Question

What is the freezing point of a solution of 9.04 g of\({\bf{I_2}}\)in 75.5 g of benzene?

1. Outline the steps necessary to answer the following question.
2. Answer the question.

Short answer

The freezing point of 9.04 g of \({\bf{I_2}}\) in 75.5 g of benzene is 350.8 K.

Step-by-step solution

Freezing Point

The temperature is also defined because the point of temperature where liquid start changes into liquid in a very given atmospheric pressure.

Subpart (a)

Firstly, find the molality of the \({\bf{I_2}}\)

\({\bf{Molality = }}\frac{{{\bf{Number}}\,{\bf{of}}\,{\bf{moles}}}}{{{\bf{Mass}}\,{\bf{of}}\,{\bf{Solvent(kg)}}}}\)

This is important because the equation we use to see the boiling point is

\({\bf{\Delta }}{{\bf{T}}_{{\bf{freezing}}}}{\bf{ = i \times Kf \times m}}\),

where

Kf - the cryoscopic constant - supports on the solvent
m - themolality of the solution;
i - the van't Hoff factor - the quantity of ions per individual molecule ofsolute.
\({\bf{\Delta }}{{\bf{T}}_{{\bf{freezing}}}}\) - The temperature depression - is defined as

\({{\bf{T}}_{\bf{F}}}{\bf{(solvent)--}}{{\bf{T}}_{\bf{F}}}{\bf{(solution)}}\)

Subpart (b)

Mass of \({\bf{I2}}\) = 9.04g

Molar Mass of\({\bf{I2}}\)= 254g/mole

Number of Moles of iodine:

\(\begin{align}{n_{{I_2}}} &= \frac{{Mass\,of\,iodine}}{{Molar\,Mass}}\\ &= \frac{{9.04\,g}}{{254\,g/mol}}\\ &= 0.036\end{align}\)

\(Mass{\rm{ }}of{\rm{ }}Solvent{\rm{ }}\left( {Water} \right){\rm{ }} = {\rm{ }}75.5g = {\rm{ }}0.0755kg\)

\(\begin{align}Molality &= \frac{{Number\,of\,moles}}{{Mass\,of\,Solvent(kg)}}\\Molality &= \frac{{0.036\,mol}}{{0.0755\,kg}}\\ &= 0.471\,m\end{align}\)

\({\bf{\Delta }}{{\bf{T}}_{{\bf{freezing}}}}{\bf{ = i \times Kf \times m}}\)

The freezing point of pure water= 273.15K

\(\begin{align}\Delta Tf &= i \times {K_f} \times m\\353.25 - Tf &= 1 \times 5.12K.kg.mo{l^{ - 1}} \times 0.471\\Tf &= \left( {353.25 - 2.41} \right)K\\Tf &= 350.84K\end{align}\)