Question

A solution is made containing 14.8 g of CH3OH in 186 g H2O.

a. Calculate the mole fraction of CH3OH.

b. Calculate the mass percent of CH3OH.

c. Calculate the molality of CH3OH. (mol/kg)

Answer 1

Answer:-

Given:-

weight or mass of CH₃OH (m_CH3OH) = 14.8 g

weight or mass of H₂O (m_H2O) = 186 g

As we know that

molar mass of  CH₃OH = 32 g mol^-1

molar mass of H₂O = 18 g mol^-1

(a)-

As we know that

no. of moles of compound = wt. or mass of compound / molar mass of compound

therefore

no. of moles of CH₃OH (n_CH3OH) = weight or mass of CH₃OH (m_CH3OH) / molar mass of CH₃OH

no. of moles of CH₃OH (n_CH3OH) = 14.8 g / 32 g mol^-1

no. of moles of CH₃OH (n_CH3OH) = 0.4625 mol

similarly

no. of moles of H₂O (n_H2O) = weight or mass of H₂O (m_H2O) / molar mass of H₂O

no. of moles of H₂O (n_H2O) = 186 g / 18 g mol^-1

no. of moles of H₂O (n_H2O) = 10.33 mol

so

total no. of moles present in solution (n_total) = no. of moles of H₂O (n_H2O) + no. of moles of CH₃OH (n_CH3OH)

total no. of moles present in solution (n_total) = 10.33 mol + 0.4625 mol

total no. of moles present in solution (n_total) = 10.7925 mol

also we know that

mole fraction of CH₃OH = no. of moles of CH₃OH (n_CH3OH) / total no. of moles present in solution (n_total)

mole fraction of CH₃OH = 0.4625 mol / 10.7925 mol

mole fraction of CH₃OH = 0.04285 (i.e the answer)

(b)-

As we know that

total mass of solution (m_total) = weight or mass of H₂O (m_H2O) + weight or mass of CH₃OH (m_CH3OH)

total mass of solution (m_total) =  186 g + 14.8 g

total mass of solution (m_total) = 200.8 g

Also we know that

mass percent of CH₃OH (%) = weight or mass of CH₃OH (m_CH3OH) 100 / total mass of solution (m_total)

mass percent of CH₃OH (%) = 14.8 g 100 / 200.8 g

mass percent of CH₃OH (%) = 1480 / 200.8

mass percent of CH₃OH (%) = 7.371 % ( i.e the answer)

(c)-

As we know that

weight or mass of CH₃OH (m_CH3OH) = 14.8 g

weight or mass of H₂O (m_H2O) = 186 g = 186 / 1000 = 0.186 Kg

As we know that

molar mass of  CH₃OH = 32 g mol^-1

As we know that

molality of the compound = wt. of compound / molar mass of compound wt of solvent in Kg

molality of the CH₃OH = wt. of CH₃OH / molar mass of CH₃OH wt of H₂O in Kg

molality of the CH₃OH = 14.8 g / 32 ​​​​​​​ 0.186 Kg

molality of the CH₃OH = 0.4625 mol / 0.186 Kg

molality of the CH₃OH = 2.4866 mol/ Kg (i.e the answer)