What is the mass ratio of ethylene glycol `(C_(2)H_(6)O_(2)`,molar mass `=62g//mol` ) required for making `500g` of `0.25` molal aqueous solution and `250mL` of `0.25` molal aqueous solution?

Options

A. `1:2`
B. `1:1`
C. `3:1`
D. `2:1`

Answer

To solve the problem of finding the mass ratio of ethylene glycol `(C₂H₆O₂)` required for making `500g` of `0.25 molal` aqueous solution and `250mL` of `0.25 molal` aqueous solution, we can follow these steps:

### Step 1: Understand the concept of molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent. The formula is:
\[ m = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} \]

### Step 2: Calculate the number of moles of ethylene glycol for the first solution
Given:
- Mass of the solution = 500 g
- Molality (m) = 0.25 molal

Convert the mass of the solution to kilograms:
\[ \text{Mass of solution in kg} = \frac{500 \text{ g}}{1000} = 0.5 \text{ kg} \]

Now, since the mass of the solute is small compared to the mass of the solvent, we can approximate that the mass of the solvent is also approximately 500 g.

Using the definition of molality:
\[ 0.25 = \frac{\text{Number of moles of ethylene glycol}}{0.5 \text{ kg}} \]

Rearranging gives:
\[ \text{Number of moles of ethylene glycol} = 0.25 \times 0.5 = 0.125 \text{ moles} \]

### Step 3: Calculate the mass of ethylene glycol required for the first solution
Using the molar mass of ethylene glycol (62 g/mol):
\[ \text{Mass of ethylene glycol} = \text{Number of moles} \times \text{Molar mass} \]
\[ \text{Mass of ethylene glycol} = 0.125 \text{ moles} \times 62 \text{ g/mol} = 7.75 \text{ g} \]

### Step 4: Calculate the number of moles of ethylene glycol for the second solution
Given:
- Volume of the solution = 250 mL (which is approximately 250 g since the density of water is about 1 g/mL)
- Molality (m) = 0.25 molal

Using the same reasoning as before, we can consider the mass of the solvent to be approximately 250 g:
\[ 0.25 = \frac{\text{Number of moles of ethylene glycol}}{0.25 \text{ kg}} \]

Rearranging gives:
\[ \text{Number of moles of ethylene glycol} = 0.25 \times 0.25 = 0.0625 \text{ moles} \]

### Step 5: Calculate the mass of ethylene glycol required for the second solution
Using the molar mass of ethylene glycol:
\[ \text{Mass of ethylene glycol} = \text{Number of moles} \times \text{Molar mass} \]
\[ \text{Mass of ethylene glycol} = 0.0625 \text{ moles} \times 62 \text{ g/mol} = 3.875 \text{ g} \]

### Step 6: Calculate the mass ratio of the two solutions
Now we have:
- Mass of ethylene glycol for 500 g solution = 7.75 g
- Mass of ethylene glycol for 250 mL solution = 3.875 g

The mass ratio is:
\[ \text{Mass ratio} = \frac{7.75 \text{ g}}{3.875 \text{ g}} = 2:1 \]

### Final Answer:
The mass ratio of ethylene glycol required for making `500g` of `0.25 molal` aqueous solution to `250mL` of `0.25 molal` aqueous solution is **2:1**.