Question

how many liters of a 5.00 m solution of na₃po₄ are needed to have 4.00 osmoles ? first we need the equation for dissolving the salt, na₃po₄ → 3 na⁺¹ + po₄⁻³ an osmole is a mole of dissolved particles the conversion factor is 1 mol = 1 osmol first convert osmoles into moles 1 mol (2 osmol)(----------) = 3 mol 4 osmol then calculate the liters 1 l (5 mol)(----------) = 6 l 7 mol a. 0.00 b. 1.00 c. 2.00 d. 3.00 e. 4.00 f. 5.00 g. 6.00 h. 8.00 i. 10.00 j. 2000 k. 10² l. 10³ m. 10⁶ n. 10⁹ o. 10¹² p. 10⁻² q. 10⁻³ r. 10⁻⁶ s. 10⁻⁹ t. 10⁻¹² u. 10⁸ v. 10⁻¹ w. 10¹ x. 30.0 y. 0.100 z. 0.200 aa. 20.0 bb. 40.0 cc. 1.33 dd. 0.267

Explanation:

Step1: Determine moles from osmoles

From the dissociation equation \( \text{Na}_3\text{PO}_4 ightarrow 3\text{Na}^+ + \text{PO}_4^{3-} \), each mole of \( \text{Na}_3\text{PO}_4 \) gives \( 3 + 1 = 4 \) moles of particles (osmoles). So the conversion factor is \( 1 \, \text{mol} = 4 \, \text{osmol} \) (wait, no—wait, the problem says "the conversion factor is 1 mol = __1__ osmol" but actually, from dissociation, 1 mol of \( \text{Na}_3\text{PO}_4 \) produces 4 osmoles (3 \( \text{Na}^+ \) and 1 \( \text{PO}_4^{3-} \)). Wait, the problem's first part: "First convert osmoles into moles". So we have 4.00 osmoles. The number of moles of \( \text{Na}_3\text{PO}_4 \) is \( \text{moles} = \frac{\text{osmoles}}{\text{osmoles per mole}} \). Since 1 mol of \( \text{Na}_3\text{PO}_4 \) gives 4 osmoles, then \( \text{moles} = \frac{4.00 \, \text{osmol}}{4 \, \text{osmol/mol}} = 1.00 \, \text{mol} \). Wait, let's check the problem's blanks:

Blank 2: osmoles is 4.00 (option E: 4.00)

Blank 4: osmoles per mole is 4 (but wait, the options: looking at options, maybe I misread. Wait, the problem says "the conversion factor is 1 mol = __1__ osmol"—maybe the problem has a typo, but according to the dissociation, 1 mol \( \text{Na}_3\text{PO}_4 \) gives 4 osmoles. So to convert osmoles to moles: \( \text{moles} = \frac{\text{osmoles}}{4} \). So for 4.00 osmoles, \( \frac{4.00}{4} = 1.00 \, \text{mol} \). So blank 2: 4.00 (E), blank 4: 4 (but no option 4? Wait, options are A-Z, AA-BB, CC-DD. Wait, maybe the problem's "conversion factor" is 1 mol = 4 osmoles, so to convert osmoles to moles: \( \text{moles} = \frac{\text{osmoles}}{4} \). So (4.00 osmol) * (1 mol / 4 osmol) = 1.00 mol. So blank 2: E (4.00), blank 4: maybe 4, but no option 4. Wait, maybe the problem's "1 mol = __1__ osmol" is a mistake, and actually, the number of particles is 4, so 1 mol = 4 osmoles. So blank 1: 4? But options don't have 4 as a single digit, but E is 4.00. Wait, maybe the first step:

Blank 2: 4.00 (E), blank 4: 4 (but no option 4, maybe the problem means that 1 mol of \( \text{Na}_3\text{PO}_4 \) gives 4 osmoles, so the conversion factor is 1 mol = 4 osmoles. So (4.00 osmol) * (1 mol / 4 osmol) = 1.00 mol. So blank 3: 1.00 (B), blank 2: E (4.00), blank 4: 4 (but no option 4, maybe the problem's "1 mol = __1__ osmol" is wrong, and it's 1 mol = 4 osmoles, so blank 1: 4, but no option. Wait, maybe I misread. Let's proceed to the second step: calculate liters. Molarity \( M = \frac{\text{moles}}{V} \), so \( V = \frac{\text{moles}}{M} \). Molarity is 5.00 M, moles is 1.00 mol (from step 1). So \( V = \frac{1.00 \, \text{mol}}{5.00 \, \text{mol/L}} = 0.200 \, \text{L} \) (option Z: 0.200). Wait, let's re-examine:

Step 1: Convert osmoles to moles.

Given: 4.00 osmoles.

From dissociation: 1 mol \( \text{Na}_3\text{PO}_4 \) → 4 osmoles (3 \( \text{Na}^+ \), 1 \( \text{PO}_4^{3-} \)).

So moles of \( \text{Na}_3\text{PO}_4 = \frac{\text{osmoles}}{\text{osmoles per mole}} = \frac{4.00}{4} = 1.00 \, \text{mol} \).

So blank 2: 4.00 (E), blank 4: 4 (but no option, maybe the problem's "conversion factor" is 1 mol = 4 osmoles, so blank 1: 4, but options have E as 4.00. Then blank 3: 1.00 (B).

Step 2: Calculate liters. Molarity \( M = 5.00 \, \text{mol/L} \), moles = 1.00 mol.

\( V = \frac{\text{moles}}{M} = \frac{1.00 \, \text{mol}}{5.00 \, \text{mol/L}} = 0.200 \, \text{L} \) (option Z: 0.200).

So blank 5: 1.00 (B), blank 7: 5.00 (F), blank 6: 0.200 (Z).

Wait, let's fill the blanks:

Blank 2: E (4.00)

Blank 4: 4 (but no option, maybe the problem's "1 mol = __1__ osmol" is a mistake, and it's 1 mol = 4 osmoles, so blank 1: 4, but no option. Alternatively, maybe the dissociation is 3 ions, but no: \( \text{Na}_3\text{PO}_4 \) dissociates into 3 \( \text{Na}^+ \) and 1 \( \text{PO}_4^{3-} \), so 4 particles. So osmolarity is 4 times molarity. So to get moles from osmoles: moles = osmoles / 4. So 4.00 osmoles / 4 = 1.00 mol. Then liters: 1.00 mol / 5.00 M = 0.200 L.

So:

Blank 2: E (4.00)

Blank 3: B (1.00)

Blank 4: 4 (but no option, maybe the problem meant 1 mol = 1 osmol, which is wrong, but maybe the question has a mistake. Alternatively, maybe the conversion factor is 1 mol = 4 osmoles, so blank 4: 4, but no option. Then blank 5: B (1.00), blank 7: F (5.00), blank 6: Z (0.200).

Step2: Calculate volume from moles and molarity

Molarity formula: \( M = \frac{n}{V} \) → \( V = \frac{n}{M} \).

n (moles) = 1.00 mol (from step 1), M = 5.00 mol/L.

\( V = \frac{1.00 \, \text{mol}}{5.00 \, \text{mol/L}} = 0.200 \, \text{L} \).

Answer:

Blank 2: E. 4.00
Blank 3: B. 1.00
Blank 4: (No option, but intended 4)
Blank 5: B. 1.00
Blank 7: F. 5.00
Blank 6: Z. 0.200

(Assuming the main answer for the final liters is Z. 0.200)