Questions & Answers: "Some Basic Concepts of Chemistry"

The number of atoms is calculated by (Mass / Atomic Mass) × Avogadro's Number. Since the mass (1.0 g) is constant for all, the sample with the lowest atomic mass will have the highest number of atoms. Atomic masses are: Au ≈ 197, Na = 23, Li = 7, Cl = 35.5 (Cl2 = 71). Lithium (Li) has the smallest atomic mass, hence 1/7 moles contains more atoms than 1/197, 1/23, or 1/71.

Step 1: Calculate moles: H=4.07/1=4.07, C=24.27/12=2.02, Cl=71.65/35.5=2.02. Ratio C:H:Cl = 1:2:1. Empirical formula is CH2Cl (mass = 12+2+35.5 = 49.5). Step 2: n = Molar Mass / Empirical Formula Mass = 98.96 / 49.5 ≈ 2. Molecular formula = 2 × (CH2Cl) = C2H4Cl2. CH2Cl is the empirical formula, not molecular. CHCl3 and C2H2Cl4 do not match the percentage composition.

According to significant figure rules: 1) Leading zeros (zeros to the left of the first non-zero digit) are not significant. 2) Trailing zeros to the right of the decimal point are significant. Thus, in 0.00250, '2', '5', and the final '0' are significant. The three zeros after the decimal but before '2' are placeholders.

Molarity (M) = Moles of solute / Volume of solution in Liters. Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol. Moles of NaOH = 4.0 / 40 = 0.1 mol. Volume = 250 mL = 0.250 L. Molarity = 0.1 / 0.250 = 0.4 M. Option 0.1 M is the number of moles, not molarity. 1.0 M and 0.25 M are incorrect calculations.

Molarity depends on the volume of the solution, and volume changes with temperature due to thermal expansion/contraction. Molality, mole fraction, and mass percentage are based on mass, which remains constant regardless of temperature changes. Therefore, Molarity is temperature-dependent while the others are not.

According to the balanced equation, 1 mole of N2 requires 3 moles of H2. For 2.0 moles of N2, we need 2 × 3 = 6.0 moles of H2. Since we only have 3.0 moles of H2, H2 will be consumed first and is the limiting reagent. N2 is in excess. NH3 is a product, not a reagent.

Using the dilution formula M1V1 = M2V2: (5 M) × (500 mL) = M2 × (1500 mL). M2 = (5 × 500) / 1500 = 2500 / 1500 = 1.666... M. Option 1.5 M is a rounding error, 2.5 M would occur if volume doubled, and 0.5 M is a calculation error.

The Law of Multiple Proportions states that if two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers. Carbon and Oxygen form CO and CO2. In CO, 12g C reacts with 16g O. In CO2, 12g C reacts with 32g O. The ratio of oxygen (16:32) is 1:2. NaCl/NaBr involve different elements (Cl vs Br). H2O/D2O involve isotopes.

The mass of one mole of C-12 atoms is 12 g. One mole contains 6.022 × 10^23 atoms. Therefore, the mass of one atom = 12 / (6.022 × 10^23) ≈ 1.9926 × 10^-23 g. 12g is the molar mass. 6.022 x 10^23 is Avogadro's number itself. 1.66 x 10^-24 g is the mass of 1 amu.

The conversion formula is °F = (9/5 × °C) + 32. Plucking in the value: °F = (9/5 × 25) + 32 = (9 × 5) + 32 = 45 + 32 = 77°F. 32°F is the freezing point of water (0°C). 298 is the value in Kelvin (273 + 25), not Fahrenheit.

Molarity is defined as the number of moles of solute dissolved in one liter of solution (M = n/V). Since the volume (V) of a liquid solution expands or contracts with changes in temperature, the molarity changes accordingly. In contrast, molality, mole fraction, and mass percentage are based on the mass of the solvent or solution. Since mass is independent of temperature, these concentration terms remain constant when temperature changes.

Step 1: Find moles of each element in 100g. C: 40/12 = 3.33 mol; H: 6.67/1 = 6.67 mol; O: (100 - 40 - 6.67) = 53.33/16 = 3.33 mol. Step 2: Simplest molar ratio is C:H:O = 3.33/3.33 : 6.67/3.33 : 3.33/3.33 = 1:2:1. Thus, the empirical formula is CH2O. Step 3: Empirical formula mass = 12 + 2(1) + 16 = 30 g/mol. Step 4: n = (Molar Mass / Empirical Mass) = 180/30 = 6. Step 5: Molecular formula = (CH2O) × 6 = C6H12O6. CH2O is the empirical formula, not the molecular formula.

