Question

la concentration
19 remplissez le tableau ci-dessous.
soluté | masse du soluté (en g) | volume de solution (en l) | concentration (en g/l) | concentration (en % m/v) | concentration (en mol/l)
na₃po₄ | 25,0 | 1,5 | | |
kcl | 2,80 | | 11,2 | |
h₂so₄ | | 0,0320 | | |
agno₃ | | 1,20 | | 1,05 | 1,47

Answer:

To solve for the missing values in the table, we use the formulas for concentration:

1. For $\boldsymbol{\ce{Na3PO4}}$:

Step 1: Concentration (g/L)

Concentration (g/L) = $\frac{\text{Mass of solute (g)}}{\text{Volume of solution (L)}}$
$= \frac{25.0\ \text{g}}{1.5\ \text{L}} \approx 16.67\ \text{g/L}$

Step 2: Concentration (% m/V)

% m/V = $\frac{\text{Mass of solute (g)}}{\text{Volume of solution (mL)}} \times 100$ (Note: $1.5\ \text{L} = 1500\ \text{mL}$)
$= \frac{25.0\ \text{g}}{1500\ \text{mL}} \times 100 \approx 1.67\%$

Step 3: Concentration (mol/L, Molarity)

Molar mass of $\ce{Na3PO4}$: $3(23.0) + 31.0 + 4(16.0) = 164.0\ \text{g/mol}$
Molarity (mol/L) = $\frac{\text{Mass (g)}}{\text{Molar mass (g/mol)} \times \text{Volume (L)}}$
$= \frac{25.0\ \text{g}}{164.0\ \text{g/mol} \times 1.5\ \text{L}} \approx 0.102\ \text{mol/L}$

2. For $\boldsymbol{\ce{KCl}}$:

Step 1: Volume of solution (L)

From concentration (g/L) = $\frac{\text{Mass (g)}}{\text{Volume (L)}}$, rearrange:
Volume (L) = $\frac{\text{Mass (g)}}{\text{Concentration (g/L)}}$
$= \frac{2.80\ \text{g}}{11.2\ \text{g/L}} = 0.25\ \text{L}$

Step 2: Concentration (% m/V)

Volume = $0.25\ \text{L} = 250\ \text{mL}$
% m/V = $\frac{2.80\ \text{g}}{250\ \text{mL}} \times 100 = 1.12\%$

Step 3: Molarity (mol/L)

Molar mass of $\ce{KCl}$: $39.1 + 35.5 = 74.6\ \text{g/mol}$
Molarity = $\frac{2.80\ \text{g}}{74.6\ \text{g/mol} \times 0.25\ \text{L}} \approx 0.147\ \text{mol/L}$

3. For $\boldsymbol{\ce{H2SO4}}$:

Step 1: Mass of solute (g)

From concentration (g/L) = $\frac{\text{Mass (g)}}{\text{Volume (L)}}$, but concentration (g/L) is not given. Wait, the table has volume = $0.0320\ \text{L}$. Wait, maybe we assume concentration (g/L) is same as KCl? No, better: Wait, the table’s $\ce{H2SO4}$ row has volume $0.0320\ \text{L}$. Let’s check the pattern. Wait, maybe the concentration (g/L) is missing, but let's use molarity? Wait, no, the table has empty cells. Wait, perhaps the original problem expects using the given columns. Let's re-express:

Wait, the table for $\ce{H2SO4}$: Volume = $0.0320\ \text{L}$. Let's assume we need to find mass (g) from concentration (g/L) (but it’s empty). Wait, maybe a typo, but let's proceed with molarity. Molar mass of $\ce{H2SO4}$: $2(1.0) + 32.1 + 4(16.0) = 98.1\ \text{g/mol}$. If we assume molarity is missing, but the table has no value. Wait, maybe the original problem has more context. Alternatively, perhaps the user expects filling the table with these calculations.

4. For $\boldsymbol{\ce{AgNO3}}$:

Step 1: Mass of solute (g)

From % m/V = $\frac{\text{Mass (g)}}{\text{Volume (mL)}} \times 100$, Volume = $1.20\ \text{L} = 1200\ \text{mL}$
Mass (g) = $\frac{\% \text{ m/V} \times \text{Volume (mL)}}{100}$
$= \frac{1.05 \times 1200\ \text{mL}}{100} = 12.6\ \text{g}$

Step 2: Concentration (g/L)

$= \frac{12.6\ \text{g}}{1.20\ \text{L}} = 10.5\ \text{g/L}$

Step 3: Molarity (mol/L) is given as $1.47\ \text{mol/L}$. Verify:

Molar mass of $\ce{AgNO3}$: $107.9 + 14.0 + 3(16.0) = 169.9\ \text{g/mol}$
Molarity = $\frac{12.6\ \text{g}}{169.9\ \text{g/mol} \times 1.20\ \text{L}} \approx 0.061\ \text{mol/L}$ – Wait, discrepancy! Maybe the % m/V is $1.05\%$, so:

Mass (g) = $1.05\% \times 1200\ \text{mL} = 12.6\ \text{g}$ (correct). Molarity = $\frac{12.6\ \text{g}}{169.9\ \text{g/mol} \times 1.20\ \text{L}} \approx 0.061\ \text{mol/L}$, but the table says $1.47\ \text{mol/L}$. So maybe the % m/V is not $1.05$ for molarity. Wait, the table has two columns: % m/V (1.05) and molarity (1.47). Let's use molarity to find mass:

Molarity = $\frac{\text{Mass (g)}}{\text{Molar mass (g/mol)} \times \text{Volume (L)}}$
Mass (g) = Molarity $\times$ Molar mass $\times$ Volume
$= 1.47\ \text{mol/L} \times 169.9\ \text{g/mol} \times 1.20\ \text{L} \approx 299\ \text{g}$ (too big). Clearly, a mistake in the table, but we proceed with the given % m/V:

Filled Table (Key Values):

Soluté	Masse (g)	Volume (L)	Concentration (g/L)	Concentration (% m/V)	Concentration (mol/L)
$\ce{KCl}$	2.80	$0.25$	$11.2$	$1.12$	$0.147$
$\ce{H2SO4}$	(depends)	$0.0320$	(depends)	(depends)	(depends)
$\ce{AgNO3}$	$12.6$	$1.20$	$10.5$	$1.05$	(discrepancy)

Final Answers (Key Calculations):

• $\ce{Na3PO4}$: g/L = $\boldsymbol{16.67}$, % m/V = $\boldsymbol{1.67}$, mol/L = $\boldsymbol{0.102}$
• $\ce{KCl}$: Volume = $\boldsymbol{0.25\ \text{L}}$, % m/V = $\boldsymbol{1.12}$, mol/L = $\boldsymbol{0.147}$
• $\ce{AgNO3}$: Mass = $\boldsymbol{12.6\ \text{g}}$, g/L = $\boldsymbol{10.5}$

(Note: Some values may vary slightly due to rounding. The table’s $\ce{AgNO3}$ molarity has a discrepancy, likely a typo.)