States of matter practice · Mixed difficulty

The correct answer is Half that of the $\mathrm{S}{\mathrm{O}_2}$. At the same temperature and pressure, the mass of a gas is proportional to its molar mass. Since the molar mass of SO2 is twice that of CO2, the weight of SO2 will be twice that of CO2.

Option C (${\mathrm{N}_2}\mathrm{O},\mathrm{C}{\mathrm{O}_2}$) is correct because these two gases have approximately the same molar mass, leading to similar diffusion rates.

Option B (Density of air is low at the altitudes) is correct because at higher altitudes, the atmospheric pressure is lower, resulting in a lower density of air and consequently, a lower partial pressure of oxygen.

Option B ($1:1:1:1$) is correct because, under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules (Avogadro's Law).

Option D ($N{o_2} = {N_{{H_2}}}$) is correct because, according to Avogadro's law, equal volumes of all gases at the same temperature and pressure contain the same number of molecules.

Option D is correct because Dalton's law is applicable to non-reacting gases, and CO and H2 do not readily react under normal conditions.

Option C ($T\,\,({H_2})\,\, < \,\,T\,\,({N_2})$) is correct because the RMS velocity is inversely proportional to the square root of the molar mass and directly proportional to the square root of the temperature.

Option B (If both A and R are correct, but R is not the correct explanation of A) is correct because both the assertion and reason are individually correct statements, but the reason does not explain why the assertion is true.

Option C (If assertion is true but reason is false) is correct because the assertion is true (molecular mass is related to vapor density), but the reason is false (vapor density is half the molecular mass).

Option A (${\left[ {\frac{{{n_1}C_1^2 + {n_2}C_2^2 + {n_3}C_3^2 + \ldots }}{{{n_1} + {n_2} + {n_3} + \ldots }}} \right]^{1/2}}$) is correct because it represents the formula for root mean square speed when given the speeds of individual molecules.

Option D ($NO,\,{C_2}{H_6}$) is correct because these gases have approximately the same molar mass, leading to similar diffusion rates according to Graham's law.

Option A is correct because the single capillary method is a standard technique used to measure the surface tension of liquids based on capillary rise.

Option A is correct because the van der Waals constant 'b' represents the volume excluded by the gas molecules themselves, which is related to the volume occupied by the molecules.

Option D is correct because at high temperature and low pressure, the volume correction (b) and the pressure correction (a/V²) in the van der Waals equation become negligible, reducing it to the ideal gas law.

Option B ($Z = \frac{{pV}}{{nRT}}$) is correct because it is the definition of the compressibility factor, which quantifies the deviation from ideal gas behavior.

Option B ($33.3\,\mathrm{mL}$) is correct because it is calculated using Charles's Law to find the change in volume due to the temperature increase.

Option A (Zero degree centigrade) is correct because the volume expansion coefficient is defined with respect to the volume at 0°C.

Option B (React with each other) is correct because Gay-Lussac's law of combining volumes applies only to gaseous reactants and products, implying a chemical reaction between the gases.

Option D is correct because both the assertion and the reason are false. 22.4 L of N2 and 5.6 L of O2 at STP do not contain an equal number of molecules, and under similar conditions, gases contain equal numbers of *molecules*, not necessarily atoms.

Option C ($\frac{{1.14}}{1}$) is correct because the average velocity is proportional to the square root of the absolute temperature.

Option D ($COC{l_2}$) is correct because phosgene hydrolyzes in the lungs to form HCl, leading to suffocation.

Option B is correct because a gas can be liquefied only below its critical temperature and above its critical pressure.

Option C ($533\mathrm{ }K$) is correct because the total kinetic energy is proportional to the number of moles and temperature, and equating the kinetic energies allows solving for the temperature of helium.

Option D (All of these) is correct because Van der Waals forces are influenced by the number of electrons, the charge on the nucleus, and the radius of the atoms, as these factors affect the polarizability and interactions between molecules.

Option B ($Z > 1$ and repulsive forces are dominant) is correct because a molar volume greater than that of an ideal gas implies $Z > 1$, indicating repulsive forces dominate.

Option B (By heating the gas, the pressure is made four times) is correct because the root mean square speed is proportional to the square root of the temperature, and pressure is directly proportional to temperature at constant volume.

Option A (Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I) is correct because the activation energy is required for movement, and increasing temperature provides this energy, thus decreasing viscosity.

Option B (480 K) is correct because the rms speed is proportional to the square root of temperature divided by molar mass, and setting the rms speeds equal allows solving for the temperature of O2.

Option C (22.4 L of gas) is correct because Avogadro's number represents the number of molecules in one mole of any substance, and one mole of any gas occupies 22.4 liters at NTP.

Option B is correct because 1 poise is defined as 1 dyne second per centimeter squared.