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# An aqueous solution of NaOH having density 1.1 $\dfrac{{kg}}{{d{m^3}}}$ contains a 0.02-mole fraction of NaOH. The molality and molarity of NaOH solution respectively, are :(A) 0.986, 1.134(B) 1.134, 1.193 (C) 1.134, 1.02 (D) 1.034, 1.134

Last updated date: 01st Mar 2024
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Hint: As a mole fraction of NaOH i.e. solute is given, we can find the mole fraction of solvent and can use the values in terms of moles of the substances to find Molality and Molarity.
Molality is defined as the ratio of the number of moles of solute (mole) to that of the weight of
solvent ( kg ) and
Molarity is defined as the ratio of the number of moles of solute ( mole ) to that of the volume of
solution ( Litre ). The number of moles is the fraction of the given mass by the molar weight of the molecule. $No{{ }}of{{ }}moles = \dfrac{{given\;weight}}{{molar\;weight}}$

Density ($\rho$)$=$ 1.1 $\dfrac{{kg}}{{d{m^3}}}$
Mole fraction of NaOH $({X_{NaOH}}) = 0.2$
To find –
( A ) Molality ( m )

( B ) Molarity ( M )
Now, we know the value of mole fraction of NaOH is 0.02
And also,
${X_{solute}} + {X_{solvent}} = {X_{solution}}$
${X_{solute}} + {X_{solvent}} = 1$
${X_{NaOH}} + {X_{{H_2}O}} = 1$
$0.02 + {X_{{H_2}O}} = 1$
On rearranging,
${X_{{H_2}O}} = 1 - 0.02$
${X_{{H_2}O}} = 0.98$
Hence, the mole fraction of solvent i.e. water is 0.98
Now, to find the number of moles of solute i.e. NaOH,
We know that the number of moles (n ) $= \dfrac{{given\;weight}}{{molar\;weight}}$
$0.02 = \dfrac{{{w_{NaOH}}}}{{40}}$ ( $\because$ molar weight of NaOH is 40$\dfrac{g}{{mol}}$)
${w_{NaOH}} = \left( {0.02} \right)\left( {40} \right)$
${w_{NaOH}} = 0.8g$
Similarly, for the weight of solvent i.e. water,
Number of moles (n ),
$0.98 = \dfrac{{{w_{{H_2}O}}}}{{18}}$
${w_{{H_2}O}} = \left( {0.98} \right)\left( {18} \right)$
${w_{{H_2}O}} = 17.64g$
Now, the weight of the solution $=$ weight of the solute $+$ weight of solvent
$= 0.8 + 17.64$
Hence, the weight of the solution $=$ 18.44 $g$
As given, the density of the solution is 1.1 $\dfrac{{kg}}{{d{m^3}}}$
From the formula, $\rho = \dfrac{m}{V}$
We can find out the volume of the solution,
Substituting the values,
$m = 18.44g$ and $\rho = 1.1\dfrac{{kg}}{{d{m^3}}}$
$1.1 = \dfrac{{18.44}}{V}$
$V = \dfrac{{18.44}}{{1.1}}$
$V = 16.76 \times {10^{ - 3}}L$
Hence, the volume of the solution is $16.76 \times {10^{ - 3}}L$
Now, we can find the molality and molarity of the solution,
MOLALITY
The formula for molality is given as –
$Molality = \dfrac{{number\;of\;moles\;of\;solute}}{{weight\;of\;solvent}}(mole/kg)$
$m = \dfrac{{0.02}}{{17.64 \times {{10}^{ - 3}}}}(mole/kg)$
$m = 0.0011337 \times {10^3}$
$m = 1.1337$
$m = 1.134(mole/kg)$
Hence, the molality of the solution is $1.134(mole/kg)$
(B) MOLARITY
The formula for molarity is given as –
$Molarity(M) = \dfrac{{number\;of\;moles\;of\;solute}}{{volume\;of\;solution}}(mole/L)$
$M = \dfrac{{0.02}}{{16.76 \times {{10}^{ - 3}}}}(mole/L)$
$M = 0.001193 \times {10^3}(mole/L)$
$M = 1.193(mole/L)$
Hence, the molarity of the solution is $1.193(mole/L)$

So, the correct answer is Option B.

Note: We easily found out the mole fraction of water and deduced further simplifications because the relation between mole fractions of solute and solvent were known.
Mole fraction is nothing but the ratio of the number of moles of solute or solvent to the number of moles of solution.