The balanced chemical equation is H2(g) + Cl2(g) → 2HCl(g). At STP, 1 mole of any gas occupies 22.4 L. Here, moles of H2 = 22.4/22.4 = 1.0 mol, and moles of Cl2 = 11.2/22.4 = 0.5 mol. According to the stoichiometry, 1 mole of H2 requires 1 mole of Cl2. Since we only have 0.5 mol of Cl2, Cl2 is the limiting reagent. The amount of product is determined by the limiting reagent: 1 mole of Cl2 produces 2 moles of HCl, so 0.5 mole of Cl2 will produce (0.5 × 2) = 1.0 mole of HCl.

To solve for molality ($m$), we follow these steps:

1. Definitions: Molarity ($M$) is moles of solute per liter of solution. Molality ($m$) is moles of solute per kilogram of solvent.

2. Assumption: Assume we have $1\text{ L}$ ($1000\text{ mL}$) of the solution.

   • Moles of solute ($NaOH$) = $2.0\text{ moles}$ (since it is a $2.0\text{ M}$ solution).
   • Mass of solute = $\text{moles} \times \text{molar mass} = 2.0\text{ mol} \times 40\text{ g/mol} = 80\text{ g}$.

3. Mass of Solution: Using density ($d = 1.28\text{ g/mL}$),

   • Mass of $1000\text{ mL}$ solution = $1000\text{ mL} \times 1.28\text{ g/mL} = 1280\text{ g}$.

4. Mass of Solvent:

   • Mass of solvent (water) = Total mass of solution - Mass of solute
   • Mass of solvent = $1280\text{ g} - 80\text{ g} = 1200\text{ g} = 1.2\text{ kg}$.

5. Molality Calculation:

   • $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{2.0\text{ mol}}{1.2\text{ kg}} \approx 1.67\text{ mol/kg}$.

Alternative Formula: $m = \frac{1000 \times M}{1000 \times d - M \times M_2} = \frac{1000 \times 2}{1000 \times 1.28 - 2 \times 40} = \frac{2000}{1200} = 1.67\text{ m}$.

Other options are incorrect because they arise from common errors: $2.00\text{ m}$ wrongly assumes molality equals molarity; $1.56\text{ m}$ or $1.20\text{ m}$ result from calculation errors such as forgetting to subtract the solute mass from the total solution mass.

First, write the balanced chemical equation: $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$. Calculate the number of moles: Moles of $H_2 = 10 \text{ g} / 2 \text{ g/mol} = 5 \text{ mol}$; Moles of $O_2 = 64 \text{ g} / 32 \text{ g/mol} = 2 \text{ mol}$. According to the stoichiometry, 1 mole of $O_2$ reacts with 2 moles of $H_2$. Therefore, 2 moles of $O_2$ require 4 moles of $H_2$. Since we have 5 moles of $H_2$, $O_2$ is the limiting reagent and $H_2$ is in excess. The amount of product formed depends on the limiting reagent ($O_2$). From the equation, 1 mole of $O_2$ gives 2 moles of $H_2O$. Thus, 2 moles of $O_2$ will produce 4 moles of $H_2O$. Mass of $H_2O = 4 \text{ mol} \times 18 \text{ g/mol} = 72 \text{ g}$.

Molality ($m$) = (moles of solute) / (mass of solvent in kg). Molarity ($M$) = 3 mol/L. Mass of 1 L (1000 mL) of solution = density $\times$ volume = $1.25 \text{ g/mL} \times 1000 \text{ mL} = 1250 \text{ g}$. Mass of $NaCl$ (solute) in 1 L = moles $\times$ molar mass = $3 \text{ mol} \times 58.5 \text{ g/mol} = 175.5 \text{ g}$. Mass of solvent (water) = mass of solution - mass of solute = $1250 \text{ g} - 175.5 \text{ g} = 1074.5 \text{ g} = 1.0745 \text{ kg}$. Molality ($m$) = $3 \text{ mol} / 1.0745 \text{ kg} \approx 2.79 \text{ mol/kg}$ or $2.79 \text{ m}$. Option 2 and 3 are incorrect due to calculation errors, and Option 4 incorrectly assumes molarity equals molality when density is not $1 \text{ g/mL}$.

Percentage of Oxygen = $100 - (40 + 6.67) = 53.33\%$. Calculate the relative number of atoms: $C = 40/12 = 3.33$, $H = 6.67/1 = 6.67$, $O = 53.33/16 = 3.33$. The simple whole number ratio $C:H:O$ is $1:2:1$. Thus, the empirical formula is $CH_2O$. Empirical formula mass = $12 + (2 \times 1) + 16 = 30$. To find the molecular formula, calculate $n = \text{Molar mass} / \text{Empirical formula mass} = 180 / 30 = 6$. Molecular formula = $n \times (\text{Empirical formula}) = 6 \times (CH_2O) = C_6H_{12}O_6$. Option 2 is the empirical formula, and Options 3 and 4 have incorrect molar masses ($60$ and $90$ respectively).