An ethylene glycol solution contains 27.6 g of ethylene glycol (C₂H₆O₂) in 99.8 ml of water. (Assume a density of 1.00 g/ml for water.) What is the molarity of the solution?

A) 0.976 M
B) 1.486 M
C) 2.174 M
D) 0.694 M

Question

High School

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Answer :

Oriz999 Oriz999 · Answer 1 · 2024-01-10T04:30:14-05:00

An ethylene glycol solution contains 27.6 g of ethylene glycol (C₂H₆O₂) in 99.8 ml of water, the molarity is A) 0.976 M.

To determine the molarity of the ethylene glycol solution, we need to follow these steps:

1. Calculate the moles of ethylene glycol:

- Given mass of ethylene glycol (C₂H₆O₂) = 27.6 g

- Calculate the molar mass of ethylene glycol (C₂H₆O₂):

Molar mass = 2(12.01 g/mol) + 6(1.01 g/mol) + 2(16.00 g/mol) = 62.07 g/mol

2. Calculate the volume of the solution in liters:

- Given volume of water = 99.8 ml = 0.0998 L (since 1 ml = 0.001 L)

- Since the density of water is 1.00 g/ml, the mass of water is equivalent to its volume in ml.

3. Calculate the molarity of the ethylene glycol solution:

- Molarity (M) = Moles of solute / Volume of solution in liters

- Substitute the values calculated in steps 1 and 2 into the formula to find the molarity.

Calculating the molarity:

Moles of ethylene glycol:

= 27.6 g / 62.07 g/mol

= 0.0998 L

Molarity:

= (27.6 g / 62.07 g/mol) / 0.0998 L

≈ 0.976 M

Therefore, the molarity of the ethylene glycol solution is approximately 0.976 M, which corresponds to option A.

The correct answer is A) 0.976 M.