Question

Urea , ${(N{H_2})_2}CO$ is dissolved in $100g$ of water. The solution freezes at $- {0.085^ \circ }C$. How many grams of urea were dissolved to make this solution ? (Given ${K_b} = 0.512$ ,${K_f}$ of water $= 1.858$ , ${K_b} = 0.512$ )

Answer 1

To deal with this question we have to know about the formula of depression in freezing point and the value is also given. Formula of depression in freezing point includes quantity of molality and since the value of ${K_f}$ is given we can easily detect molality from that. So , we will be getting molality per kilogram of urea which helps us to calculate the mass of urea in a given $100g$ of water.

Formula used: Formula for freezing point depression is given by,
$\Delta {T_f} = i \times {K_f} \times m$
Here, $\Delta {T_f} =$ depression in freezing point
$i =$ Van’t Hoff factor
$m =$ molality of solution
${K_f} =$ cryoscopic constant of given solvent

Complete step-by-step answer:
Before proceeding to the calculations we have to note down some basic point on depression in freezing point and later we can find out the molality from it. Actually, we have to know that freezing point depression gives out the difference between freezing point of solvent and solution. Next, we have to write down both freezing point depressions. Therefore,
Given freezing point depression of solution $= - {0.085^ \circ }C$
Here the solvent is pure water which has a normal freezing point of ${0^ \circ }C$ .

So, freezing point depression , that is , $\Delta {T_f} = {0^ \circ }C - ( - {0.085^ \circ }C)$ $= {0.085^ \circ }C$
Hence, from this we can understand that the freezing point of solution is ${0.085^ \circ }C$ which is lower than normal freezing point of solvent.

Next our duty is finding molality from the equation of depression in freezing point. The equation can be written once more below,
$\Delta {T_f} = i \times {K_f} \times m$
This can be again modify to a new equation to calculate molality which is as follows,
$m = \dfrac{{\Delta {T_f}}}{{i \times {K_f}}}$
$\Delta {T_f} = {0.085^ \circ }C$ , which we find above from a given value
$i = 1$ , since urea is non electrolyte it does not dissociate in aqueous forms and so the van’t hoff factor is one.
${K_f} = 1.858$ , which is given in question.

Now, let us substitute the values in the above equation. Then it will be as follows,
$m = \dfrac{{{{0.085}^ \circ }C}}{{1 \times {{1.858}^ \circ }Ckgmo{l^{ - 1}}}}$. Therefore, $m = 0.04575mol/kg$

So, from this answer we can understand that for $1kg$ of urea, we have $0.04575moles$. Since here we have $100g$ of water , the calculation of number of moles of urea in it will be as follows,
$100g$ water $\times \dfrac{{0.04575moles({{(N{H_2})}_2}CO)}}{{1000g({H_2}O)}}$ $= 0.004575moles$ of urea

Next we have to do an important process of conversion of grams to moles. So, for doing this we use molar mass of urea. So let us proceed the conversion as follows,
$0.004575moles$ of urea $\times \dfrac{{60.06g}}{{1mole{{(N{H_2})}_2}CO}}$ $= 0.27g$

Hence, $0.27g$ of urea were dissolved to make the solution.

Note: When dealing with this type of questions we have to clearly know about the particular formula which is to be used and its respective quantities as here it is about freezing point depression. We have to keep remember a point in mind always as in this question , we cannot directly find out the final answer instead we have to modify the particular formula to find out molality which in turn finds out the mass of urea. These methods can be followed in this type of any other questions. Also, here there is a greater chance to be wrong as to directly use the given ${T_f}$in the equation. It should not be done as we need to find the difference between freezing point of solvent and solution. Because the given ${T_f}$ will definitely change as here in this question the sign is reversed since the freezing point of solvent is ${0^ \circ }C$ .