Chemistry, 14.07.2019 04:30 Connor20000006

Estimate the boiling point of a 0.6 molal solution of nacl in water. the boiling point elevation constant of water is 0.51 k/molal. give your answer in °c. 99.7˚c 373.45 ˚c 373.75 ˚c 100.6 ˚c 100.3˚c

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Answer from: loganhowington26

Boiling point of 0.6 molar solution of NaCl in water is $\boxed{100.6{\text{ }}^\circ {\text{C}}}$ .

Further Explanation:

Some of the properties of substances depend only on concentration of solute particles and not depending upon identities of solutes. Such properties are termed as colligative properties.

Following four are colligative properties:

Relative lowering of vapor pressure

Elevation in boiling point

Depression in freezing point

Osmotic pressure

The formula to calculate the elevation in boiling point elevation is as follows:

$\Delta {{\text{T}}_{\text{b}}} = {\text{i}}{{\text{k}}_{\text{b}}}{\text{m}}$ …… (1)

Here,

$\Delta {{\text{T}}_{\text{b}}}$ is boiling point elevation.

i is the vont Hoff factor.

${{\text{k}}_{\text{b}}}$ is boiling point elevation constant.

m is molality of solution.

NaCl is an ionic compound and dissociates to form ions as follows:

${\text{NaCl}} \rightleftharpoons {\text{N}}{{\text{a}}^ + } + {\text{C}}{{\text{l}}^ - }$

Since NaCl dissociates to give two ions, its vont Hoff factor becomes two.

Substitute 2 for i, 0.6 m for m and $0.51{\text{ K/m}}$ for ${{\text{k}}_{\text{b}}}$ in equation (1).

$\begin{aligned}\Delta {{\text{T}}_{\text{b}}} &= 2\left( {0.51{\text{ K/m}}} \right)\left( {0.6{\text{ m}}} \right) \\&= 0.612{\text{ K}} \\\end{aligned}$

Boiling point of pure water is $100{\text{ }}^\circ {\text{C}}$ . It is to be converted into Kelvin with help of following conversion factor:

$0{\text{ }}^\circ {\text{C}} = 273{\text{ K}}$

Therefore boiling point of pure water can be calculated as follows:

$\begin{aligned}{\text{Boiling point of pure water}} &= \left( {100 + 273} \right){\text{ K}} \\&= {\text{373 K}} \\\end{aligned}$

The formula to calculate the temperature change in the NaCl solution is as follows:

${{\Delta }}{{\text{T}}_{\text{b}}} = {\text{Boiling}}\;{\text{point}}\;{\text{of }}{\text{NaCl}} - {\text{Boiling}}\;{\text{point}}\;{\text{of }}{\text{pure water}}$ …… (2)

Rearrange equation (2) to calculate boiling point of NaCl.

${\text{Boiling point of NaCl}} = {{\Delta }}{{\text{T}}_{\text{b}}} + {\text{Boiling point of pure water}}$ …… (3)

Substitute 373 K for boiling point of pure water and 0.612 K for $\Delta {{\text{T}}_{\text{b}}}$ in equation (3).

$\begin{aligned}{\text{Boiling point of NaCl}} &= 0.612{\text{ K}} + 373{\text{ K}} \\&= 373.612{\text{ K}} \\\end{aligned}$

Boiling point of NaCl is to be converted into degrees Celsius as follows:

$\begin{aligned} {\text{Boiling point of NaCl}} &= \left( {373.612 - 273} \right){\text{ }}^\circ {\text{C}} \\ &={\text{ 100}}{\text{.612 }}^\circ {\text{C}} \\&\approx 100.6{\text{ }}^\circ {\text{C}} \\ \end{aligned}$

Learn more:

Choose the solvent that would produce the greatest boiling point elevation: What is the molarity of the stock solution?

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Colligative properties

Keywords: colligative properties, boiling point, elevation in boiling point, boiling point of NaCl, NaCl, concentration, solute particles, relative lowering of vapor pressure, depression in freezing point, osmotic pressure, 100.6 degrees Celsius, 0.612 K.

Answer from: Quest

The answer would be

Answer from: Quest

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