Question

$10g$ of $NaOH$ is mixed with $36g$ of ${H_2}O$ .What is the mole fraction of $NaOH$ in solution?
$(1)0.125$
$(2)0.111$
$(3)0.72$
$(4)0.5$

Answer 1

Hint : To calculate the number of moles of any substance present in the sample, we divide the given weight of the substance by its molar mass. The mole fraction is the number of moles of a particular component in the solution divided by the total number of moles in the given solution.
Formula used:
Number of moles = MassMolar mass
Mole fraction = \dfrac{\text{Number of moles of NaOH }}{\text{Number of moles of NaOH + $Number of moles of$ {H_2}O }}

Complete Step By Step Answer:
Given:
Mass of $NaOH = 10g$
Mass of ${H_2}O = 36g$
To find: Mole fraction of $NaOH$
Molar mass of $NaOH = 40g$
Molar mass of ${H_2}O = 18g$
Using the formula, we can calculate the number of moles,
Number of moles = MassMolar mass
Substituting the values, we get,
Number of moles of $NaOH = \dfrac{{10g}}{{40g}}$
Number of moles of $NaOH = 0.25mol$
Number of moles of ${H_2}O = \dfrac{{36g}}{{18g}}$
Number of moles of ${H_2}O = 2mol$
Now, using the formula of mole fraction we can calculate the mole fraction of $NaOH$ ,
Mole fraction = \dfrac{\text{Number of moles of NaOH }}{\text{Number of moles of NaOH + $Number of moles of$ {H_2}O }}
Substituting the values, we get,
Mole fraction $= \dfrac{{0.25}}{{0.25 + 2}}$
Mole fraction $= 0.111$
Hence, mole fraction of $NaOH$ is $0.111$
Therefore, the correct option is $(2)0.111$ .

Note :
Mole fraction is not dependent on the temperature. The information about the density of the phase is not necessary to calculate the mole fraction. In the case of an ideal gas mixture, the mole fraction is represented by the ratio of partial pressure to total pressure of the mixture. Mole fraction is said to be a unitless and dimensionless expression.