Question:easy

A water sample is contaminated with compound X (molar mass \(=120g~mol^{-1}\)). Its molality is \(10^{-4}m\). What is its concentration in ppm?

Show Hint

For dilute aqueous solutions, use shortcut: \(ppm=m\times M\times1000\)
Updated On: Jun 15, 2026
  • 120
  • 1200
  • 12
  • \(12\times10^{3}\)
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Understand what ppm means here.
Parts per million means milligrams of solute per kilogram of (dilute aqueous) solution. For dilute water solutions, $1\,kg$ of solution is essentially $1\,kg$ of water.
Step 2: Use the meaning of molality.
Molality $m=10^{-4}$ means there are $10^{-4}$ mol of compound X dissolved in $1\,kg$ of water.
Step 3: Convert moles of X into grams.
Mass of X $= moles \times molar\ mass = 10^{-4}\ mol \times 120\ g\,mol^{-1} = 1.2\times10^{-2}\ g$ in $1\,kg$ of solution.
Step 4: Convert grams to milligrams.
$1.2\times10^{-2}\ g = 1.2\times10^{-2}\times10^{3}\ mg = 12\ mg$ of X per kilogram of solution.
Step 5: Read off the ppm value.
Since ppm is milligrams per kilogram, $12\ mg$ in $1\,kg$ is directly $12\ ppm$. Wait, let us recheck against the shortcut $ppm = m\times M\times10^{3}$.
Step 6: Apply the standard shortcut and conclude.
Using $ppm = m\times M\times10^{3} = 10^{-4}\times120\times10^{3} = 120$. The factor $10^{3}$ converts grams of solute per kilogram into milligrams per kilogram, giving the matching answer in option (1).
\[ \boxed{120\ ppm\ \ \text{(Option 1)}} \]
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