Question

Given that the molecular weight of ethyl alcohol ($$C{H}_{3}C{H}_{2}OH$$) is 46, and that of water is 18, how many grams of ethyl alcohol must be mixed with 100ml of water for the mole fraction (X) of ethyl alcohol to be 0.2?
(A) 74.4g
(B) 54.4g
(C) 44.4g
(D) 64.4g

Answer 1

Hint: Mole fraction is defined as the number of moles of a particular component in a mixture divided by the total number of moles in that solute-solvent mixture. Ratio of given mass to molar mass gives the value of number of moles.

Complete step-by-step solution:
Mole fraction is the ratio of moles of one substance in a solution mixture to the total number of moles of all substances in the same mixture. For a mixture of two substances, A and B, we can write the mole fractions of each as follows:
$$molefractionofA={\displaystyle \frac{n_A}{n_A+n_B}}$$
We can calculate number of moles of each substance using the below formula:
$$moles={\displaystyle \frac{givenmass}{molarmass}}$$
Number of moles of water are
$$moles={\displaystyle \frac{100}{18}}=5.55moles$$
We are provided that the mole fraction of ethyl alcohol is 0.2 and molecular weight of ethyl alcohol is 46 grams, putting these values in mole fraction formula, we get
$$0.2={\displaystyle \frac{m_x/46}{m_x/46+5.55}}$$
Solving this equation, we get the value of $$m_x$$= 64.4 grams.
Therefore, 64.4 grams of ethyl alcohol must be mixed with 100ml of water for the mole fraction (X) of ethyl alcohol to be 0.2.

Hence, the correct option is (D).

Note: The sum of mole fraction of all the components in the solution mixture is always equal to unity. Since it is a fraction of moles i.e. the ratio of same quantity, it is unitless. Other concentration terms can be calculated using mole fraction such as molarity, molality, etc.