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Answer :
Final answer:
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.
Explanation:
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.
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