Two methanol-water mixtures are contained in separate flasks. the first mixture is 40.0 wt % methanol, and the second is 70.0 wt % methanol. if 200 g of the first mixture is combined with 150 g of the second, what will be the mass and composition of the resulting mixture?

The mass of the final mixture is 350 grams and composition of methanol in the final mixture is 52.9 %

Explanation:

In first mixture:

Mass of first mixture taken, = 200 g

Percentage of methanol, = 40 % = 0.4

Percentage of water, = (100 - 40)% = 60% = 0.6

In second mixture:

Mass of second mixture, = 150 g

Percentage of methanol, = 70 % = 0.7

Percentage of water, = (100 - 70)% = 30% = 0.3

Let us consider that no chemical reaction is taking place while mixing the solutions.

Calculating the mass of methanol and water in the final mixture:

Putting values in above equation, we get:

Putting values in above equation, we get:

Total mass of the final mixture =

To calculate the percentage composition of methanol in final mixture, we use the equation:

Mass of final mixture = 350 g

Mass of methanol = 185 g

Putting values in above equation, we get:

Hence, the mass of the final mixture is 350 grams and composition of methanol in the final mixture is 52.9 %

answer:

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an example of solubility is salt or sugar dissolving in water.

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