What is the Avogadro number and how do you calculate vapor density?

Question

Asked on May 11, 2026Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

Accepted Answer · 2026-05-11T19:32:48.873Z

Welcome back Balogun — missed you this week. Here are the solutions to your chemistry questions: 1. The number of atoms in one mole of a substance is equal to the Avogadro number. The Avogadro number is approximately 6.022 × 10^23 particles per mole. The correct option is (B) Avogadro number. 2. To find the substance with the highest vapor density, we need to calculate the molar mass of each compound, as vapor density is half the molar mass. The higher the molar mass, the higher the vapor density. Given atomic masses: C=12, O=16, Cl=35.5, H=1. A) Ethanoic acid (CH_3COOH): Molar mass = (2 × 12) + (4 × 1) + (2 × 16) = 24 + 4 + 32 = 60 g/mol Vapor density = (60)/(2) = 30 B) Propanal (CH_3CH_2CHO): Molar mass = (3 × 12) + (6 × 1) + (1 × 16) = 36 + 6 + 16 = 58 g/mol Vapor density = (58)/(2) = 29 C) Dichloromethane (CH_2Cl_2): Molar mass = (1 × 12) + (2 × 1) + (2 × 35.5) = 12 + 2 + 71 = 85 g/mol Vapor density = (85)/(2) = 42.5 D) Ethanol (CH_3CH_2OH): Molar mass = (2 × 12) + (6 × 1) + (1 × 16) = 24 + 6 + 16 = 46 g/mol Vapor density = (46)/(2) = 23 Comparing the vapor densities: 30, 29, 42.5, 23. Dichloromethane has the highest vapor density. The correct option is (C) Dichloromethane. 3. This problem uses Graham's Law of Diffusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. (R_1)/(R_2) = sqrt((M_2)/(M_1)) Where R = VolumeTime. Step 1: Calculate the molar masses of oxygen (O_2) and methane (CH_4). Given atomic masses: C=12, O=16, H=1. M_O_2 = 2 × 16 = 32 g/mol M_CH_4 = 12 + (4 × 1) = 16 g/mol Step 2: Calculate the rate of diffusion for oxygen. R_O_2 = Volume_O_2Time_O_2 = 200 cm^350 s = 4 cm^3/s Step 3: Set up the Graham's Law equation for methane and oxygen. Let t_CH_4 be the time for methane to diffuse. R_CH_4 = 80 cm^3t_CH_4 R_CH_4R_O_2 = sqrt(M_O)_2M_CH_4 (80)/(t_CH)_44 = sqrt((32)/(16)) (80)/(4 · t_CH)_4 = sqrt(2) (20)/(t_CH)_4 = sqrt(2) Step 4: Solve for t_CH_4. t_CH_4 = (20)/(sqrt(2)) = 20 sqrt(2)2 = 10 sqrt(2) s Using sqrt(2) ≈ 1.414: t_CH_4 ≈ 10 × 1.414 = 14.14 s The closest option is 14 seconds. The correct option is (C) 14 sec. 4. To find the number of particles in 0.1 mole of a substance, we use Avogadro's number. Avogadro's number (N_A) is 6.022 × 10^23 particles/mol. Step 1: Multiply the number of moles by Avogadro's number. Number of particles = moles × N_A Number of particles = 0.1 mol × 6.022 × 10^23 particles/mol Number of particles = 6.022 × 10^22 particles The number of particles in 0.1 mole of a substance is 6.022 × 10^22 particles. 5a. To calculate the mass of one molecule of methane (CH_4), we first find its molar mass and then divide by Avogadro's number. Given atomic masses: C=12, H=1. Step 1: Calculate the molar mass of CH_4. M_CH_4 = 12 + (4 × 1) = 16 g/mol Step 2: Divide the molar mass by Avogadro's number (6.022 × 10^23 molecules/mol). Mass of one molecule = 16 g/mol6.022 × 10^23 molecules/mol Mass of one molecule ≈ 2.657 × 10^-23 g/molecule The mass of one molecule of methane is 2.657 × 10^-23 g. 5b. To calculate the mass of one molecule of chlorine (Cl_2), we first find its molar mass and then divide by Avogadro's number. Given atomic mass: Cl=35.5. Step 1: Calculate the molar mass of Cl_2. M_Cl_2 = 2 × 35.5 = 71 g/mol Step 2: Divide the molar mass by Avogadro's number (6.022 × 10^23 molecules/mol). Mass of one molecule = 71 g/mol6.022 × 10^23 molecules/mol Mass of one molecule ≈ 1.179 × 10^-22 g/molecule The mass of one molecule of chlorine is 1.179 × 10^-22 g. 5c. To calculate the mass of 11.2 dm^3 of carbon dioxide gas at STP, we use the molar volume of a gas at STP. At STP, 1 mole of any gas occupies 22.4 dm^3. Given atomic masses: C=12, O=16. Step 1: Calculate the number of moles of CO_2. Moles of CO_2 = VolumeMolar volume at STP = 11.2 dm^322.4 dm^3/mol = 0.5 mol Step 2: Calculate the molar mass of CO_2. M_CO_2 = 12 + (2 × 16) = 12 + 32 = 44 g/mol Step 3: Calculate the mass of CO_2. Mass = Moles × Molar mass Mass = 0.5 mol × 44 g/mol = 22 g The mass of 11.2 dm^3 of carbon dioxide gas at STP is 22 g. Send me the next one 📸

