Question

how many equivalents in 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 equivalent is a mole of charges, either + or -, but not both, they are not added
the conversion factor is 1 mol = 1 eq
first calculate the moles
2 mol
(3 l)(---------) = 1.00 x 4 mol
1 l
then convert into equivalents
5 eq
(1.00 x 6 mol)(---------)= 7 eq
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. 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 the conversion factor for equivalents

For \( \text{Na}_2\text{SO}_4 \), when it dissociates: \( \text{Na}_2\text{SO}_4 ightarrow 2\text{Na}^+ + \text{SO}_4^{2-} \). The total positive charge per mole of \( \text{Na}_2\text{SO}_4 \) is \( 2\times1 = 2 \) (from \( \text{Na}^+ \)) and negative charge is \( 2 \) (from \( \text{SO}_4^{2-} \)). But an equivalent is a mole of charge (either + or -). For \( \text{Na}_2\text{SO}_4 \), the number of equivalents per mole: for cations, each \( \text{Na}^+ \) is +1, so 2 \( \text{Na}^+ \) give 2 equivalents of positive charge. So 1 mol of \( \text{Na}_2\text{SO}_4 \) has 2 eq (based on cation charge) or 2 eq (based on anion charge, since \( \text{SO}_4^{2-} \) is -2, so 1 mol of \( \text{SO}_4^{2-} \) is 2 eq). Wait, the problem says "the conversion factor is 1 mol = __1__ eq" – maybe I misread. Wait, no, let's recalculate moles first.

Step2: Calculate moles of \( \text{Na}_2\text{SO}_4 \)

Molarity \( M = \frac{\text{moles}}{\text{volume (L)}} \), so moles \( n = M \times V \). Given \( M = 5.00 \, \text{M} \), \( V = 2.00 \, \text{L} \). So \( n = 5.00 \, \text{mol/L} \times 2.00 \, \text{L} = 10.00 \, \text{mol} \). Wait, the problem has a typo? Wait, the problem's step says "(__3__ L)(---------) = 1.00 x __4__ mol" – maybe the given M is 5.00, but the calculation in the problem is written as (2.00 L)(5.00 mol/L) = 10.00 mol, but the problem wrote "1.00 x __4__" – maybe a typo, but let's proceed.

Step3: Determine equivalents per mole

From dissociation: \( \text{Na}_2\text{SO}_4 ightarrow 2\text{Na}^+ + \text{SO}_4^{2-} \). The charge per \( \text{Na}_2\text{SO}_4 \): for \( \text{Na}^+ \), each is +1, so 2 \( \text{Na}^+ \) give 2 equivalents of positive charge. So 1 mol of \( \text{Na}_2\text{SO}_4 \) has 2 equivalents (since 2 moles of \( \text{Na}^+ \) with +1 each: 2 eq). So conversion factor: 1 mol \( \text{Na}_2\text{SO}_4 \) = 2 eq.

Step4: Convert moles to equivalents

If moles of \( \text{Na}_2\text{SO}_4 \) is 10.00 mol (from 5.00 M * 2.00 L = 10.00 mol), then equivalents = moles * equivalents per mole. Since 1 mol \( \text{Na}_2\text{SO}_4 \) = 2 eq, then equivalents = 10.00 mol * 2 eq/mol = 20.0 eq? Wait, no, the problem's options have AA. 20.0. Wait, let's check the problem's steps again.

Wait, the problem's first step: "the conversion factor is 1 mol = __1__ eq" – maybe the problem is considering that for \( \text{Na}_2\text{SO}_4 \), the number of equivalents is based on the total charge. Wait, no, let's re-express:

Wait, the problem's step: "First calculate the moles" – (__3__ L)(---------) = 1.00 x __4__ mol. So __3__ is 2.00 L, the concentration is 5.00 mol/L, so (2.00 L)(5.00 mol/L) = 10.00 mol, so __4__ is 10^1 (since 10.00 = 1.00 x 10^1). Then "Then convert into equivalents" – (1.00 x __6__ mol)(---------) = __7__ eq. __6__ is 10^1 (since moles are 10.00 = 1.00 x 10^1). Then the conversion factor: 1 mol = __5__ eq. From dissociation, 1 mol \( \text{Na}_2\text{SO}_4 \) gives 2 eq (as 2 \( \text{Na}^+ \) or 2 eq from \( \text{SO}_4^{2-} \)). So __5__ is 2.00 (option C). Then (1.00 x 10^1 mol)(2.00 eq/1 mol) = 2.00 x 10^1 = 20.0 eq (option AA).

