Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water....
Question:
Calculate the freezing point of a solution containing {eq}\rm 5.0 {/eq} grams of {eq}\rm KCl {/eq} and {eq}\rm 550.0 {/eq} grams of water. The molal-freezing-point-depression constant {eq}\rm (K_f) {/eq} for water is {eq}\rm 1.86\ ^\circ C/m {/eq}. (Assume {eq}\rm 100 \% {/eq} dissociation)
A. {eq}\rm -0.45 ^\circ C {/eq}.
B. {eq}\rm -0.23 ^\circ C {/eq}.
C. {eq}\rm -1.23 ^\circ C {/eq}.
D. {eq}\rm +0.23 ^\circ C {/eq}.
E. {eq}\rm +0.45 ^\circ C {/eq}.
Freezing Point Depression:
Normally, the solvent particles form a lattice structure when freezing. This allows the solvent to be converted to solid. However, when a non-volatile solute is added to the solvent, the solute particles will disrupt the lattice structure requiring the system a much lower temperature for the complete arrangement of the solvent particles to occur.
Answer and Explanation: 1
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- The answer is A. {eq}\rm -0.45^\circ C {/eq}.
The equation that will be used is:
{eq}\Delta Tf = iKfm {/eq}
where:
- {eq}\Delta Tf {/eq} -...
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Using Colligative Properties to Determine Molar Mass
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Chapter 8 / Lesson 7
6.6K
Colligative properties are important to determine molar mass as related to vapor pressure, boiling point, freezing point, and osmotic pressure. Examine freezing point depression to understand the process for finding molar mass.
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