1. Given data

• Mass of organic compound $= 0.172\,\text{g}$
• Mass of bromine present $= 0.08\,\text{g}$

We first calculate the percentage of bromine in the compound.

$\%\,\text{Br} = \frac{0.08}{0.172}\times 100$

$\%\,\text{Br} \approx 46.51\%$

So, the compound contains about $46.5\%$ bromine by mass.

2. Check option A: $\mathrm{CH_3CH_2Br}$ (bromoethane)

Molar mass of $\mathrm{CH_3CH_2Br}$:

$M = 2(12) + 5(1) + 80 = 24 + 5 + 80 = 109$

Percentage of bromine:

$\%\,\text{Br} = \frac{80}{109}\times 100 \approx 73.39\%$

This does not match $46.5\%$.

So, Option A is incorrect.

3. Check option B: $\mathrm{CH_3Br}$ (bromomethane)

Molar mass of $\mathrm{CH_3Br}$:

$M = 12 + 3 + 80 = 95$

Percentage of bromine:

$\%\,\text{Br} = \frac{80}{95}\times 100 \approx 84.21\%$

This also does not match $46.5\%$.

So, Option B is incorrect.

4. Find likely molecular mass from bromine percentage

If the compound contains one bromine atom, then

$\frac{80}{M}\times 100 = 46.51$

$M = \frac{80\times 100}{46.51} \approx 172$

So the molar mass of the compound should be about $172$.

A monobromo compound with molar mass $\approx 172$ would be:

$\mathrm{C_6H_5Br}$

since its molar mass is

$6(12) + 5(1) + 80 = 72 + 5 + 80 = 157$

Not enough.

Try dibromo compound with two bromines:

$\frac{160}{M}\times 100 = 46.51$

$M \approx 344$

This is unlikely for the simple options shown.

Now check a compound like $\mathrm{C_6H_5CH_2Br}$:

$M = 7(12) + 7(1) + 80 = 84 + 7 + 80 = 171$

Percentage of bromine:

$\%\,\text{Br} = \frac{80}{171}\times 100 \approx 46.78\%$

This matches the observed value very closely.

So the compound is benzyl bromide, $\mathrm{C_6H_5CH_2Br}$.

5. Conclusion

The correct structure must be the one corresponding to benzyl bromide.

Hence, the correct option is C.

6. Comparison with stored correct answer

Stored correct answer: C

My derived answer: C

So, they agree.