Question

The density of a 56.0% by weight aqueous solution of 1-propanol is 0.8975g/cm3. What is the mole fraction of the compound?

Answer 1

Answer on Question #43843 - Chemistry - Other

Question:

The density of a 56.0% by weight aqueous solution of 1-propanol is 0.8975 g/cm³. What is the mole fraction of the compound?

Solution:

Let us consider 1 cm³ of the solution. Mass of 1 cm³ of the solution $m_s = 0.8975$ g.

Mass of 1-propanol in this amount of the solution:

$$m_p = m_s \cdot \omega_p = 0.8975 \cdot 0.560 = 0.5026 \text{ g}$$

Mass of water:

$$m_w = m_s - m_p = 0.8975 - 0.5026 = 0.3949 \text{ g}$$

Number of moles of 1-propanol:

$$n_p = m_p / M_p = 0.5026 / 60.095 = 0.0084 \text{ mole}$$

where $M_p$ – molar weight of 1-propanol.

Number of moles of water:

$$n_w = m_w / M_w = 0.3949 / 18.015 = 0.0219 \text{ mole}$$

where $M_w$ – molar weight of water.

Mole fraction of 1-propanol:

$$x_p = n_p / n_{\text{total}} = n_p / (n_p + n_w) = 0.0084 / (0.0084 + 0.0219) = 0.2772$$

Answer: 0.2772

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