Question

A solution is prepared by mixing 631 mLmL of methanol with 501 mLmL of water. The molarity of methanol in the resulting solution is 14.29 MM. The density of methanol at this temperature is 0.792 g/mLg/mL. Calculate the difference in volume between this solution and the total volume of water and methanol that were mixed to prepare the solution

Answer 1

The difference in volume is 40 mL or 0.040 L

Step-by-step explanation:

Step 1: Data given

Volume of methanol = 631 mL = 0.631 L

Volume of water = 501 mL = 0.501 L

Molarity of methanol = 14.29 M

Density of methanol = 0.792 g/mL

Step 2: Calculate mass methanol

Mass = density * volume

Mass methanol = 0.792 g/mL * 631 mL methanol = 499.8 g methanol

Step 3: Calculate moles methanol

Moles = mass / molar mass

Moles methanol = 499.8 g methanol / 32.04 g/mol

Moles methanol = 15.6 moles methanol

Step 4: Calculate volume

We should have 15.6 moles of methanol, and we need 14.29 M solution

Volume = moles / molarity

Volume = 15.6 moles / 14.29 M

Volume = 1.092 L

Step 5: Calculate the difference in volume

So we should have 1.092 Liters of solution, but we haf 1.132 Liters of solution:

1.132 L - 1.092 L = 0.040 L or 40 mL

The difference in volume is 40 mL or 0.040 L

So the difference is 39 mL.

Answer 2

Answer: The difference between the volume of solution and mixture of methanol and water is 0.039 L

Step-by-step explanation:

We are given:

Volume of methanol = 631 mL

Volume of water = 501 mL

Volume of methanol + water = [631 + 501] mL = 1132 mL = 1.132 L (Conversion factor: 1 L = 1000 mL)

To calculate the mass of methanol, we use the equation:

`\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}`

Density of methanol = 0.792 g/mL

Volume of methanol = 631 mL

Putting values in above equation, we get:

`0.792g/mL=\frac{\text{Mass of methanol}}{631mL}\\\\\text{Mass of methanol}=(0.792g/mL* 631mL)=499.75g`

• To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

`\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}* \text{Volume of solution (in L)}}`

Given mass of methanol = 499.75 g

Molar mass of methanol = 32 g/mol

Molarity of solution = 14.29 M

Putting values in above equation, we get:

`14.29mol/L=(499.75g)/(32g/mol* V)\\\\V=(499.75)/(32* 14.29)=1.093L`

Calculating the difference between the volumes:

`\Delta V=V_(mixture)+V_(solution)`

Putting values in above equation, we get:

`\Delta V=[1.132-1.093]L=0.039L`

Hence, the difference between the volume of solution and mixture of methanol and water is 0.039 L