Units of parts per million (ppm) or per billion (ppb) are often used to describe the concentrations of solutes in very dilute solutions. The units are defined as the number of grams of solution per million or per billion gram of solvent. Bay of Bengal has 2.1 ppm of lithium ions. What is the molality of `Li^(+)` in this water ? (Li = 7)

Answer

To find the molality of lithium ions (Li⁺) in the Bay of Bengal water, we can follow these steps:

### Step 1: Understand the definition of ppm
Parts per million (ppm) means that there are 2.1 grams of lithium ions in 1,000,000 grams of water (solvent).

### Step 2: Convert grams of solvent to kilograms
Since molality is defined as moles of solute per kilogram of solvent, we need to convert 1,000,000 grams of solvent to kilograms:
\[ 1,000,000 \text{ grams} = 1000 \text{ kg} \]

### Step 3: Calculate the mass of lithium in 1 kg of solvent
Since 2.1 ppm means that in 1,000,000 grams of solvent there are 2.1 grams of lithium ions, we can find the mass of lithium in 1 kg (1000 grams) of solvent:
\[ \text{Mass of Li in 1 kg of solvent} = \frac{2.1 \text{ grams}}{1,000,000 \text{ grams}} \times 1000 \text{ grams} = 2.1 \times 10^{-3} \text{ grams} \]

### Step 4: Convert mass of lithium to moles
To find the number of moles of lithium ions, we use the formula:
\[ \text{Moles of Li} = \frac{\text{mass of Li}}{\text{molar mass of Li}} = \frac{2.1 \times 10^{-3} \text{ grams}}{7 \text{ g/mol}} \]
Calculating this gives:
\[ \text{Moles of Li} = \frac{2.1 \times 10^{-3}}{7} \approx 3 \times 10^{-4} \text{ moles} \]

### Step 5: Calculate molality
Now that we have the moles of lithium ions and the mass of the solvent in kilograms, we can calculate the molality:
\[ \text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{3 \times 10^{-4} \text{ moles}}{1 \text{ kg}} = 3 \times 10^{-4} \text{ mol/kg} \]

### Final Answer
The molality of lithium ions (Li⁺) in the Bay of Bengal water is:
\[ \text{Molality} = 3 \times 10^{-4} \text{ mol/kg} \]

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