Answer :
Sure! Let's go through the solution step-by-step to understand how to calculate the mass of urea needed and the molality of the solution.
### Step 1: Understanding the Vapour Pressure Decrease
The problem states that the vapor pressure of water decreases by 25%. This indicates that the vapor pressure of the solution is 75% of the pure solvent's vapor pressure. Using Raoult's law, we can express this as:
[tex]\[
\frac{p_{\text{solv}}^0 - p_{\text{soln}}}{p_{\text{solv}}^0} = \frac{1}{4}
\][/tex]
Where:
- [tex]\( p_{\text{solv}}^0 \)[/tex] is the vapor pressure of the pure solvent (water).
- [tex]\( p_{\text{soln}} \)[/tex] is the vapor pressure of the solution.
### Step 2: Setting up the Equations
Based on Raoult's law, the above expression can be rewritten as:
[tex]\[
\frac{1}{4} = \frac{m_{\text{urea}} / M_{\text{urea}}}{(m_{\text{urea}} / M_{\text{urea}}) + (m_{\text{water}} / M_{\text{water}})}
\][/tex]
Rearranging the equation to solve for the mass of urea:
[tex]\[
4 = 1 + \frac{m_{\text{water}} / M_{\text{water}}}{m_{\text{urea}} / M_{\text{urea}}}
\][/tex]
Which simplifies to:
[tex]\[
3 = \frac{m_{\text{water}}}{M_{\text{water}}} \times \frac{M_{\text{urea}}}{m_{\text{urea}}}
\][/tex]
### Step 3: Solving for the Mass of Urea
Given:
- [tex]\( m_{\text{water}} = 100 \)[/tex] grams
- [tex]\( M_{\text{urea}} = 60 \)[/tex] g/mol
- [tex]\( M_{\text{water}} = 18 \)[/tex] g/mol
We can solve for [tex]\( m_{\text{urea}} \)[/tex]:
[tex]\[
m_{\text{urea}} = \frac{m_{\text{water}} \times M_{\text{urea}}}{M_{\text{water}} \times 3} = \frac{100 \times 60}{18 \times 3} = 111.11 \, \text{grams}
\][/tex]
### Step 4: Calculating the Molality of the Solution
Molality is defined as the moles of solute per kilogram of solvent:
[tex]\[
\text{Molality of urea} = \frac{m_{\text{urea}} / M_{\text{urea}}}{m_{\text{water}} / 1000}
\][/tex]
Substituting the known values:
[tex]\[
\text{Molality of urea} = \frac{111.11 / 60}{100 / 1000} = 18.52 \, \text{mol/kg}
\][/tex]
### Conclusion
Therefore, to achieve a 25% decrease in the vapor pressure of water, you would need to dissolve approximately 111.11 grams of urea in 100 grams of water, resulting in a molality of approximately 18.52 mol/kg.
### Step 1: Understanding the Vapour Pressure Decrease
The problem states that the vapor pressure of water decreases by 25%. This indicates that the vapor pressure of the solution is 75% of the pure solvent's vapor pressure. Using Raoult's law, we can express this as:
[tex]\[
\frac{p_{\text{solv}}^0 - p_{\text{soln}}}{p_{\text{solv}}^0} = \frac{1}{4}
\][/tex]
Where:
- [tex]\( p_{\text{solv}}^0 \)[/tex] is the vapor pressure of the pure solvent (water).
- [tex]\( p_{\text{soln}} \)[/tex] is the vapor pressure of the solution.
### Step 2: Setting up the Equations
Based on Raoult's law, the above expression can be rewritten as:
[tex]\[
\frac{1}{4} = \frac{m_{\text{urea}} / M_{\text{urea}}}{(m_{\text{urea}} / M_{\text{urea}}) + (m_{\text{water}} / M_{\text{water}})}
\][/tex]
Rearranging the equation to solve for the mass of urea:
[tex]\[
4 = 1 + \frac{m_{\text{water}} / M_{\text{water}}}{m_{\text{urea}} / M_{\text{urea}}}
\][/tex]
Which simplifies to:
[tex]\[
3 = \frac{m_{\text{water}}}{M_{\text{water}}} \times \frac{M_{\text{urea}}}{m_{\text{urea}}}
\][/tex]
### Step 3: Solving for the Mass of Urea
Given:
- [tex]\( m_{\text{water}} = 100 \)[/tex] grams
- [tex]\( M_{\text{urea}} = 60 \)[/tex] g/mol
- [tex]\( M_{\text{water}} = 18 \)[/tex] g/mol
We can solve for [tex]\( m_{\text{urea}} \)[/tex]:
[tex]\[
m_{\text{urea}} = \frac{m_{\text{water}} \times M_{\text{urea}}}{M_{\text{water}} \times 3} = \frac{100 \times 60}{18 \times 3} = 111.11 \, \text{grams}
\][/tex]
### Step 4: Calculating the Molality of the Solution
Molality is defined as the moles of solute per kilogram of solvent:
[tex]\[
\text{Molality of urea} = \frac{m_{\text{urea}} / M_{\text{urea}}}{m_{\text{water}} / 1000}
\][/tex]
Substituting the known values:
[tex]\[
\text{Molality of urea} = \frac{111.11 / 60}{100 / 1000} = 18.52 \, \text{mol/kg}
\][/tex]
### Conclusion
Therefore, to achieve a 25% decrease in the vapor pressure of water, you would need to dissolve approximately 111.11 grams of urea in 100 grams of water, resulting in a molality of approximately 18.52 mol/kg.
