Q. \(5\,\text{M}\) \(\text{NaCl}\) molarity.

Q: If I have 5\text{ M NaCl}, what does “molarity 5\text{ M}” mean?

It means 5 moles of NaCl dissolved per 1\text{ L} of solution: M=\dfrac{n}{V}. So n=5 mol when V=1 L.

Q: How many grams of NaCl are in 1\text{ L} of 5\text{ M} solution?

Use m=n\cdot M_r. With M_r(\text{NaCl})\approx 58.44\text{ g/mol}: m=5\times 58.44\approx 292.2\text{ g}.

Q: How many moles of NaCl are needed to make 250\text{ mL} of 5\text{ M} NaCl?

n=M\cdot V. Convert 250\text{ mL}=0.250\text{ L}: n=5\times 0.250=1.25\text{ mol}.

Q: What mass of NaCl is required for 250\text{ mL} of 5\text{ M} solution?

First find moles: 1.25\text{ mol}. Then m=n\cdot M_r=1.25\times 58.44\approx 73.05\text{ g}.

Q: What is the total molar concentration of ions in 5\text{ M} NaCl (assuming complete dissociation)?

NaCl dissociates as \text{NaCl}\rightarrow \text{Na}^+ + \text{Cl}^-. So [\text{Na}^+]=5\text{ M} and [\text{Cl}^-]=5\text{ M}; total ions \approx 10\text{ M}.

Q: What is the mass percent of NaCl if the solution density is 1.25\text{ g/mL}?

For 1\text{ L}, mass of solution =1.25\text{ g/mL}\times 1000\text{ mL}=1250\text{ g}. NaCl mass \approx 292.2\text{ g}. Mass percent =\dfrac{292.2}{1250}\times 100\approx 23.4\%.

Answer

Problem: “5 m NaCl molarity.”

Assuming “5 m” means a 5 molal solution (5 m NaCl), convert molality (m) to molarity (M).

For NaCl, take molar mass \(=58.44\ \text{g/mol}\). Also use the usual density of a 5 m NaCl solution (commonly about \( \rho \approx 1.19\ \text{g/mL}\)). Start with \(1.00\ \text{kg}\) of water:

\[
\text{moles NaCl} = 5
\]
\[
\text{mass NaCl} = 5 \times 58.44 = 292.2\ \text{g}
\]

Total mass of solution \(= 1000 + 292.2 = 1292.2\ \text{g}\).

Volume of solution:
\[
V = \frac{1292.2\ \text{g}}{1.19\ \text{g/mL}} \approx 1086.4\ \text{mL} = 1.0864\ \text{L}
\]

Molarity:
\[
M = \frac{5\ \text{mol}}{1.0864\ \text{L}} \approx 4.60\ \text{M}
\]

Final result: \(M \approx 4.6\ \text{M NaCl}\).

Detailed Explanation

Step 1: Clarify what “5M NaCl” means

\( \text{“5M NaCl”}\) means sodium chloride has a molarity of \(5\) moles per liter.

Step 2: State the definition of molarity

Molarity is defined as:

\[
M = \frac{n}{V}
\]

where:

\(M\) is molarity (units: \(\text{mol/L}\))
\(n\) is the number of moles of solute
\(V\) is the volume of the solution in liters

Step 3: Plug in the given molarity

Since the problem gives \(M = 5\), we interpret that as:

\[
5 = \frac{n}{V}
\]

So the relationship between moles and volume is:

\[
n = 5V
\]

with \(V\) in liters.

Step 4: Compute an example if the question is asking for moles

If you want to know how many moles of NaCl are in a certain volume, use \(n = 5V\).

If \(V = 1.0\ \text{L}\), then:

\[
n = 5(1.0) = 5.0\ \text{mol}
\]
If \(V = 0.50\ \text{L}\), then:

\[
n = 5(0.50) = 2.5\ \text{mol}
\]

Step 5: If the real goal is grams of NaCl, convert using molar mass

If you need the mass required to make a \(5.0\ \text{M}\) NaCl solution, use the molar mass of NaCl.

The molar mass of NaCl is:

\[
M_r(\text{NaCl}) = 58.44\ \text{g/mol}
\]

Mass is calculated by:

\[
m = n \times 58.44
\]

and since \(n = 5V\):

\[
m = (5V)(58.44)
\]

So:

\[
m = 292.2V
\]

where \(m\) is in grams and \(V\) is in liters.

To make \(1.0\ \text{L}\) of \(5.0\ \text{M}\) NaCl:

\[
m = 292.2(1.0) = 292.2\ \text{g}
\]
To make \(0.50\ \text{L}\) of \(5.0\ \text{M}\) NaCl:

\[
m = 292.2(0.50) = 146.1\ \text{g}
\]

Final Answer

\( \text{5 M NaCl means } 5\ \text{mol of NaCl per 1.0 L of solution.} \)

Equivalently, for any volume \(V\) (in liters): \( \ n = 5V\).

If you need mass: \( \ m = 292.2V\ \text{g}\) (since molar mass of NaCl is \(58.44\ \text{g/mol}\)).

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General Chemistry FAQs

If I have \(5\text{ M NaCl}\), what does “molarity \(5\text{ M}\)” mean?

It means \(5\) moles of NaCl dissolved per \(1\text{ L}\) of solution: \(M=\dfrac{n}{V}\). So \(n=5\) mol when \(V=1\) L.

How many grams of NaCl are in \(1\text{ L}\) of \(5\text{ M}\) solution?

Use \(m=n\cdot M_r\). With \(M_r(\text{NaCl})\approx 58.44\text{ g/mol}\): \(m=5\times 58.44\approx 292.2\text{ g}\).

How many moles of NaCl are needed to make \(250\text{ mL}\) of \(5\text{ M}\) NaCl?

\(n=M\cdot V\). Convert \(250\text{ mL}=0.250\text{ L}\): \(n=5\times 0.250=1.25\text{ mol}\).

What mass of NaCl is required for \(250\text{ mL}\) of \(5\text{ M}\) solution?

First find moles: \(1.25\text{ mol}\). Then \(m=n\cdot M_r=1.25\times 58.44\approx 73.05\text{ g}\).

What is the total molar concentration of ions in \(5\text{ M}\) NaCl (assuming complete dissociation)?

NaCl dissociates as \(\text{NaCl}\rightarrow \text{Na}^+ + \text{Cl}^-\). So \([\text{Na}^+]=5\text{ M}\) and \([\text{Cl}^-]=5\text{ M}\); total ions \(\approx 10\text{ M}\).

What is the mass percent of NaCl if the solution density is \(1.25\text{ g/mL}\)?

For \(1\text{ L}\), mass of solution \(=1.25\text{ g/mL}\times 1000\text{ mL}=1250\text{ g}\). NaCl mass \(\approx 292.2\text{ g}\). Mass percent \(=\dfrac{292.2}{1250}\times 100\approx 23.4\%\).

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