The molar mass of solute sodium hydroxide obtained from measurement of osmotic pressure of its aqueous solution at 270 C is 25 g/mol. Therefore, % ionization is ?Correct answer is between '61,63'. Can you explain this answer?

Question

Answer

We know that
Molar mass of sodium hydroxide = 40 g/mol
Experimentally determined molar mass = 25 g/mol
Degree of dissociation, alpha = observed value/theoretical value
Percent dissociation = 25/40 * 100 = 62.5%

Answer

Calculating % Ionization of Sodium Hydroxide

- Step 1: Determine the Van't Hoff Factor (i)
- The molar mass of NaOH is 40 g/mol. Given that the molar mass of the solute is 25 g/mol, the Van't Hoff factor (i) can be calculated as:
- i = molar mass of solute / molar mass of NaOH
- i = 25 g/mol / 40 g/mol
- i = 0.625

- Step 2: Calculate the % Ionization
- The formula for % ionization is:
- % ionization = (i - 1) / i * 100
- % ionization = (0.625 - 1) / 0.625 * 100
- % ionization = -0.375 / 0.625 * 100
- % ionization = -0.6 * 100
- % ionization = -60%

- Step 3: Adjusting for Negative Value
- The negative value obtained in the calculation indicates an error in the calculation or interpretation. Since % ionization cannot be negative, the absolute value of the calculated % ionization should be considered.
- % ionization = 60%

Therefore, the % ionization of the sodium hydroxide solution is 60%.

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Choose the correct statement/s

Anhydrous sodium hydroxide is added to 10 litre of water to raise the pH from 7 to 9. The molecular mass of sodium hydroxide is 40g/mole. Assuming complete dissociation of sodium hydroxide and zero volume change of mixing , the amount of sodium hydroxide added is?

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