Vessel-1 contains w2 g of a non-volatile solute X dissolved in w1 g of water. Vessel-2 contains w2 g of another non-volatile solute Y dissolved in w1 g of water. Both the vessels are at the same temperature and pressure. The molar mass of X is 80% of that of Y. The van’t Hoff factor for X is 1.2 times of that of Y for their respective concentrations.
The elevation of boiling point for solution in Vessel-1 is _____ % of the solution in Vessel-2.
The elevation of boiling point for solution in Vessel-1 is _____ % of the solution in Vessel-2.
Correct Answer: 150
Approach Solution - 1
To solve the problem, we need to determine the ratio of the boiling point elevation for two solutions in Vessel-1 (solute X) and Vessel-2 (solute Y), expressed as a percentage.
1. Define Variables:
- \(w_1\): Mass of water (solvent) in both vessels
- \(w_2\): Mass of solute in both vessels
- \(M_x\): Molar mass of solute X
- \(M_y\): Molar mass of solute Y
- \(i_x\): van't Hoff factor for solute X
- \(i_y\): van't Hoff factor for solute Y
- \(\Delta T_{b1}\): Boiling point elevation for Vessel-1
- \(\Delta T_{b2}\): Boiling point elevation for Vessel-2
2. Given Information:
\[
M_x = 0.8 M_y, \quad i_x = 1.2 i_y
\]
3. Boiling Point Elevation Formula:
The elevation of boiling point is given by:
\[
\Delta T_b = i \cdot K_b \cdot m
\]
where \(i\) is the van't Hoff factor, \(K_b\) is the ebullioscopic constant (same for both vessels since the solvent is the same), and \(m\) is the molality (moles of solute per kg of solvent).
4. Calculate Molality:
- Molality for Vessel-1 (solute X):
\[
m_x = \frac{\text{moles of X}}{\text{kg of solvent}} = \frac{w_2 / M_x}{w_1 / 1000} = \frac{1000 w_2}{M_x w_1}
\]
- Molality for Vessel-2 (solute Y):
\[
m_y = \frac{w_2 / M_y}{w_1 / 1000} = \frac{1000 w_2}{M_y w_1}
\]
5. Calculate Boiling Point Elevations:
- For Vessel-1:
\[
\Delta T_{b1} = i_x \cdot K_b \cdot m_x = i_x \cdot K_b \cdot \frac{1000 w_2}{M_x w_1}
\]
- For Vessel-2:
\[
\Delta T_{b2} = i_y \cdot K_b \cdot m_y = i_y \cdot K_b \cdot \frac{1000 w_2}{M_y w_1}
\]
6. Find the Ratio \(\Delta T_{b1} / \Delta T_{b2}\):
\[
\frac{\Delta T_{b1}}{\Delta T_{b2}} = \frac{i_x \cdot K_b \cdot \frac{1000 w_2}{M_x w_1}}{i_y \cdot K_b \cdot \frac{1000 w_2}{M_y w_1}} = \frac{i_x}{i_y} \cdot \frac{M_y}{M_x}
\]
The \(K_b\), \(w_1\), \(w_2\), and 1000 terms cancel out.
7. Substitute Given Relationships:
\[
i_x = 1.2 i_y, \quad M_x = 0.8 M_y
\]
\[
\frac{i_x}{i_y} = 1.2, \quad \frac{M_y}{M_x} = \frac{M_y}{0.8 M_y} = \frac{1}{0.8} = 1.25
\]
\[
\frac{\Delta T_{b1}}{\Delta T_{b2}} = 1.2 \cdot 1.25 = 1.5
\]
8. Express as a Percentage:
\[
\frac{\Delta T_{b1}}{\Delta T_{b2}} = 1.5 \implies \Delta T_{b1} = 1.5 \Delta T_{b2}
\]
As a percentage:
\[
1.5 \times 100\% = 150\%
\]
Final Answer:
The elevation of boiling point for the solution in Vessel-1 is \(150\%\) of that in Vessel-2.
Approach Solution -2
To solve the problem, we need to find the percentage ratio of elevation in boiling points of solutions in Vessel-1 and Vessel-2.
1. Given data:
- Mass of solute in both vessels: \( w_2 \) g
- Mass of solvent (water) in both vessels: \( w_1 \) g
- Molar mass of solute \( X = 0.8 \times \) molar mass of solute \( Y \)
- Van’t Hoff factor \( i_X = 1.2 \times i_Y \)
- Temperature and pressure are the same for both solutions.
2. Elevation in boiling point formula:
\[
\Delta T_b = i K_b m
\]
where
\( i \) = van't Hoff factor,
\( K_b \) = ebullioscopic constant (same for both as solvent and conditions same),
\( m \) = molality of solution.
3. Calculate molality for both solutions:
\[
m_X = \frac{\text{moles of } X}{\text{kg of solvent}} = \frac{\frac{w_2}{M_X}}{\frac{w_1}{1000}} = \frac{1000 w_2}{M_X w_1}
\]
\[
m_Y = \frac{\frac{w_2}{M_Y}}{\frac{w_1}{1000}} = \frac{1000 w_2}{M_Y w_1}
\]
4. Ratio of elevation in boiling points:
\[
\frac{\Delta T_{b,X}}{\Delta T_{b,Y}} = \frac{i_X m_X}{i_Y m_Y} = \frac{(1.2 i_Y) \times \frac{1000 w_2}{M_X w_1}}{i_Y \times \frac{1000 w_2}{M_Y w_1}} = 1.2 \times \frac{M_Y}{M_X}
\]
5. Substitute \( M_X = 0.8 M_Y \):
\[
\frac{\Delta T_{b,X}}{\Delta T_{b,Y}} = 1.2 \times \frac{M_Y}{0.8 M_Y} = 1.2 \times \frac{1}{0.8} = 1.5
\]
6. Convert ratio to percentage:
\[
\text{Elevation of boiling point for Vessel-1} = 150\% \text{ of that in Vessel-2}
\]
Final Answer:
The elevation of boiling point for solution in Vessel-1 is \(\boxed{150\%}\) of that of the solution in Vessel-2.
Top Questions on Solutions
According to the generally accepted definition of the ideal solution there are equal interaction forces acting between molecules belonging to the same or different species. (This is equivalent to the statement that the activity of the components equals the concentration.) Strictly speaking, this concept is valid in ecological systems (isotopic mixtures of an element, hydrocarbons mixtures, etc.). It is still usual to talk about ideal solutions as limiting cases in reality since very dilute solutions behave ideally with respect to the solvent. This law is further supported by the fact that Raoult’s law empirically found for describing the behaviour of the solvent in dilute solutions can be deduced thermodynamically via the assumption of ideal behaviour of the solvent.
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OR
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