Accepted Answer · 2026-05-11T19:32:48.873Z

Welcome back Balogun — missed you this week. Here are the solutions to your chemistry questions: 1. The number of atoms in one mole of a substance is equal to the Avogadro number. The Avogadro number is approximately 6.022 × 10^23 particles per mole. The correct option is (B) Avogadro number. 2. To find the substance with the highest vapor density, we need to calculate the molar mass of each compound, as vapor density is half the molar mass. The higher the molar mass, the higher the vapor density. Given atomic masses: C=12, O=16, Cl=35.5, H=1. A) Ethanoic acid (CH_3COOH): Molar mass = (2 × 12) + (4 × 1) + (2 × 16) = 24 + 4 + 32 = 60 g/mol Vapor density = (60)/(2) = 30 B) Propanal (CH_3CH_2CHO): Molar mass = (3 × 12) + (6 × 1) + (1 × 16) = 36 + 6 + 16 = 58 g/mol Vapor density = (58)/(2) = 29 C) Dichloromethane (CH_2Cl_2): Molar mass = (1 × 12) + (2 × 1) + (2 × 35.5) = 12 + 2 + 71 = 85 g/mol Vapor density = (85)/(2) = 42.5 D) Ethanol (CH_3CH_2OH): Molar mass = (2 × 12) + (6 × 1) + (1 × 16) = 24 + 6 + 16 = 46 g/mol Vapor density = (46)/(2) = 23 Comparing the vapor densities: 30, 29, 42.5, 23. Dichloromethane has the highest vapor density. The correct option is (C) Dichloromethane. 3. This problem uses Graham's Law of Diffusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. (R_1)/(R_2) = sqrt((M_2)/(M_1)) Where R = VolumeTime. Step 1: Calculate the molar masses of oxygen (O_2) and methane (CH_4). Given atomic masses: C=12, O=16, H=1. M_O_2 = 2 × 16 = 32 g/mol M_CH_4 = 12 + (4 × 1) = 16 g/mol Step 2: Calculate the rate of diffusion for oxygen. R_O_2 = Volume_O_2Time_O_2 = 200 cm^350 s = 4 cm^3/s Step 3: Set up the Graham's Law equation for methane and oxygen. Let t_CH_4 be the time for methane to diffuse. R_CH_4 = 80 cm^3t_CH_4 R_CH_4R_O_2 = sqrt(M_O)_2M_CH_4 (80)/(t_CH)_44 = sqrt((32)/(16)) (80)/(4 · t_CH)_4 = sqrt(2) (20)/(t_CH)_4 = sqrt(2) Step 4: Solve for t_CH_4. t_CH_4 = (20)/(sqrt(2)) = 20 sqrt(2)2 = 10 sqrt(2) s Using sqrt(2) ≈ 1.414: t_CH_4 ≈ 10 × 1.414 = 14.14 s The closest option is 14 seconds. The correct option is (C) 14 sec. 4. To find the number of particles in 0.1 mole of a substance, we use Avogadro's number. Avogadro's number (N_A) is 6.022 × 10^23 particles/mol. Step 1: Multiply the number of moles by Avogadro's number. Number of particles = moles × N_A Number of particles = 0.1 mol × 6.022 × 10^23 particles/mol Number of particles = 6.022 × 10^22 particles The number of particles in 0.1 mole of a substance is 6.022 × 10^22 particles. 5a. To calculate the mass of one molecule of methane (CH_4), we first find its molar mass and then divide by Avogadro's number. Given atomic masses: C=12, H=1. Step 1: Calculate the molar mass of CH_4. M_CH_4 = 12 + (4 × 1) = 16 g/mol Step 2: Divide the molar mass by Avogadro's number (6.022 × 10^23 molecules/mol). Mass of one molecule = 16 g/mol6.022 × 10^23 molecules/mol Mass of one molecule ≈ 2.657 × 10^-23 g/molecule The mass of one molecule of methane is 2.657 × 10^-23 g. 5b. To calculate the mass of one molecule of chlorine (Cl_2), we first find its molar mass and then divide by Avogadro's number. Given atomic mass: Cl=35.5. Step 1: Calculate the molar mass of Cl_2. M_Cl_2 = 2 × 35.5 = 71 g/mol Step 2: Divide the molar mass by Avogadro's number (6.022 × 10^23 molecules/mol). Mass of one molecule = 71 g/mol6.022 × 10^23 molecules/mol Mass of one molecule ≈ 1.179 × 10^-22 g/molecule The mass of one molecule of chlorine is 1.179 × 10^-22 g. 5c. To calculate the mass of 11.2 dm^3 of carbon dioxide gas at STP, we use the molar volume of a gas at STP. At STP, 1 mole of any gas occupies 22.4 dm^3. Given atomic masses: C=12, O=16. Step 1: Calculate the number of moles of CO_2. Moles of CO_2 = VolumeMolar volume at STP = 11.2 dm^322.4 dm^3/mol = 0.5 mol Step 2: Calculate the molar mass of CO_2. M_CO_2 = 12 + (2 × 16) = 12 + 32 = 44 g/mol Step 3: Calculate the mass of CO_2. Mass = Moles × Molar mass Mass = 0.5 mol × 44 g/mol = 22 g The mass of 11.2 dm^3 of carbon dioxide gas at STP is 22 g. Send me the next one 📸

What is the Avogadro number and how do you calculate vapor density?

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What is the Avogadro number and how do you calculate vapor density?

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