in water : B > C > A

Solution

Π = CRT Slope = RT 25.73 = 0.083 × T 25.73T 309.47 310 K0.083= = ≈ ∴ Temperature in °C = 310 – 273 = 37°C