Courses
Courses for Kids
Free study material
Offline Centres
More
Store

Calculate the mass of ascorbic acid $\left( {Vitamin \;C,\;{C_6}{H_8}{O_6}} \right)$ to be dissolved in $75\;g$ of acetic acid to lower its melting point by $1.5^\circ C$. ${K_f} = 3.9\;K \cdot kg \cdot mo{l^{ - 1}}$Given: - The formula of ascorbic acid is: ${C_6}{H_8}{O_6}$- The mass of acetic acid is: ${m_{C{H_3}COOH}} = 75\;g$- Lowering in the melting point is: $\Delta {T_f} = 1.5^\circ C$- The molal depression constant for acetic acid is: ${K_f} = 3.9\;K \cdot kg \cdot mo{l^{ - 1}}$

Last updated date: 20th Jun 2024
Total views: 405k
Views today: 4.05k
Verified
405k+ views
Hint:
We can use the relationship between the molality of the solution and lowering of freezing point to determine the required mass.

Complete step by step solution:
We know that the numerical value of freezing point is equal to that of melting point as both are the temperature at which liquid and solid states are in equilibrium with each other. So, we can say that the given lowering in the melting point can be taken as depression in freezing point.

For depression in freezing point of a solvent, we have a relationship between the depression in the freezing point and molality of the solution that can be written as follows:
$\Delta {T_f} = {K_f}m$

Here, $\Delta {T_f}$ is the depression in the freezing point of the solvent, ${K_f}$ is the molal depression constant for the given solvent and $m$ is the molality of the solution.

We can write the molality of a solution in terms of moles of solute $\left( {{n_{solute}}} \right)$ and mass of solvent $\left( {{m_{solvent}}} \right)$ as follows:
$m = \dfrac{{{n_{solute}}}}{{{m_{solvent}}\;in\;kg}}$

Let’s rewrite the above expression in terms of masses only as follows:
$m = \dfrac{{\left( {{m_{solute}}/{M_{solute}}} \right)}}{{{m_{solvent}}\;in\;kg}}$
Here, ${M_{solute}}$ is the molar mass of the solute.

Now we can use the above expression to rewrite equation for depression in the freezing point and molality of the solution as follows:
$\Delta {T_f} = {K_f}\left\{ {\dfrac{{\left( {{m_{solute}}/{M_{solute}}} \right)}}{{{m_{solvent}}\;in\;kg}}} \right\}$
Let’s rearrange this equation for mass of solute:
${K_f}\left\{ {\dfrac{{\left( {{m_{solute}}/{M_{solute}}} \right)}}{{{m_{solvent}}\;in\;kg}}} \right\} = \Delta {T_f}\\ \dfrac{{\left( {{m_{solute}}/{M_{solute}}} \right)}}{{{m_{solvent}}\;in\;kg}} = \dfrac{{\Delta {T_f}}}{{{K_f}}}\\ \left( {{m_{solute}}/{M_{solute}}} \right) = \left( {\dfrac{{\Delta {T_f}}}{{{K_f}}}} \right) \times {m_{solvent}}\;in\;kg\\ {m_{solute}} = \left( {\dfrac{{\Delta {T_f}}}{{{K_f}}}} \right) \times {m_{solvent}}\;in\;kg \times {M_{solute}}$

Here, we have ascorbic acid as solute and acetic acid as solvent, so we can write the expression for them:
${m_{{C_6}{H_8}{O_6}}} = \left( {\dfrac{{\Delta {T_f}}}{{{K_f}}}} \right) \times {m_{C{H_3}COOH}}\;in\;kg \times {M_{{C_6}{H_8}{O_6}}}$

Now, let’s convert the units of mass of acetic acid as follows:
${m_{C{H_3}COOH}} = 75\;g \times \left( {\dfrac{{1\;kg}}{{1000\;g}}} \right)\\ = {\rm{0}}{\rm{.075}}\;kg$

Finally, we will calculate the required mass of ascorbic acid by substituting values in the derived expression as follows:
${m_{{C_6}{H_8}{O_6}}} = \left( {\dfrac{{{\rm{1}}{\rm{.5}}\;K}}{{3.9\;K \cdot kg \cdot mo{l^{ - 1}}}}} \right) \times {\rm{0}}{\rm{.075}}\;kg \times 176.12\;g \cdot mo{l^{ - 1}}\\ = {\rm{5}}{\rm{.08}}\;g$

Hence, the required mass of ascorbic acid is ${\rm{5}}{\rm{.08}}\;g$.

Note:

We have to keep in mind that depression in freezing point is a change in temperature so it will be the same on the Kelvin scale as well.