Respuesta :
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, [tex]M_1[/tex] = 200 g
Percentage of methanol, [tex]\chi_m(1)[/tex] = 40 % = 0.4
Percentage of water, [tex]\chi_w(1)[/tex] = (100 - 40)% = 60% = 0.6
In second mixture:
Mass of second mixture, [tex]M_2[/tex] = 150 g
Percentage of methanol, [tex]\chi_m(2)[/tex] = 70 % = 0.7
Percentage of water, [tex]\chi_w(2)[/tex] = (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:
[tex]m_f=[M_1\chi_m(1)+M_2\chi_m(2)][/tex]
Putting values in above equation, we get:
[tex]m_f=[(200\times 0.4)+(150\times 0.7)]=185g[/tex]
- Mass of water:
[tex]w_f=[M_1\chi_w(1)+M_2\chi_w(2)][/tex]
Putting values in above equation, we get:
[tex]w_f=[(200\times 0.6)+(150\times 0.3)]=165g[/tex]
Total mass of the final mixture = [tex]m_f+w_f=185+165=350g[/tex]
To calculate the percentage composition of methanol in final mixture, we use the equation:
[tex]\%\text{ composition of methanol}=\frac{\text{Mass of methanol}}{\text{Mass of final mixture}}\times 100[/tex]
Mass of final mixture = 350 g
Mass of methanol = 185 g
Putting values in above equation, we get:
[tex]\%\text{ composition of methanol in final mixture}=\frac{185g}{350g}\times 100=52.9\%[/tex]
Hence, the mass of the final mixture is 350 grams and composition of methanol in the final mixture is 52.9 %
Each of the methanol–water mixture contribute the amount by weight of
methanol given in their composition.
- The mass and composition of the resulting mixture is 350 g, and approximately 52.9 wt% methanol.
Reasons:
The given parameters are;
Number of methanol–water mixtures = 2
Weight percentage of methanol in the first mixture = 40.0 wt %
Weight percentage in the second mixture = 70.0 wt %
Mass of the first mixture in the resulting mixture = 200 g
Mass of the second mixture in the resulting mixture = 150 g
Therefore;
- Mass of the resulting mixture = 200 g + 150 g = 350 g
[tex]\displaystyle Methanol \ contribution \ from \ the \ first \ mixture = \frac{40}{100} \times 200 \, g= \mathbf{80 \, g}[/tex]
[tex]\displaystyle Methanol \ contribution \ from \ the \ second \ mixture = \frac{70}{100} \times 150\, g= \mathbf{105 \, g}[/tex]
Mass of methanol in the resulting mixture, M = 80 g + 105 g = 185 g
[tex]\displaystyle Percentage \ by \ weight \ of \ methanol \ in \ resulting \ mixture = \frac{185}{350} \times 100 \approx 52.9 \%[/tex]
- The composition of the resulting mixture is approximately 52.9 wt % methanol
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