When $10^{-3} \mathrm{M}$ solution of glucose in water, freezes at $-0.0186^{\circ} \mathrm{C}$, then at…
When $10^{-3} \mathrm{M}$ solution of glucose in water, freezes at $-0.0186^{\circ} \mathrm{C}$, then at what temperature $10^{-3} \mathrm{M}$ solution of $\mathrm{NaCl}$ will freeze?
$0{ }^{\circ} \mathrm{C}$
$0.186^{\circ} \mathrm{C}$
$-0.186^{\circ} \mathrm{C}$
$-0.0372^{\circ} \mathrm{C}$
Solution
$\begin{aligned} & \Delta \mathrm{T}_{\mathrm{f}}=0.0^{\circ} \mathrm{C}-\left(-0.0186^{\circ} \mathrm{C}\right)=0.086^{\circ} \mathrm{C} \\ & \text { (glucose) }\end{aligned}$
$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \times \mathrm{m} \times \mathrm{k}_{\mathrm{f}}$
Both solutions have the same molar concentrations so their molal concentration will also be the same.
$\begin{aligned}
& \text { Also, } \text { i for } \mathrm{NaCl}=2 \\
& \begin{aligned}
\Rightarrow \quad & \Delta \mathrm{T}_{\mathrm{f}}(\mathrm{NaCl})=2 \times \Delta \mathrm{T}_{\mathrm{f}} \text { (glucose) } \\
& =2 \times 0.0186{ }^{\circ} \mathrm{C}=0.0372{ }^{\circ} \mathrm{C}
\end{aligned}
\end{aligned}$
Thus, the $\mathrm{NaCl}$ solution will freeze at $-0.0372{ }^{\circ} \mathrm{C}$