Question

each solute is dissolved in 1 l of solvent. which solution will exhibit the largest increase in boiling point compared to plain water? 0.5 moles koh (dissociates into 2 particles) 0.5 moles al(no₃)₃ (dissociates into 4 particles) 4.0 moles ch₂o (dissociates into 1 particle)

Explanation:

Step1: Recall Boiling Point Elevation

Boiling point elevation depends on the number of solute particles. The formula for the number of particles is moles × dissociation particles.

Step2: Calculate Particles for KOH

For KOH: \( 0.5 \, \text{mol} \times 2 = 1 \, \text{mol} \) of particles.

Step3: Calculate Particles for \( \text{Al(NO}_3\text{)}_3 \)

For \( \text{Al(NO}_3\text{)}_3 \): \( 0.5 \, \text{mol} \times 4 = 2 \, \text{mol} \) of particles.

Step4: Calculate Particles for \( \text{CH}_2\text{O} \)

For \( \text{CH}_2\text{O} \): \( 4.0 \, \text{mol} \times 1 = 4.0 \, \text{mol} \)? Wait, no, wait. Wait, \( \text{CH}_2\text{O} \) (formaldehyde) is a molecular solute, doesn't dissociate. Wait, but wait, the problem says "each solute is dissolved in 1 L". Wait, no, wait, let's recheck. Wait, KOH: 0.5 mol, dissociates into 2, so 0.5*2=1. \( \text{Al(NO}_3\text{)}_3 \): 0.5*4=2. \( \text{CH}_2\text{O} \): 4.0 mol, dissociates into 1, so 4.0*1=4.0? Wait, but that can't be. Wait, no, maybe I misread. Wait, the options: 0.5 moles KOH, 0.5 moles \( \text{Al(NO}_3\text{)}_3 \), 4.0 moles \( \text{CH}_2\text{O} \). Wait, but boiling point elevation is \( \Delta T_b = iK_bm \), where \( i \) is van't Hoff factor (number of particles), \( m \) is molality (moles per kg solvent, here 1 L water is ~1 kg, so molality ≈ molarity). So for KOH: \( i=2 \), \( m=0.5 \), so \( i \times m = 2 \times 0.5 = 1 \). For \( \text{Al(NO}_3\text{)}_3 \): \( i=4 \), \( m=0.5 \), so \( 4 \times 0.5 = 2 \). For \( \text{CH}_2\text{O} \): \( i=1 \), \( m=4.0 \), so \( 1 \times 4.0 = 4.0 \). Wait, but that would mean \( \text{CH}_2\text{O} \) has more particles? But that contradicts. Wait, no, maybe the problem has a typo? Wait, no, wait, \( \text{CH}_2\text{O} \) is a molecular solute, so \( i=1 \), but 4.0 moles in 1 L. Wait, but 4.0 moles of \( \text{CH}_2\text{O} \) would give 4.0 moles of particles, while \( \text{Al(NO}_3\text{)}_3 \) gives 2.0, KOH 1.0. But that can't be, because usually ionic compounds have higher \( i \), but if the molarity is high for molecular, it can be higher. Wait, but maybe I made a mistake. Wait, no, let's check again. Wait, the options: 0.5 moles KOH (i=2, so 1.0 mol particles), 0.5 moles \( \text{Al(NO}_3\text{)}_3 \) (i=4, so 2.0 mol particles), 4.0 moles \( \text{CH}_2\text{O} \) (i=1, so 4.0 mol particles). But that would mean \( \text{CH}_2\text{O} \) has more particles. But that seems odd. Wait, maybe the problem is that \( \text{CH}_2\text{O} \) is a nonelectrolyte, but 4.0 moles is a lot. Wait, but maybe the question is miswritten, or I misread. Wait, no, the original problem: "Which solution will exhibit the largest increase in boiling point". Boiling point elevation is proportional to the number of solute particles. So calculate \( i \times n \) (moles) for each:

- KOH: \( 2 \times 0.5 = 1 \)
- \( \text{Al(NO}_3\text{)}_3 \): \( 4 \times 0.5 = 2 \)
- \( \text{CH}_2\text{O} \): \( 1 \times 4.0 = 4 \)

Wait, but that would mean \( \text{CH}_2\text{O} \) has the most particles. But that seems counterintuitive. Wait, maybe the \( \text{CH}_2\text{O} \) is a typo, maybe 0.4 moles? Or maybe I misread the moles. Wait, the user's image: "4.0 moles \( \text{CH}_2\text{O} \) (dissociates into 1 particle)". So according to that, \( \text{CH}_2\text{O} \) has 4.0*1=4.0 moles of particles, \( \text{Al(NO}_3\text{)}_3 \) has 0.5*4=2.0, KOH 0.5*2=1.0. So \( \text{CH}_2\text{O} \) would have the largest \( \Delta T_b \). But that seems wrong. Wait, no, maybe the problem is that \( \text{CH}_2\text{O} \) is a molecular solute, but 4.0 moles in 1 L is a very high concentration, but mathematically, the number of particles is 4.0, which is higher than 2.0. But maybe the question has a mistake. Alternatively, maybe I misread the moles for \( \text{CH}_2\text{O} \). Wait, the options: 0.5 moles KOH, 0.5 moles \( \text{Al(NO}_3\text{)}_3 \), 4.0 moles \( \text{CH}_2\text{O} \). So according to the calculation, \( \text{CH}_2\text{O} \) has 4.0 particles, \( \text{Al(NO}_3\text{)}_3 \) has 2.0, KOH 1.0. So \( \text{CH}_2\text{O} \) would have the largest increase. But that seems odd. Wait, maybe the \( \text{CH}_2\text{O} \) is supposed to be 0.4 moles? Or maybe the question is different. Wait, no, the user's problem is as given. So according to the calculation, \( \text{CH}_2\text{O} \) has the most particles. But that can't be right. Wait, no, wait, boiling point elevation is \( \Delta T_b = iK_bm \), where \( m \) is molality (moles per kg solvent). Since 1 L water is approximately 1 kg, molality ≈ molarity. So for \( \text{CH}_2\text{O} \), \( m = 4.0 \, \text{mol/kg} \), \( i=1 \), so \( \Delta T_b = 1 \times K_b \times 4.0 \). For \( \text{Al(NO}_3\text{)}_3 \), \( m = 0.5 \, \text{mol/kg} \), \( i=4 \), so \( \Delta T_b = 4 \times K_b \times 0.5 = 2 \times K_b \). For KOH, \( m = 0.5 \, \text{mol/kg} \), \( i=2 \), so \( \Delta T_b = 2 \times K_b \times 0.5 = 1 \times K_b \). So \( \text{CH}_2\text{O} \) has \( \Delta T_b = 4K_b \), which is larger than \( \text{Al(NO}_3\text{)}_3 \)'s \( 2K_b \). So the answer would be 4.0 moles \( \text{CH}_2\text{O} \). But that seems surprising. Wait, maybe the problem has a typo, and \( \text{CH}_2\text{O} \) is 0.4 moles? Or maybe I misread the moles. Wait, the user's image: "4.0 moles \( \text{CH}_2\text{O} \) (dissociates into 1 particle)". So as per the given data, that's correct. So the solution with the largest number of particles is \( \text{CH}_2\text{O} \) with 4.0 moles of particles. But that seems odd. Alternatively, maybe the question is wrong, but according to the given numbers, that's the case.

Answer:

4.0 moles \( \text{CH}_2\text{O} \) (dissociates into 1 particle)