Osmotic pressure (π) = (n/V)RT = (0.5/2) * 0.0821 * 298 = 6.14 atm.

Osmotic pressure (π) = nRT/V = (0.5 moles * 0.0821 * 298) / 2 = 12.21 atm.

Osmotic pressure (π) = nRT/V = (1 mol)(0.0821 L·atm/(K·mol))(298 K) / 2 L = 0.82 atm.

Osmotic pressure is directly proportional to both the temperature and the concentration of solute.

In NaClO3, chlorine has an oxidation state of +5.

In NH4+, nitrogen has an oxidation state of -3.

In NO3^-, nitrogen has an oxidation state of +5.

In H2SO4, the oxidation state of sulfur is +6.

Percentage by mass = (mass of solute / mass of solution) * 100 = (20 g / (20 g + 180 g)) * 100 = 20%.

Percentage by mass = (mass of solute / total mass) × 100 = (5 g / (5 g + 95 g)) × 100 = 5%.

Molar mass of glucose = 6(12) + 12(1) + 6(16) = 180 g. Percentage of carbon = (72/180) x 100 = 40%.

The pH of a 0.01 M HCl solution is 2, as HCl is a strong acid and fully dissociates.

For NaOH, which is a strong base, pOH = -log[OH-]. [OH-] = 0.01 M, so pOH = 2. Therefore, pH = 14 - pOH = 14 - 2 = 12.

Sodium acetate is a salt of a weak acid (acetic acid) and a strong base (sodium hydroxide). The pH can be calculated using the formula pH = 7 + 0.5(pKa - log[C]), where pKa of acetic acid is 4.76.

pH = 14 - pOH; pOH = -log[OH-] = -log(0.01) = 2; pH = 14 - 2 = 12.

H2SO4 is a strong acid and dissociates completely, so pH = -log(0.01) = 2.

H2SO4 is a strong acid and dissociates completely; [H+] = 0.05 M, so pH = -log(0.05) ≈ 1.3

pOH = -log[OH-] = -log(0.1) = 1; pH = 14 - pOH = 14 - 1 = 13

Using the formula for weak acids, pH = 0.5(pKa - log[C]) = 0.5(4.74 - log(0.1)) = 4.76.

NH4Cl is a salt of a weak base and a strong acid; it hydrolyzes to give H+, resulting in a pH < 7.

The pH of a strong acid like HCl is calculated using the formula pH = -log[H+]. For 0.1 M HCl, [H+] = 0.1 M, so pH = -log(0.1) = 1.

KHP is a weak acid; its pKa is approximately 5.0, so the pH of a 0.1 M solution is around 5.0.

pH = 7 + 0.5(pKa + log[C]) = 7 + 0.5(4.76 + log(0.1)) = 9.24

NaHCO3 is a weak base; its pH is around 8.4.

Using the Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]). Here, pKa ≈ 4.74, so pH = 4.74 + log(1) = 4.74.

Using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) = 4.76 + log(0.1/0.2) = 5.74

Using the Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]) = 4.76.

At 25°C, the pH of a neutral solution is 7.

HCl and NaOH neutralize each other completely, resulting in a neutral solution with a pH of 7.

pH is calculated as pH = -log[H+]. For [H+] = 1 x 10^-5 M, pH = -log(1 x 10^-5) = 5.