Question

[1.1.7.
3. A solution is 0.120 m methanol dissolved in ethanol. Calculate the mole-fractions
methanol, CH3OH, and ethanol, C2H5OH, in the solution.​

Answer 1

The mole-fraction of methanol, $(CH_3OH)$, and ethanol, $(C_2H_5OH)$ in the solution. is $5.50\times 10^{-3}$ and 0.99 respectively.

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

$Molality=\frac{n\times 1000}{W_s}$

where,

n = moles of solute

$W_s$ = weight of solvent in g

Given : 0.120 molal methanol solution in ethanol, i.e. 0.120 moles of methanol in 1 kg or 1000 g of ethanol

Moles of methanol = 0.120

moles of ethanol = $\frac{\text {given mass}}{\text {molar mass}}=\frac{1000g}{46g/mol}=21.7mol$

${\text {mole fraction of ethanol}=\frac{\text {moles of ethanol}}{\text {moles of ethanol+moles of methanol}}=\frac{21.7}{21.7+0.120}=0.994$

${\text {mole fraction of methanol}=\frac{\text {moles of methanol}}{\text {moles of ethanol+moles of methanol}}=\frac{0.120}{21.7+0.120}=5.50\times 10^{-3}$

Learn More about mole fraction

https://brainly.in/question/12853195

https://brainly.com/question/12154684