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The density of a 2.45 M aqueous solution of methanol (CH₃OH) is 0.976 g mLâ»Â¹. Calculate the molarity of the solution (MM=32.04g)

a) 1.23 M
b) 2.45 M
c) 3.67 M
d) 4.89 M

User Famf
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Final answer:

The molarity of the methanol solution given its density of 0.976 g/mL is calculated to be 2.45 M. This calculation confirms the molarity provided in the question, indicating that the concentration has not changed.

Step-by-step explanation:

The question asks to calculate the molarity of an aqueous solution of methanol with a given density and concentration in moles per liter (M).

To find the new molarity, we need to first determine the mass of 1 liter (or 1000 mL) of the solution using the given density:

Mass of solution = density × volume = 0.976 g/mL × 1000 mL = 976 g

Since we know the molarity (Moles/Liter) of the solution and have calculated the mass of 1 liter of the solution, we can find the moles of methanol in 1 liter:

Moles of CH₃OH = molarity × volume = 2.45 M × 1 L = 2.45 moles

Next, we calculate the mass of methanol using its molar mass:

Mass of CH₃OH = moles × molar mass = 2.45 moles × 32.04 g/mole = 78.498 g

With the total mass of the solution and the mass of methanol, we can calculate the mass percentage of methanol:

Mass percentage of CH₃OH = (mass of CH₃OH / mass of solution) × 100 = (78.498 g / 976 g) × 100 = 8.041%

Since the concentration of methanol is not changing, the molarity also remains the same. Thus, the molarity of the solution is 2.45 M, matching the given value.

User Bertone
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