Question

the diagram on the left represents molecular - level particle drawings illustrating four aqueous solutions prepared from the same unknown ionic salt. each particle represents 1 mole of the solute. arrange the solutions from highest molarity to lowest molarity (most dilute).
a iii, iv, ii, i
b i, ii, iii, iv
c iv, iii, ii, i
d iv, ii, iii, i

Explanation:

Step1: Recall Molarity Formula

Molarity ($M$) is defined as moles of solute ($n$) divided by volume of solution in liters ($V$), so $M = \frac{n}{V}$.

Step2: Determine Moles and Volume for Each Solution

- **Solution I (A)**: From diagram, moles of solute ($n_I$) = 3, volume ($V_I$) = 3.0 L. So $M_I = \frac{3}{3.0} = 1$ M.
- **Solution II (B)**: Moles ($n_{II}$) = 3, volume ($V_{II}$) = 1.0 L. So $M_{II} = \frac{3}{1.0} = 3$ M.
- **Solution III (C)**: Moles ($n_{III}$) = 9, volume ($V_{III}$) = 4.0 L. So $M_{III} = \frac{9}{4.0} = 2.25$ M. Wait, no, wait the diagram: Wait, Solution III (C) has 9 particles? Wait no, looking back: Solution A (I): 3 particles, 3.0 L; Solution B (II): 3 particles, 1.0 L; Solution C (III): 9 particles, 4.0 L? Wait no, original diagram: Let's re-express:

Wait, the problem says "each particle represents 1 mole of the solute". Let's list each:

- Solution I (A): Volume = 3.0 L, number of solute particles (moles) = 3. So $M_I = \frac{3}{3.0} = 1$ M.
- Solution II (B): Volume = 1.0 L, moles = 3. So $M_{II} = \frac{3}{1.0} = 3$ M.
- Solution III (C): Volume = 4.0 L, moles = 9? Wait no, the diagram for Solution C (III) has 9 particles? Wait the user's diagram: Solution A (I) has 3 particles, 3.0 L; Solution B (II) has 3 particles, 1.0 L; Solution C (III) has 9 particles, 4.0 L? Wait no, maybe I misread. Wait Solution D (IV): Volume = 2.0 L, moles = 6? Wait no, let's check again:

Wait the options: Let's recalculate correctly.

Wait, let's list each solution:

- Solution I (A): moles (n) = 3, volume (V) = 3.0 L. $M = 3/3 = 1$ M.
- Solution II (B): n = 3, V = 1.0 L. $M = 3/1 = 3$ M.
- Solution III (C): n = 9, V = 4.0 L. $M = 9/4 = 2.25$ M.
- Solution IV (D): n = 6, V = 2.0 L. $M = 6/2 = 3$ M? Wait no, maybe the particles: Wait Solution D (IV) has 6 particles? Wait the diagram: Solution D (IV) has 6 particles (blue with white dots), volume 2.0 L. So n=6, V=2.0 L: M=6/2=3 M. Solution C (III) has 9 particles, volume 4.0 L: M=9/4=2.25 M. Solution II (B) has 3 particles, volume 1.0 L: M=3/1=3 M. Solution I (A) has 3 particles, volume 3.0 L: M=3/3=1 M.

Wait, no, maybe I messed up the particles. Let's re-express:

Wait the problem says "four aqueous solutions prepared from the same unknown ionic salt. Each particle represents 1 mole of the solute." So:

- Solution I (A): 3 particles (moles), 3.0 L. Molarity = 3/3 = 1 M.
- Solution II (B): 3 particles, 1.0 L. Molarity = 3/1 = 3 M.
- Solution III (C): 9 particles, 4.0 L. Molarity = 9/4 = 2.25 M.
- Solution IV (D): 6 particles, 2.0 L. Molarity = 6/2 = 3 M.

Now, we need to arrange from highest molarity to lowest. Let's list the molarities:

- II: 3 M
- IV: 3 M
- III: 2.25 M
- I: 1 M

Wait, but the options: Let's check the options again. The options are:

A. III, IV, II, I → No, because II and IV are 3 M, III is 2.25, I is 1.

B. I, II, III, IV → No.

C. IV, III, II, I → No.

D. IV, II, III, I → Wait, II and IV both have 3 M? Wait maybe I miscounted the particles. Let's look at the diagram again (as per user's image):

- Solution I (A): 3 particles, 3.0 L.
- Solution II (B): 3 particles, 1.0 L.
- Solution III (C): 9 particles, 4.0 L.
- Solution IV (D): 6 particles, 2.0 L.

Wait, no, maybe Solution IV (D) has 6 particles? Wait 6/2=3, Solution II (B) has 3/1=3, so II and IV have same molarity? But the options have D as IV, II, III, I. Wait maybe I made a mistake in particles. Let's re-express:

Wait, maybe Solution III (C) has 9 particles? No, maybe the correct count is:

Wait, let's calculate molarity for each:

- Solution I: n=3, V=3L → M=1
- Solution II: n=3, V=1L → M=3
- Solution III: n=9, V=4L → M=2.25
- Solution IV: n=6, V=2L → M=3

So the molarities are: II (3), IV (3), III (2.25), I (1). But the options: Let's check the options again. The options are:

A. III, IV, II, I → No.

B. I, II, III, IV → No.

C. IV, III, II, I → No.

D. IV, II, III, I → Wait, maybe the particles in Solution IV are 6, Solution II is 3, Solution III is 9, Solution I is 3. Wait, no, maybe I misread the volume. Wait Solution C (III) has volume 4.0 L, Solution D (IV) has 2.0 L, Solution B (II) has 1.0 L, Solution A (I) has 3.0 L.

Wait, let's recalculate:

- Solution I: 3 moles / 3 L = 1 M
- Solution II: 3 moles / 1 L = 3 M
- Solution III: 9 moles / 4 L = 2.25 M
- Solution IV: 6 moles / 2 L = 3 M

So the order from highest to lowest: II (3), IV (3), III (2.25), I (1). But the options don't have that. Wait maybe the particles are different. Wait maybe Solution III (C) has 3 particles? No, the diagram shows Solution C with more particles. Wait, maybe the user's diagram:

Wait, Solution A (I): 3 particles, 3.0 L.

Solution B (II): 3 particles, 1.0 L.

Solution C (III): 9 particles, 4.0 L.

Solution D (IV): 6 particles, 2.0 L.

Wait, maybe the question is about the number of solute particles (ions or molecules) but since it's an ionic salt, maybe each particle dissociates, but the problem says "each particle represents 1 mole of the solute", so we can take moles as number of particles.

Wait, maybe I made a mistake in the options. Let's check the options again:

Option D: IV, II, III, I.

Let's calculate molarity for each in Option D order:

IV: 6 moles / 2 L = 3 M

II: 3 moles / 1 L = 3 M

III: 9 moles / 4 L = 2.25 M

I: 3 moles / 3 L = 1 M

So IV and II have same molarity (3 M), then III (2.25 M), then I (1 M). So the order from highest to lowest is IV (3), II (3), III (2.25), I (1), which is option D.

Answer:

D. IV, II, III, I