QuestionAugust 16, 2025

How many grams of sodium hydroxide are required to dissolve in 232 g of water to make a 2 .88 m solution? NaOH:39.99g/mol [?]gNaOH

How many grams of sodium hydroxide are required to dissolve in 232 g of water to make a 2 .88 m solution? NaOH:39.99g/mol [?]gNaOH
How many grams of sodium
hydroxide are required to
dissolve in 232 g of water to
make a 2 .88 m solution?
NaOH:39.99g/mol
[?]gNaOH

Solution
4.4(253 votes)

Answer

26.72 g NaOH Explanation 1. Understand Molality Formula Molality (m) is defined as moles of solute per kilogram of solvent. **Molality = \frac{\text{moles of solute}}{\text{kg of solvent}}** 2. Convert Solvent Mass to Kilograms 232 g of water is equivalent to 0.232 kg. 3. Calculate Moles of NaOH Required Using the molality formula: 2.88 = \frac{\text{moles of NaOH}}{0.232}. Therefore, moles of NaOH = 2.88 \times 0.232. 4. Calculate Grams of NaOH Moles of NaOH = 2.88 \times 0.232 = 0.66816. Grams of NaOH = moles \times molar mass = 0.66816 \times 39.99.

Explanation

1. Understand Molality Formula<br /> Molality (m) is defined as moles of solute per kilogram of solvent. **Molality = \frac{\text{moles of solute}}{\text{kg of solvent}}**<br /><br />2. Convert Solvent Mass to Kilograms<br /> 232 g of water is equivalent to 0.232 kg.<br /><br />3. Calculate Moles of NaOH Required<br /> Using the molality formula: $2.88 = \frac{\text{moles of NaOH}}{0.232}$. Therefore, moles of NaOH = $2.88 \times 0.232$.<br /><br />4. Calculate Grams of NaOH<br /> Moles of NaOH = $2.88 \times 0.232 = 0.66816$. Grams of NaOH = moles $\times$ molar mass = $0.66816 \times 39.99$.
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