Question

If 4 g of $NaOH$ dissolves in 36 g of H20, calculate the mole fraction of each component in the solution. Also, determine the molarity of solution (specific gravity of solution is 1 g $mL^{-1}$ ).

Answer 1

Hint : Initially you must be aware of the term’s mole, mole fraction and molarity. From the definition you'll be able to find the mole fraction as its formula is hidden in the definition part.

Complete Step By Step Answer:
Given: Amount of $NaOH$ is 4gm
Amount of Water is 36gm
Specific gravity is 1 g m/L
Formula used:
Mole fraction: ${x_i} = \dfrac{{{n_i}}}{{{n_{{\text{total}}}}}}$
Molarity: $m = \dfrac{{{n_{solute}}}}{{m{\;_{solvent}}\;}} = {\text{ }}\dfrac{{{m_{solute}}\;}}{{\left( {W{\;_{solute}}\; \times {\text{ }}m{\;_{solvent}}} \right)}}$
Calculation:
Mole fraction of ${H_2}O$ = No. of moles of ${H_2}O$ / Total No. of moles ( ${H_2}O$ + $NaOH$ )
No. of moles of ${H_2}O$ = 36/18 = 2 moles
Number of moles $NaOH = \dfrac{4}{{40}} = 0.1mole$
Total number of moles = 2+ 0.1 = 2.1
Mole fraction of ${H_2}O = \dfrac{2}{{2.1}}{\text{ }} = {\text{ }}0.952$
Mole fraction of $NaOH = \dfrac{{0.1}}{{2.1}} = {\text{ }}0.048$
Mass of solution= Mass of ${H_2}O$ + Mass of $NaOH$ = 36 + 4 = 40 g
Volume of solution $= \dfrac{{40}}{1} = 40$ L
$Molarity{\text{ }} = {\text{ }}\dfrac{{No.{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ }}solute}}{{Volume{\text{ }}of{\text{ }}solution{\text{ }}}} = \dfrac{{0.1}}{{0.04L}} = {\text{ }}2.5M\;$

Note :
A mole of a substance or a mole of particles is described as it contains exactly $6.02214076 \times {10^{23}}$ particles, which may be atoms, molecules, ions, or electrons. The mole fraction or molar fraction is defined as a unit of the quantity of a constituent (expressed in moles), divided by the entire amount of all constituents in a mixture also expressed in moles. The Molarity is defined as the moles of a solute per litres of a solution. Molarity is additionally referred to as the ratio of the amount of mole and total solution.