Question:
What is the mass ratio of ethylene glycol \((C _2H _6O_2\), molar mass \(62\ g/mol)\) required for making \(500\ g\) of \(0.25\) molal aqueous solution and \(250\ ml\) of \(0.25\) molal aqueous solution?
What is the mass ratio of ethylene glycol \((C _2H _6O_2\), molar mass \(62\ g/mol)\) required for making \(500\ g\) of \(0.25\) molal aqueous solution and \(250\ ml\) of \(0.25\) molal aqueous solution?
Updated On: Mar 19, 2025
- $1: 2$
- $1: 1$
- $3:1$
- $2:1$
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The Correct Option is D
Approach Solution - 1
Assume : Mass of solvent Mass of solution
Case I :-
Case II :-
So, the correct option is (D): \(2:1\)
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Approach Solution -2
For the 500 g solution:
First, calculate the moles of ethylene glycol: \[ \text{Moles of ethylene glycol} = 0.25 \frac{\text{mol}}{\text{L}} \times 0.5 \text{L} = 0.125 \text{mol} \] Next, calculate the mass of ethylene glycol using its molar mass (62 g/mol): \[ \text{Mass of ethylene glycol} = 0.125 \text{mol} \times 62 \text{g/mol} = 7.75 \text{g} \] For the 250 mL solution:
Calculate the moles of ethylene glycol: \[ \text{Moles of ethylene glycol} = 0.25 \frac{\text{mol}}{\text{L}} \times 0.25 \text{L} = 0.0625 \text{mol} \] Then, calculate the mass of ethylene glycol: \[ \text{Mass of ethylene glycol} = 0.0625 \text{mol} \times 62 \text{g/mol} = 3.875 \text{g} \] Thus, the mass ratio of ethylene glycol needed for the two solutions is: \[ 7.75 \text{g} : 3.875 \text{g} \approx 2:1 \]
First, calculate the moles of ethylene glycol: \[ \text{Moles of ethylene glycol} = 0.25 \frac{\text{mol}}{\text{L}} \times 0.5 \text{L} = 0.125 \text{mol} \] Next, calculate the mass of ethylene glycol using its molar mass (62 g/mol): \[ \text{Mass of ethylene glycol} = 0.125 \text{mol} \times 62 \text{g/mol} = 7.75 \text{g} \] For the 250 mL solution:
Calculate the moles of ethylene glycol: \[ \text{Moles of ethylene glycol} = 0.25 \frac{\text{mol}}{\text{L}} \times 0.25 \text{L} = 0.0625 \text{mol} \] Then, calculate the mass of ethylene glycol: \[ \text{Mass of ethylene glycol} = 0.0625 \text{mol} \times 62 \text{g/mol} = 3.875 \text{g} \] Thus, the mass ratio of ethylene glycol needed for the two solutions is: \[ 7.75 \text{g} : 3.875 \text{g} \approx 2:1 \]
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Concepts Used:
Solutions
A solution is a homogeneous mixture of two or more components in which the particle size is smaller than 1 nm.
For example, salt and sugar is a good illustration of a solution. A solution can be categorized into several components.
Types of Solutions:
The solutions can be classified into three types:
- Solid Solutions - In these solutions, the solvent is in a Solid-state.
- Liquid Solutions- In these solutions, the solvent is in a Liquid state.
- Gaseous Solutions - In these solutions, the solvent is in a Gaseous state.
On the basis of the amount of solute dissolved in a solvent, solutions are divided into the following types:
- Unsaturated Solution- A solution in which more solute can be dissolved without raising the temperature of the solution is known as an unsaturated solution.
- Saturated Solution- A solution in which no solute can be dissolved after reaching a certain amount of temperature is known as an unsaturated saturated solution.
- Supersaturated Solution- A solution that contains more solute than the maximum amount at a certain temperature is known as a supersaturated solution.