Question

The freezing point of aqueous solution
containing 20% w/w urea and 20% w/w
NaOH also, will be - X°C. The value of x
is (if Kf of H20 is 1.8 K kg mol-1)​

Answer 1

Final answer: $X= 25.002 \°C$

Given that: We are given the freezing point of aqueous solution containing $20$ % $\frac{w}{w}$ urea and $20$ % $\frac{w}{w}$ $NaOH$ will be $- X\°C$ and $K_{f}$ of $H_{2}O=1.8 K kg mol^{-1}$.

To find: We have to find the value of $X$.

Explanation:

• Suppose the weight of the solution containing $20$ % $\frac{w}{w}$ urea and $20$ % $\frac{w}{w}$ $NaOH$ $= 100$ $g$

• Weight of urea $= 20$ $g$

• Weight of $NaOH$ $=20$ $g$

• Weight of $H_{2}O$ $= 100-(20+20)=100-40=60$ $g$

• $Number of moles of urea= \frac{Given mass of urea}{Molar mass of urea}$

           Given the mass of urea $= 20$ $g$

           Molar mass of urea $=60$ $\frac{g}{mol}$

          Number of moles of urea $=\frac{20}{60}$ $=0.3333$ $mol$

• $Number of moles of NaOH= \frac{Given mass of NaOH}{Molar mass of NaOH}$

          Given the mass of $NaOH$ $= 20$ $g$

          Molar mass of $NaOH$ $=40$ $\frac{g}{mol}$

          Number of moles of $NaOH$ $=\frac{20}{40}$ $=0.5$ $mol$

$Total number of moles of solute = Number of moles of urea+ Number of moles of NaOH$

                                            $=0.3333+0.5$

                                            $=0.8333$ $mol$

• Depression in freezing point, $\triangle T_{f}=K_{f} \times m$

Where,

$K_{f}$ = Molal depression constant

Given that $K_{f}$ of $H_{2}O=1.8 K kg mol^{-1}$

$m$ = Molality of the solution

• $Molality, m = \frac{Number of moles of solute}{Mass of solvent in kg}$

• Here, $H_{2}O$ is the solvent.

         Mass of $H_{2}O$ $=60$ $g$

         Mass of $H_{2}O$ in $kg$ $=0.06$ $=0.06$ $kg$

• Number of moles of solute $=0.8333$ $mol$

• The molality of the solution containing  $20$ % $\frac{w}{w}$ urea and $20$ % $\frac{w}{w}$ $NaOH$, $m= \frac{0.8333}{0.06}=13.89 \frac{mol}{kg}$

• Substitute the value of $K_{f}$ and $m$ in, $\triangle T_{f}=K_{f} \times m$

           $\triangle T_{f}=(1.8 K kg mol^{-1}) \times (13.89\frac{mol}{kg})$

           $\triangle T_{f} =25.002$ $K$

• The freezing point of the solution $= T\°_{f}- \triangle T_{f}$

                                                                   $=273 K-25.002K=247.998 K$

• The freezing point of the solution $= 247.998 K$

           $247.998 K= (247.998-273) \°C=-25.002 \°C$

• The freezing point of the solution $=-25.002 \°C$

Given that, the freezing point of the solution $=- X\°C$

By calculation, we get the freezing point of solution $=-25.002 \°C$

Comparing these two we get the value of $X$.

• $X= 25.002 \°C$

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