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olganol [36]
2 years ago
11

3.01×10 to the 23rd molecules H2O

Chemistry
1 answer:
lisov135 [29]2 years ago
7 0

Answer: The factor is (1 mol/6.02 x 10^23 anythings). 3.01 x 10^23 molecules H2O x (1 mole/6.02 x 10^23 molecules H2O)= 0.500 mole H2O.Mar 16, 2009

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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