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A solution is prepared by dissolving 20.2 ml of methanol (ch3oh) in 100.0 ml of water at 25oc. the final volume of the solution is 118 ml. the densities of methanol and water at this temperature are 0.782 g/ml and 1.00 g/ml, respectively. for this solution, calculate the molarity, molality, percent by mass, mole fraction, and mole percent.

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Dejanras
m(CH₃OH) = 20,2 ml · 0,782 g/ml.
m(CH₃OH) = 15,8 g.
n(CH₃OH) = 15,8 g ÷ 32 g/mol.
n(CH₃OH) = 0,493 mol.
m(H₂O) = 100 ml · 1 g/ml.
m(H₂O) = 100 g.
n(H₂O) = 100 g ÷ 18 g/mol.
n(H₂O) = 5,55 mol.
1) c(CH₃OH) = 0,493 mol ÷ 0,118 dm³.
c(CH₃OH) = 4,18 mol/dm³.
2) b(CH₃OH) = 0,493 mol ÷ 0,1 kg.
b(CH₃OH) = 4,93 mol/kg.
3) moll fraction = 0,493 mol ÷ 6,043 mol = 0,081.
4) percent by mass = 15,8 g ÷ 115,8 g = 0,136 = 13,6%.
5) 0,493 mol ÷ 6,043 mol = 8,1%.
sebassandin

Answer:

Molarity: [tex]M=4.18M[/tex]

Molality: [tex]m=4.94m[/tex]

Percent by mass: %[tex]m/m=13.6[/tex]%

[tex]x_{CH_3OH}=0.0816[/tex]

Mole percent: [tex]MolPercent=8.16[/tex]%

Explanation:

Hello,

a.) Molarity: in this case, we first must compute the moles of methanol:

[tex]n_{CH_3OH}=20.2mL*\frac{0.782g}{1mL}*\frac{1mol}{32g}=0.494molCH_3OH[/tex]

Then the volume of the solution:

[tex]V_{solution}=118mL*\frac{1L}{1000mL}=0.118L[/tex]

And the molarity:

[tex]M=\frac{n_{CH_3OH}}{V_{solution}}=\frac{0.494mol}{0.118L}=4.18M[/tex]

b.) Molality: in this case, we already computed the moles of methanol, so we just divide by the water's mass in kilograms:

[tex]m=\frac{n_{CH_3OH}}{m_{H_2O}}=\frac{0.494mol}{100mL*\frac{1L}{1000mL}*\frac{1kg}{1L} } \\m=4.94m[/tex]

c.) Percent by mass: here, we compute the mass for both methanol and water as follows:

[tex]m_{CH_3OH}=20.2mL*\frac{0.782g}{1mL}=15.8g\\m_{H_2O}=100.0mL*\frac{1g}{1mL} =100.0g\\[/tex]

%[tex]m/m=\frac{15.8g}{15.8g+100.0g} *100[/tex]%

%[tex]m/m=13.6[/tex]%

d.) Mole fraction: based on the densities and the measured volumes, one computes the moles of water because the methanol moles were already computed:

[tex]n_{H_2O}=100mL*\frac{1g}{1mL}*\frac{1mol}{18g}=5.56molH_2O[/tex]

[tex]x_{CH_3OH}=\frac{0.494mol}{0.494mol+5.56mol}= 0.0816[/tex]

e.) Mole percent: here, we just multiply the mole fraction by 100% as follows:

[tex]MolPercent=0.0816*100[/tex]%

[tex]MolPercent=8.16[/tex]%

Best regards.

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