Comprehension # 1
Potash is any potassium mineral that is used for its potassium content. Most of the potash produced in the United States goes into fertilizer. The major sources of potash are potassium choride `(KCI)` and potassium sulphate `(K_(2)SO_(4))`. Potash production is often reported as the potassium oxide `(K_(2)O)` equivalent or the amount of `K_(2)O` that could be made from a given mineral. `KCI` cost `Rs 50 per kg`
For what price must `K_(2)SO_(4)` be sold in order to supply the same amount of potassium as in `KCI` ?

Options

A. `"Rs". 58.40 kg^(-1)`
B. `"Rs". 50.00 kg^(-1)`
C. `"Rs". 42.82 kg^(-1)`
D. `"Rs". 25.00 kg^(-1)`

Explanation

To determine the price at which potassium sulfate `(K₂SO₄)` must be sold to supply the same amount of potassium as potassium chloride `(KCl)`, we need to follow these steps:

### Step-by-Step Solution:

1. **Calculate the Molecular Weight of Potassium Chloride `(KCl)`**:
- Atomic mass of Potassium (K) = 39 g/mol
- Atomic mass of Chlorine (Cl) = 35.5 g/mol
- Molecular weight of `KCl` = K + Cl = 39 + 35.5 = **74.5 g/mol**

2. **Calculate the Molecular Weight of Potassium Sulfate `(K₂SO₄)`**:
- There are 2 Potassium (K) atoms, 1 Sulfur (S) atom, and 4 Oxygen (O) atoms.
- Molecular weight of `K₂SO₄` = (2 × 39) + 32 + (4 × 16)
- = 78 + 32 + 64 = **174 g/mol**

3. **Determine the Amount of Potassium in Each Compound**:
- In `KCl`, there is 1 atom of K per molecule, so the amount of K in 1 mole of `KCl` is 39 g.
- In `K₂SO₄`, there are 2 atoms of K per molecule, so the amount of K in 1 mole of `K₂SO₄` is 2 × 39 = 78 g.

4. **Calculate the Cost of Potassium from `KCl`**:
- The cost of `KCl` is Rs 50 per kg (1000 g).
- Cost per gram of `KCl` = 50 / 1000 = Rs 0.05 per gram.
- Therefore, the cost of 39 g of K from `KCl` = 39 × 0.05 = Rs **1.95**.

5. **Calculate the Cost of Potassium from `K₂SO₄`**:
- Since `K₂SO₄` provides 78 g of K, we need to find out how much `K₂SO₄` is needed to provide the same amount of potassium (39 g).
- The cost of `K₂SO₄` per kg = Rs X (what we need to find).
- Cost per gram of `K₂SO₄` = X / 1000.
- Therefore, the cost for 78 g of K from `K₂SO₄` = 78 × (X / 1000).

6. **Set the Costs Equal**:
- Since we want the cost of potassium from `K₂SO₄` to equal the cost from `KCl`:
\[ 78 \times \frac{X}{1000} = 1.95 \]

7. **Solve for X**:
\[ X = \frac{1.95 \times 1000}{78} = \frac{1950}{78} \approx 25 \]

8. **Final Calculation**:
- The price at which `K₂SO₄` must be sold to supply the same amount of potassium as in `KCl` is approximately Rs **25 per kg**.