Question

Aqueous solution of $NaOH$ is marked $10\% \left( {w/w} \right)$. The density of the solution is $1.070\,gc{m^{ - 3}}$. Calculate
(i) molarity
(ii) molality and
(iii) mole fraction of $NaOH$& water.
$\left[ {Na = 23,H = 10,O = 16} \right]$

Answer 1

Hint: We only need the equation of molarity, molality and mole fraction to find the solution of the given question.

Formula Used:
$Molarity\;\left( M \right) = \dfrac{{number\;of\;moles\;of\;solute}}{{Volume\;of\;solution\left( L \right)}}$
$Molality = \dfrac{{number\;of\;moles\;of\;solute\;}}{{Weight\;of\;solvent\left( {kg} \right)}}$
$Mole\;fraction,\chi = \dfrac{{number\;of\;moles\;of\;A}}{{Total\;number\;of\;moles}}$

Complete step by step solution:
We have an aqueous solution of $NaOH$, also given that it has $10\% \left( {w/w} \right)$.
First of all, we have to understand what that means. When we have 100g of solution, here it is diluted. $NaOH$ solution only 10g of $NaOH$ is present in it. Since it has only 10% of its weight by weight.
So we understood that
$mass\;of\;solute = 10g$
$mass\;of\;water = 90g$
Also, we know that
$density = \dfrac{{mass}}{{volum{e_{solution}}}}$
The density of the solution is given,
$density = 1.070gc{m^{ - 3}}$
$ \Rightarrow Volume\;of\;solution = \dfrac{{mass}}{{density}}$
$mass = 100g$
Substituting the values to the equation, we get
$Volume\;of\;solution = \dfrac{{100}}{{1.070}} = 93.46c{m^3}$
Converting it to litre we get, 0.09346L
Now, we have to find number of moles of $NaOH$
$number\;of\;moles,n = \dfrac{{weight\;of\;subs{\text{tan}}ce\left( w \right)}}{{molar\;mass\;\left( M \right)}}$
Weight of the substance Is 10g and the molar mass of $NaOH$ is $1 \times 23 + 16 + 10 = 49$
$ \Rightarrow n = \dfrac{{10}}{{49}} = 0.2gmo{l^{ - 1}}$
(i) molarity:
 Molarity is the number of moles of solute dissolved per litre of the solution.
$Molarity\;\left( M \right) = \dfrac{{number\;of\;moles\;of\;solute}}{{Volume\;of\;solution\left( L \right)}}$
$ \Rightarrow Molarity = \dfrac{{0.2}}{{0.09346}}M = 2.14M$
i.e., the molarity of the aqueous solution is 2.14M.

(ii) Molality:
Molality is the number of moles of solute dissolved per kilogram of the solvent.
$Molality = \dfrac{{number\;of\;moles\;of\;solute\;}}{{Weight\;of\;solvent\left( {kg} \right)}}$
Weight of the solvent is 90g
$ \Rightarrow molality = \dfrac{{0.2}}{{0.09\left( {kg} \right)}} = 2.22m$
i.e., the molality of an aqueous solution is 2.22m.

(iii) mole fraction:
Mole fraction of a component of a solution is defined as the ratio of the number of moles of that component to the total number of moles of the solution.
We already know the number of moles of $NaOH$. To find the number of water,
$number\;of\;moles\;of\;water = \dfrac{{weight\;of\;water\;}}{{molar\;mass\;of\;water}}$
Weight of water is given as 90g and molar of water is 18.
Substituting the values we get
$number\;of\;moles\;of\;water = \dfrac{{90}}{{18}} = 5mol$
$Mole\;fraction,{\chi _A} = \dfrac{{number\;of\;moles\;of\;A}}{{Total\;number\;of\;moles}}$
Here we have only two component hence,
$Mole\;of\;fraction\;of\;NaOH = \dfrac{{number\;of\;moles\;of\;NaOH}}{{Total\;number\;of\;moles}}$
$Total\;number\;of\;moles = 0.2 + 5 = 5.2mol$
Substituting the values,
$Mole\;of\;fraction\;of\;NaOH = \dfrac{{0.2}}{5} = 0.04$
$Mole\;of\;fraction\;of\;{H_2}O = \dfrac{5}{{5.2}} = 0.96$
Mole fraction of $NaOH$ is 0.04 and water is 0.96

Note:
Sum of mole fractions of all the components of a solution is always unity
i.e., ${\chi _A} + {\chi _B} = 1$