Question

A solution is prepared by urea and water. If mole fraction of water is \(0.8\) in the solution. Find the ratio of mass of urea \(\&\) water. (1) \(\frac{5}{6}\) (2) \(\frac{6}{5}\) (3) \(\frac{1}{1}\) (4) \(\frac{4}{1}\)

Short answer

The ratio of the mass of urea to water is \( \frac{5}{6} \).

Step-by-step solution

- Understand Mole Fraction

The mole fraction of a component in a solution is the ratio of the number of moles of that component to the total number of moles in the solution. The mole fraction of water is given as 0.8, so the mole fraction of urea will be 1 - 0.8 = 0.2.

- Define Mole Variables

Let the number of moles of water be denoted by n_w and the number of moles of urea be denoted by n_u. According to the mole fraction definition, we have: \[ \text{Mole fraction of water} = \frac{n_w}{n_w + n_u} = 0.8 \] and \[ \text{Mole fraction of urea} = \frac{n_u}{n_w + n_u} = 0.2 \]

- Setup Equations

We can now set up the equation from the given mole fraction of water: \[ \frac{n_w}{n_w + n_u} = 0.8 \] By cross-multiplying, we get the equation: \[ n_w = 0.8(n_w + n_u) \]

- Solve for Ratio of Moles

Simplify the equation to solve for the ratio of moles: \[ n_w = 0.8n_w + 0.8n_u \] Subtract 0.8n_w from both sides: \[ n_w - 0.8n_w = 0.8n_u \] \[ 0.2n_w = 0.8n_u \] Divide both sides by 0.2: \[ n_w = 4n_u \] Thus, the ratio of the number of moles of water to urea is 4:1.

- Convert to Mass Ratio

The molar mass of water (H2O) is 18 g/mol, and the molar mass of urea (CH4N2O) is 60 g/mol. Using the mole ratio of 4:1, we find the mass of water and the mass of urea. Mass of water = 4 moles * 18 g/mol = 72 g Mass of urea = 1 mole * 60 g/mol = 60 g. Therefore, the mass ratio of urea and water is \[ \frac{60}{72} = \frac{5}{6} \]

Key concepts

Mole Fraction

Mole fraction is a way to express the concentration of a component in a solution. It's practical in many areas of chemistry.
In our exercise, the solution contains water and urea. Given that the mole fraction of water is 0.8, let's clarify why this matters.
The mole fraction tells us that for every 1 mole of the solution, 0.8 moles are water molecules. The rest is urea: 0.2 moles.
This is because the sum of the mole fractions of all components in a solution always equals 1.
So, mole fraction can be written as follows:

• For water: \[ \text{Mole fraction of water} = \frac{n_w}{n_w + n_u} \]
• For urea: \[ \text{Mole fraction of urea} = \frac{n_u}{n_w + n_u} \]

With the provided mole fraction of water, we can set equations and solve for these variables, leading us to find mass values later down the line.

Mass Ratio

The concept of mass ratio is handy when comparing the masses of different substance quantities in a solution.
In our case, after finding the mole ratios, converting to mass lets us understand how much each component weighs.
The molar masses play a crucial role here.
Water, H2O, has a molar mass of 18 g/mol. Urea, CO(NH2)2, has a molar mass of 60 g/mol.
Knowing these, we take the derived mole ratio of 4:1 (water to urea), and calculate as follows:

• Mass of water = 4 * 18 g/mol = 72 g
• Mass of urea = 1 * 60 g/mol = 60 g

Thus, the mass ratio of urea to water in the solution becomes \[ \text{Mass ratio of urea to water} = \frac{60}{72} = \frac{5}{6} \]This gives us a straightforward understanding of the weights of components in the mixture.

Solution Chemistry

Solution chemistry involves studying solutions, which are uniform mixtures of two or more substances.
In our exercise, we've dealt with a solution of urea and water. Here are some core concepts:

• **Solute**: The substance being dissolved (urea in this case).
• **Solvent**: The substance that does the dissolving (water, because it's in larger quantity).
• **Mole Fraction**: As discussed, it gives us a ratio of mole quantities.
• **Mass Ratio**: Seen in mass comparisons, its calculation shows relative weights of components.
• **Concentration**: Mole fraction is one method of expressing concentration along with molarity and molality.

Solution chemistry equips us with tools to understand how different substances interact and combine in predictable ways.From our exercise, mastering these tools helps in solving more complex problems involving various solutions.