Decreasing order (first having highest and then other following it) of mass of pure NaOH in each of the aqueous solution
(P) 50 gm of `40%(w//w)` NaOH
(Q) 50 gm of `50%(w//v)` NaOH `[d_("soln.")=1.2gm//ml]`
(R) 50 gm of 20 M NaOH `[d_("soln").=1gm//ml]`

Options

A. I,ii,iii
B. iii,ii,i
C. ii,iii,i
D. ii,I,iii

Answer

To solve the problem of determining the decreasing order of mass of pure NaOH in each of the given aqueous solutions, we will analyze each solution step-by-step.

### Step 1: Calculate the mass of pure NaOH in Solution P (50 gm of 40% w/w NaOH)

1. **Understanding the percentage**: A 40% (w/w) NaOH solution means that in 100 grams of the solution, there are 40 grams of NaOH.
2. **Calculate the mass of NaOH in 50 grams of solution**:
\[ \text{Mass of NaOH} = \left( \frac{40 \text{ gm NaOH}}{100 \text{ gm solution}} \right) \times 50 \text{ gm solution} = 20 \text{ gm NaOH} \]

### Step 2: Calculate the mass of pure NaOH in Solution Q (50 gm of 50% w/v NaOH)

1. **Understanding the percentage**: A 50% (w/v) NaOH solution means that in 100 mL of solution, there are 50 grams of NaOH.
2. **Calculate the mass of the solution using density**: Given that the density of the solution is 1.2 gm/mL, we can find the mass of 100 mL of the solution:
\[ \text{Mass of solution} = \text{Volume} \times \text{Density} = 100 \text{ mL} \times 1.2 \text{ gm/mL} = 120 \text{ gm} \]
3. **Now, calculate the mass of NaOH in 50 grams of solution**:
\[ \text{Mass of NaOH} = \left( \frac{50 \text{ gm NaOH}}{120 \text{ gm solution}} \right) \times 50 \text{ gm solution} = \frac{50}{120} \times 50 = 20.83 \text{ gm NaOH} \]

### Step 3: Calculate the mass of pure NaOH in Solution R (50 gm of 20 M NaOH)

1. **Understanding molarity**: A 20 M NaOH solution means that there are 20 moles of NaOH per liter of solution.
2. **Calculate the mass of the solution**: Given the density of the solution is 1 gm/mL, the mass of 1 liter (1000 mL) of solution is:
\[ \text{Mass of solution} = 1000 \text{ mL} \times 1 \text{ gm/mL} = 1000 \text{ gm} \]
3. **Calculate the number of moles in 50 grams of solution**:
\[ \text{Moles of NaOH} = \left( \frac{20 \text{ moles}}{1000 \text{ gm solution}} \right) \times 50 \text{ gm solution} = 1 \text{ mole NaOH} \]
4. **Calculate the mass of NaOH**:
\[ \text{Mass of NaOH} = \text{Number of moles} \times \text{Molar mass} = 1 \text{ mole} \times 40 \text{ gm/mole} = 40 \text{ gm NaOH} \]

### Step 4: Compare the masses of pure NaOH in each solution

- Solution P: 20 gm NaOH
- Solution Q: 20.83 gm NaOH
- Solution R: 40 gm NaOH

### Step 5: Arrange in decreasing order

The decreasing order of mass of pure NaOH in each solution is:
1. Solution R: 40 gm
2. Solution Q: 20.83 gm
3. Solution P: 20 gm

### Final Answer

The decreasing order of mass of pure NaOH in the solutions is:
**R > Q > P**