Question

each solute is dissolved in 1 l of solvent. which solution will exhibit the smallest increase in boiling point compared to plain water?
2.0 moles k₂so₄ (dissociates into 3 particles)
2.0 moles c₆h₁₂o₆ (dissociates into 1 particle)
0.5 moles aln (dissociates into 2 particles)

Explanation:

Step1: Recall boiling point elevation formula

The boiling point elevation \(\Delta T_b = i \times K_b \times m\), where \(i\) is the van't Hoff factor (number of particles per formula unit), \(K_b\) is the ebullioscopic constant (same for all aqueous solutions here), and \(m\) is the molality. Since the volume is 1 L and we assume density of water is 1 kg/L, molality \(m\) is approximately equal to moles of solute (because mass of solvent is 1 kg). So we calculate \(i \times n\) (where \(n\) is moles of solute) for each solution.

Step2: Calculate \(i \times n\) for \(K_2SO_4\)

For \(K_2SO_4\), \(n = 2.0\) moles, \(i = 3\). So \(i \times n = 3 \times 2.0 = 6.0\).

Step3: Calculate \(i \times n\) for \(C_6H_{12}O_6\)

For \(C_6H_{12}O_6\), \(n = 2.0\) moles, \(i = 1\). So \(i \times n = 1 \times 2.0 = 2.0\).

Step4: Calculate \(i \times n\) for \(AlN\)

For \(AlN\), \(n = 0.5\) moles, \(i = 2\). So \(i \times n = 2 \times 0.5 = 1.0\).

Step5: Compare the values

The smaller the \(i \times n\) value, the smaller the \(\Delta T_b\) (since \(K_b\) is constant). Comparing 6.0, 2.0, and 1.0, 1.0 is the smallest, which corresponds to \(AlN\). Wait, wait, no, wait. Wait, let's check again. Wait, the question is which has the smallest increase. Wait, maybe I made a mistake. Wait, no: \(\Delta T_b\) is proportional to \(i \times m\). So lower \(i \times m\) means lower \(\Delta T_b\). Wait, let's recalculate:

Wait, for \(AlN\): \(n = 0.5\) mol, \(i = 2\), so \(i \times n = 1.0\)

For \(C_6H_{12}O_6\): \(n = 2.0\) mol, \(i = 1\), so \(i \times n = 2.0\)

For \(K_2SO_4\): \(n = 2.0\) mol, \(i = 3\), so \(i \times n = 6.0\)

So the smallest \(i \times n\) is 1.0 for \(AlN\)? Wait, but wait, the options: let's check the options again. Wait, the options are:

1. 2.0 moles \(K_2SO_4\) (i=3)
2. 2.0 moles \(C_6H_{12}O_6\) (i=1)
3. 0.5 moles \(AlN\) (i=2)

Wait, so \(i \times n\) for \(AlN\) is 0.5*2=1, for \(C_6H_{12}O_6\) is 2*1=2, for \(K_2SO_4\) is 2*3=6. So the smallest \(i \times n\) is 1, which is for \(AlN\)? Wait, but wait, maybe I messed up. Wait, no, the boiling point elevation is proportional to the number of solute particles. So the more particles, the higher the boiling point elevation. So the solution with the least number of particles will have the smallest \(\Delta T_b\).

Wait, let's calculate the number of moles of particles:

- \(K_2SO_4\): 2.0 mol * 3 = 6.0 mol particles
- \(C_6H_{12}O_6\): 2.0 mol * 1 = 2.0 mol particles
- \(AlN\): 0.5 mol * 2 = 1.0 mol particles

So \(AlN\) has the least number of particles, so its solution will have the smallest increase in boiling point. Wait, but let me check the options again. Wait, the question is which has the smallest increase. So the one with the least \(i \times n\) (since \(K_b\) and \(m\) related to moles) will have the smallest \(\Delta T_b\). So \(AlN\) has \(i \times n = 1.0\), which is less than \(C_6H_{12}O_6\)'s 2.0 and \(K_2SO_4\)'s 6.0. So the answer should be 0.5 moles \(AlN\) (dissociates into 2 particles). Wait, but let me confirm the formula again. \(\Delta T_b = iK_bm\), and since volume is 1 L, mass of solvent is 1 kg, so molality \(m = \frac{n}{1 kg}\), so \(m = n\) (in mol/kg). So \(\Delta T_b \propto i \times n\). So yes, the smaller \(i \times n\), the smaller \(\Delta T_b\). So 0.5*2=1, 2*1=2, 2*3=6. So 1 is the smallest, so \(AlN\) solution.

Wait, but maybe I made a mistake in the van't Hoff factor? Let's check \(AlN\): when \(AlN\) dissociates, does it form 2 particles? \(AlN\) in water? Wait, \(AlN\) reacts with water: \(AlN + 3H_2O ightarrow Al(OH)_3 + NH_3\). Oh! Wait, maybe I made a mistake here. Wait, the problem says "dissociates into 2 particles" – maybe it's a hypothetical dissociation, like if it were an electrolyte that dissociates into \(Al^{3+}\) and \(N^{3-}\), but in reality, \(AlN\) reacts with water, not dissociates. But the problem states "dissociates into 2 particles", so we have to go with the given \(i = 2\). So according to the problem's information, we use \(i = 2\) for \(AlN\). So with that, \(i \times n = 0.5 \times 2 = 1\), which is less than \(C_6H_{12}O_6\)'s \(2 \times 1 = 2\), so \(AlN\) has the smallest \(\Delta T_b\).

Answer:

0.5 moles AlN (dissociates into 2 particles)