Question

how many osmoles are in 2.00 l of a 5.00 m solution of na₂so₄ ?
first we need the equation for dissolving the salt, na₂so₄ → 2 na⁺¹ + so₄⁻²
an osmole is a mole of dissolved particles
the conversion factor is 1 mol = 1 osmol
first calculate the moles
2 mol
(3 l)(---------—) = 1.00 x 4 mol
1 l
then convert into osmoles
5 osmol
(1.00 x 6 mol)(-------------) = 7 osmoles
1 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. 12.00 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: Calculate moles of \( Na_2SO_4 \)

Molarity (\( M \)) formula: \( M = \frac{n}{V} \), so \( n = M \times V \). Given \( M = 5.00 \, M \), \( V = 2.00 \, L \).
\( n = 5.00 \, \frac{mol}{L} \times 2.00 \, L = 10.0 \, mol \) (matches \( 1.00 \times 10^1 \, mol \), so \( 4 = W \) (since \( 10^1 \) is \( W \)), \( 3 = C \) (2.00 L), \( 2 = F \) (5.00 mol/L)).

Step2: Determine osmoles per mole of \( Na_2SO_4 \)

Dissociation: \( Na_2SO_4
ightarrow 2Na^+ + SO_4^{2-} \), so 1 mol \( Na_2SO_4 \) gives \( 2 + 1 = 3 \) mol of particles. Thus, 1 mol \( Na_2SO_4 = 3 \) osmol (so \( 1 = D \)).

Step3: Convert moles to osmoles

Moles of \( Na_2SO_4 = 10.0 \, mol \) (from Step1). Osmoles = moles × osmoles per mole.
\( 10.0 \, mol \times 3 \, \frac{osmol}{mol} = 30.0 \, osmol \) (so \( 7 = X \), \( 6 = W \) (10.0 mol = \( 1.00 \times 10^1 \, mol \)), \( 5 = D \) (3.00 osmol per mol)).

Answer:

1. D (1 mol = 3 osmol)
2. F (5.00 mol/L)
3. C (2.00 L)
4. W (\( 10^1 \))
5. D (3.00 osmol/mol)
6. W (\( 10^1 \) mol)
7. X (30.0 osmoles)

(Note: The question asks for the number of osmoles, so the final answer for total osmoles is 30.0, corresponding to option X.)