Wait, let's fill the blanks:

__1__: For \( \text{Na}_2\text{SO}_4 \), the number of equivalents per mole. From \( \text{Na}_2\text{SO}_4 ightarrow 2\text{Na}^+ + \text{SO}_4^{2-} \), the charge on \( \text{Na}^+ \) is +1, so 2 \( \text{Na}^+ \) contribute 2 equivalents of positive charge. So 1 mol \( \text{Na}_2\text{SO}_4 \) = 2 eq. So __1__ is C (2.00).

__3__: Volume is 2.00 L, so __3__ is 2.00 (option C? Wait, options for __3__: the first blank. Wait, the options are A to Z, AA, BB, etc. Let's list the blanks:

Blank 1: 1 mol = __1__ eq → answer C (2.00)

Blank 3: volume is 2.00 L → so __3__ is 2.00 (option C)

Blank 4: moles = 5.00 * 2.00 = 10.00 = 1.00 x 10^1 → so __4__ is W (10^1)

Blank 6: moles are 10.00 = 1.00 x 10^1 → __6__ is W (10^1)

Blank 5: conversion factor 2.00 eq/1 mol → __5__ is C (2.00)

Blank 7: (1.00 x 10^1 mol)(2.00 eq/1 mol) = 20.0 eq → __7__ is AA (20.0)

But let's check the problem's steps again. The problem says:

"First calculate the moles

__2__ mol

(__3__ L)(---------) = 1.00 x __4__ mol

1 L

Then convert into equivalents

__5__ eq

(1.00 x __6__ mol)(---------) = __7__ eq

1 mol"

So:

- __3__: volume is 2.00 L → so __3__ is 2.00 (option C)

- The concentration is 5.00 mol/L, so (2.00 L)(5.00 mol/L) = 10.00 mol = 1.00 x 10^1 mol → so __4__ is W (10^1)

- __2__: moles are 10.00 mol → but the options for __2__? Wait, the options are 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:10^2, L:10^3, etc. So __2__ is I (10.00)

- __5__: conversion factor: 1 mol \( \text{Na}_2\text{SO}_4 \) has 2 eq (from 2 \( \text{Na}^+ \) or 2 eq from \( \text{SO}_4^{2-} \)) → so __5__ is C (2.00)

- __6__: moles are 10.00 = 1.00 x 10^1 → __6__ is W (10^1)

- __7__: (1.00 x 10^1 mol)(2.00 eq/1 mol) = 20.0 eq → __7__ is AA (20.0)

So filling the blanks:

__1__: C (2.00)

__2__: I (10.00)

__3__: C (2.00)

__4__: W (10^1)

__5__: C (2.00)

__6__: W (10^1)

__7__: AA (20.0)

But the question is "How many equivalents in are in 2.00 L of a 5.00 M solution of \( \text{Na}_2\text{SO}_4 \)?"

Moles of \( \text{Na}_2\text{SO}_4 \): \( n = 5.00 \, \text{mol/L} \times 2.00 \, \text{L} = 10.00 \, \text{mol} \)

Equivalents: For \( \text{Na}_2\text{SO}_4 \), each \( \text{Na}^+ \) is +1, so 2 \( \text{Na}^+ \) per \( \text{Na}_2\text{SO}_4 \), so equivalents from \( \text{Na}^+ \): 2 eq per mol of \( \text{Na}_2\text{SO}_4 \). So equivalents = 10.00 mol × 2 eq/mol = 20.0 eq.

Answer:

The number of equivalents is 20.0, so the answer for the final blank (7) is AA. 20.0 (AA. 20.0)