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sergij07 [2.7K]
3 years ago
11

I need help with this question please asap I would really appreciate it

Chemistry
1 answer:
yanalaym [24]3 years ago
4 0

Answer:

I think c

Explanation:

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A 3.301 mass % aqueous solution of potassium hydroxide has a density of 1.05 g/mL. Calculate the molality of the solution. Give
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<u>Answer:</u> The molality of potassium hydroxide solution is 0.608 m

<u>Explanation:</u>

We are given:

3.301 mass % of potassium hydroxide solution.

This means that 3.301 grams of potassium hydroxide is present in 100 grams of solution

Mass of solvent = Mass of solution - Mass of solute (KOH)

Mass of solvent = (100 - 3.301) g = 96.699 g

To calculate the molality of solution, we use the equation:

\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams})}

Where,

m_{solute} = Given mass of solute (KOH) = 3.301 g

M_{solute} = Molar mass of solute (KOH) = 56.1 g/mol

W_{solvent} = Mass of solvent = 96.699 g

Putting values in above equation, we get:

\text{Molality of KOH}=\frac{3.301\times 1000}{56.1\times 96.699}\\\\\text{Molality of KOH}=0.608m

Hence, the molality of potassium hydroxide solution is 0.608 m

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