Q. \( \text{molar mass } \mathrm{BaCl_2} \)

Q: What is the molar mass of \mathrm{BaCl_2} ?

\mathrm{BaCl_2} molar mass = M(\mathrm{Ba})+2M(\mathrm{Cl}) \approx 137.33+2(35.45)=208.23\ \mathrm{g/mol}.

Q: How do you calculate M(\mathrm{BaCl_2}) step-by-step?

Use atomic masses: M(\mathrm{BaCl_2})=1\cdot M(\mathrm{Ba})+2\cdot M(\mathrm{Cl}). Substitute 137.33 and 35.45\ \mathrm{g/mol}, then add: 137.33+70.90=208.23\ \mathrm{g/mol}.

Q: Which atomic mass values should be used for \mathrm{Ba} and \mathrm{Cl} ?

Common periodic table values: M(\mathrm{Ba}) \approx 137.33\ \mathrm{g/mol} and M(\mathrm{Cl}) \approx 35.45\ \mathrm{g/mol}.

Q: What is the molar mass if I use different rounding?

If you round differently (example: M(\mathrm{Ba})=137.3, M(\mathrm{Cl})=35.5), then M\approx 137.3+2(35.5)=208.3\ \mathrm{g/mol}. Small differences come from rounding.

Q: How much mass of \mathrm{BaCl_2} is in 0.500 mol?

m=nM=(0.500)(208.23)=104.12\ \mathrm{g}. Use M=208.23\ \mathrm{g/mol} for consistency.

Q: How do you find moles from a given mass of \mathrm{BaCl_2} ?

n=\dfrac{m}{M}. For example, if m=20.0\ \mathrm{g}, then n=20.0/208.23=0.0961\ \mathrm{mol}.

Q: What is the molar mass of \mathrm{Ba^{2+}} and \mathrm{Cl^-} in \mathrm{BaCl_2} ?

M(\mathrm{Ba^{2+}}) uses atomic Ba mass: 137.33\ \mathrm{g/mol}. M(\mathrm{Cl^-}) equals atomic Cl mass: 35.45\ \mathrm{g/mol}. Total for \mathrm{BaCl_2} is 137.33+2(35.45).

Answer

To find the molar mass of \( \text{BaCl}_2 \), add the molar masses of its atoms.

\[ M(\text{BaCl}_2)=M(\text{Ba})+2M(\text{Cl}) \]

Using \( M(\text{Ba})=137.33 \,\text{g/mol} \) and \( M(\text{Cl})=35.45 \,\text{g/mol} \):

\[ M(\text{BaCl}_2)=137.33+2(35.45)=137.33+70.90=208.23 \,\text{g/mol} \]

Final result: \( 208.23 \,\text{g/mol} \)

Detailed Explanation

We want the molar mass of barium chloride, which has the formula \( \text{BaCl}_2 \).

Step 1: Identify the elements in the formula.

The formula \( \text{BaCl}_2 \) contains:

\(1\) atom of \( \text{Ba} \)
\(2\) atoms of \( \text{Cl} \)

Step 2: Get the atomic masses (molar masses of elements).

Use standard atomic masses:

\( \text{Ba} \approx 137.33 \ \text{g/mol} \)
\( \text{Cl} \approx 35.45 \ \text{g/mol} \)

Step 3: Multiply each atomic mass by how many atoms there are.

Barium contribution:

\[
1 \times 137.33 = 137.33 \ \text{g/mol}
\]

Chlorine contribution (because there are \(2\) chlorines):

\[
2 \times 35.45 = 70.90 \ \text{g/mol}
\]

Step 4: Add the contributions to get the molar mass of \( \text{BaCl}_2 \).

\[
137.33 + 70.90 = 208.23 \ \text{g/mol}
\]

Final Answer:

The molar mass of \( \text{BaCl}_2 \) is \( \boxed{208.23 \ \text{g/mol}} \).

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General Chemistry FAQs

What is the molar mass of \( \mathrm{BaCl_2} \)?

\(\mathrm{BaCl_2}\) molar mass \(= M(\mathrm{Ba})+2M(\mathrm{Cl}) \approx 137.33+2(35.45)=208.23\ \mathrm{g/mol}\).

How do you calculate \( M(\mathrm{BaCl_2}) \) step-by-step?

Use atomic masses: \(M(\mathrm{BaCl_2})=1\cdot M(\mathrm{Ba})+2\cdot M(\mathrm{Cl})\). Substitute \(137.33\) and \(35.45\ \mathrm{g/mol}\), then add: \(137.33+70.90=208.23\ \mathrm{g/mol}\).

Which atomic mass values should be used for \( \mathrm{Ba} \) and \( \mathrm{Cl} \)?

Common periodic table values: \(M(\mathrm{Ba}) \approx 137.33\ \mathrm{g/mol}\) and \(M(\mathrm{Cl}) \approx 35.45\ \mathrm{g/mol}\).

What is the molar mass if I use different rounding?

If you round differently (example: \(M(\mathrm{Ba})=137.3\), \(M(\mathrm{Cl})=35.5\)), then \(M\approx 137.3+2(35.5)=208.3\ \mathrm{g/mol}\). Small differences come from rounding.

How much mass of \( \mathrm{BaCl_2} \) is in 0.500 mol?

\(m=nM=(0.500)(208.23)=104.12\ \mathrm{g}\). Use \(M=208.23\ \mathrm{g/mol}\) for consistency.

How do you find moles from a given mass of \( \mathrm{BaCl_2} \)?

\(n=\dfrac{m}{M}\). For example, if \(m=20.0\ \mathrm{g}\), then \(n=20.0/208.23=0.0961\ \mathrm{mol}\).

What is the molar mass of \( \mathrm{Ba^{2+}} \) and \( \mathrm{Cl^-} \) in \( \mathrm{BaCl_2} \)?

\(M(\mathrm{Ba^{2+}})\) uses atomic Ba mass: \(137.33\ \mathrm{g/mol}\). \(M(\mathrm{Cl^-})\) equals atomic Cl mass: \(35.45\ \mathrm{g/mol}\). Total for \( \mathrm{BaCl_2} \) is \(137.33+2(35.45)\).

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