Question

Aqueous solution of NaOH is marked 10% w/w. The density of the solution is \[{\text{1}}{\text{.070 g}}/{\text{c}}{{\text{m}}^3}\]. Calculate
A.Molarity
B.Molality
C.Mole fraction of \[{\text{NaOH}}\] in water
[\[{\text{Na}} = 23,{\text{H}} = 1,{\text{O}} = 16\]]

Answer 1

Hint: In case a different amount of reactants is given then we will proceed with respect to limiting reagent. Limiting reagent is the reagent which consumes in the reaction completely. We shall calculate the number of moles of NaOH and then use that to calculate the molarity, molality and mole fraction.

Formula used:
 \[{\text{Molarity}} = \dfrac{{{\text{number of moles}}}}{{{\text{volume of solution in L}}}}\]
\[{\text{Molality}} = \dfrac{{{\text{number of moles}}}}{{{\text{mass of solvent in Kg}}}}\]
\[{\text{Mole fraction of component A}} = \dfrac{{{\text{number of moles of A}}}}{{{\text{total number of moles}}}}\]

Complete step by step answer:
In the given question, \[{\text{NaOH}}\] is marked as \[{\text{10% w}}/{\text{w}}\] which is also known as percentage composition that means 10 g of solute is present in 100 g of solution. Density of the solution is given to us in the question. Hence the volume of the solution will be: \[\dfrac{{100}}{{{\text{1}}{\text{.070 g}}/{\text{c}}{{\text{m}}^3}}}\]
The molar mass of sodium hydroxide that is \[{\text{NaOH}}\] is 40 g. using the volume of the solution and formula for molarity and number of moles we will get:
\[{\text{Molarity }} = \dfrac{{10 \times 1.070 \times 1000}}{{40 \times 100}} = 2.14{\text{ M}}\]
We have used 1000 to convert mL into L.
For calculation of molarity we need the mass of the solvent.
\[{\text{Mass of solvent }} = {\text{ Mass of solution }} - {\text{ Mass of solvent}}\]
\[ \Rightarrow {\text{Mass of solvent}} = 100 - 10 = 90{\text{g}}\]
Using the formula we will get:
\[{\text{Molality }} = \dfrac{{10 \times 1000}}{{40 \times 90}} = 2.2{\text{ m}}\]
There are two components that are water and sodium hydroxides. Mass and molar mass of both of them is known to us hence the mole fraction of sodium hydroxide using the formula will be:
\[{\text{Mole fraction}} = \dfrac{{0.25}}{{0.25 + \dfrac{{90}}{{10}}}}\]
\[ \Rightarrow {\text{Mole fraction}} = \dfrac{{0.25}}{{5.25}} = 0.9\]

The number of moles of sodium hydroxide was calculated already and the molar mass of water is 18.

Note:

Sodium hydroxide is also known as caustic soda. It is very much soluble in water. It readily absorbs water from the air. Whenever they are left open in air they absorb water and sticks to each other. It is a strong base and undergoes complete neutralization in the presence of strong acid forming salt and water. It is used in pH regulation, to synthesise the sodium salt, in making soaps and detergents.