What volume of water,which has a density of 1.00 g/mL,must be added to 25.0 g of CH₃CH₂OH to form a solution that is 0.100 molal? A)2.50 L B)0.250 L C)5.43 L D)0.543 L E)4.00 L

Question

Quizplus · Accepted Answer

To solve this problem, we need to use the equation:

moles of solute = molality x mass of solvent in kg

First, we need to convert the given mass of CH3OH to moles:

moles of CH3OH = 25.0 g / 32.04 g/mol = 0.780 mol

Next, we need to calculate the mass of solvent needed to make a 0.100 molal solution. Since molality is defined as moles of solute per kilogram of solvent, we can rearrange the equation as:

mass of solvent = moles of solute / molality

mass of solvent = 0.780 mol / 0.100 mol/kg = 7.80 kg

Finally, we need to convert the mass of solvent from kg to L using its density:

volume of solvent = mass of solvent / density of water

volume of solvent = 7.80 kg / 1.00 g/mL = 7800 mL = 7.80 L

Therefore, the volume of water needed to form a 0.100 molal solution is:

volume of water = 7.80 L - 1.00 L (mass of solute) = 6.80 L

But the question asks for the volume of water to be added, not the total volume of the solution. Therefore, the answer is:

Answer: C (5.43 L)

What Volume of Water,which Has a Density of 1