Question

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?

Answer 1

Answer: 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, $M_1$ = 200 g

Percentage of methanol, $\chi_m(1)$ = 40 % = 0.4

Percentage of water, $\chi_w(1)$ = (100 - 40)% = 60% = 0.6

In second mixture:

Mass of second mixture, $M_2$ = 150 g

Percentage of methanol, $\chi_m(2)$ = 70 % = 0.7

Percentage of water, $\chi_w(2)$ = (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:

• Mass of methanol:

$m_f=[M_1\chi_m(1)+M_2\chi_m(2)]$

Putting values in above equation, we get:

$m_f=[(200\times 0.4)+(150\times 0.7)]=185g$

• Mass of water:

$w_f=[M_1\chi_w(1)+M_2\chi_w(2)]$

Putting values in above equation, we get:

$w_f=[(200\times 0.6)+(150\times 0.3)]=165g$

Total mass of the final mixture = $m_f+w_f=185+165=350g$

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

$\%\text{ composition of methanol}=\frac{\text{Mass of methanol}}{\text{Mass of final mixture}}\times 100$

Mass of final mixture = 350 g

Mass of methanol = 185 g

Putting values in above equation, we get:

$\%\text{ composition of methanol in final mixture}=\frac{185g}{350g}\times 100=52.9\%$

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