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Find the molality of the solution if 35.0 g of propyl alcohol (\( \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \)) are dissolved in 200 g of water.

A. 1.55 m
B. 2.92 m
C. 2.37 m
D. 0.965 m

Answer :

The molality of the solution is approximately 1.55 m. Therefore, the correct answer is A) 1.55 m.

To find the molality (m) of the solution, we need to determine the number of moles of the solute (propyl alcohol) and the mass of the solvent (water).

Given:

Mass of propyl alcohol (solute) = 35.0 g

Mass of water (solvent) = 200 g

Step 1: Convert the mass of propyl alcohol to moles.

First, we need to calculate the molar mass of propyl alcohol (C3H7OH):

C = 12.01 g/mol, H = 1.008 g/mol, O = 16.00 g/mol

Molar mass of propyl alcohol = (3 * 12.01 g/mol) + (8 * 1.008 g/mol) + 16.00 g/mol = 60.12 g/mol

Now, we can calculate the number of moles of propyl alcohol:

Number of moles of propyl alcohol = Mass of propyl alcohol / Molar mass of propyl alcohol

= 35.0 g / 60.12 g/mol

Step 2: Calculate the molality.

Molality (m) is defined as the number of moles of solute per kilogram of solvent.

Mass of water (in kg) = Mass of water (g) / 1000

Now we can calculate the molality:

Molality (m) = Number of moles of propyl alcohol / Mass of water (in kg)

= (35.0 g / 60.12 g/mol) / (200 g / 1000)

Simplifying the expression, we get:

Molality (m) = (35.0 g * 1000) / (60.12 g/mol * 200 g)

= 1.55 m (rounded to two decimal places)

Option A

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