Find the weight of sodium nitrate required to prepare 60 g pure crystals from its saturated solution at `70^@C`. Solubility of sodium nitrate is 140 g at `70^@C " and 100 g at " 25^@C`.

Explanation

To find the weight of sodium nitrate (NaNO3) required to prepare 60 g of pure crystals from its saturated solution at 70°C, we can follow these steps:

### Step 1: Understand the solubility at different temperatures
The solubility of sodium nitrate at 70°C is 140 g per 100 g of water, and at 25°C, it is 100 g per 100 g of water. This means that at 70°C, a saturated solution can hold 140 g of NaNO3, and when cooled to 25°C, it can only hold 100 g.

### Step 2: Calculate the amount of sodium nitrate that crystallizes when cooled
When the solution is cooled from 70°C to 25°C, the difference in solubility will give us the amount of sodium nitrate that crystallizes out:
\[ \text{Crystallized NaNO3} = \text{Solubility at 70°C} - \text{Solubility at 25°C} = 140 \, \text{g} - 100 \, \text{g} = 40 \, \text{g} \]

### Step 3: Set up a proportion to find the amount of NaNO3 needed for 60 g of crystals
We know that 140 g of NaNO3 yields 40 g of crystals. We need to find out how much NaNO3 is required to obtain 60 g of crystals. We can set up a proportion:
\[ \frac{140 \, \text{g NaNO3}}{40 \, \text{g crystals}} = \frac{x \, \text{g NaNO3}}{60 \, \text{g crystals}} \]

### Step 4: Solve for x
Cross-multiply to solve for x:
\[ 140 \times 60 = 40 \times x \]
\[ 8400 = 40x \]
\[ x = \frac{8400}{40} = 210 \, \text{g} \]

### Conclusion
Thus, the weight of sodium nitrate required to prepare 60 g of pure crystals from its saturated solution at 70°C is **210 g**.

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