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2 years ago

Which solution has the higher boiling point?

Chemistry

1 answer:

dedylja [7]2 years ago

7 0

NaCl has higher boiling point than C₂H₆O₂ in H₂O.

<h3>What is molality?</h3>

The number of moles of solute in a solution equal to 1 kg or 1000 g of solvent is referred to as its molality.
The definition of molarity, on the other hand, is based on a certain volume of solution.
Mol/kg is a typical molality measurement unit in chemistry.

<h3>Calculation of boiling point:</h3>

When the non-volatile solute is dissolved in a solvent, the boiling point rises along with the molality (concentration) of the solute.

Given,

Mass of C₂H₆O₂ (solute) = 15.0 g

Mass of solvent (H₂O) = 0.50 kg

Molar mass of C₂H₆O₂ = 62 g/mol

Thus, the moles of solute = 15.0 g x 1.0 mol solute/62 g/mol

= 0.2419 mol

Therefore, molality(m) of C₂H₆O₂ (solute) = Amount of solute (mol)/ Mass of solvent

= 0.2419/0.50

= 0.4838 m

Similarly,

Given ,

Mass of NaCl (solute) = 15 g

Mass of solvent (H₂O) = 0.50 kg

Molar mass of NaCl (solute) = 58.44 g/mol

Thus, the moles of NaCl (solute) = 15.0 g x 1.0 mol solute/58.44

= 0.2566 mol

Therefore, molality(m) of NaCl (solute) = Amount of solute (mol)/ Mass of solvent

= 0.2566 mol/0.50 kg

= 0.51 m

Hence, the molality of NaCl (solute) is more than the molality of C₂H₆O₂(solute).

So, with an increase in the solute's concentration (molality), the boiling point rises.

Therefore, NaCl has higher boiling point than C₂H₆O₂ in H₂O.

Learn more about boiling point here:

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A solution is made by adding 50.0 ml of 0.200 m acetic acid (ka = 1.8 x 10–5) to 50.0 ml of 1.00 x 10–3m hcl. (a) calculate the

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Final pH of the solution: 2.79.

Explanation:

What's in the solution after mixing?

$\displaystyle c = \frac{n}{V}$ ,

where

$c$ is the concentration of the solute,

$n$ is the number of moles of the solute, and

$V$ is the volume of the solution.

$V(\text{Final}) = 0.050 \;\textbf{L} + 0.050\;\textbf{L} = 0.100\;\textbf{L}$ .

Acetic (ethanoic) acid:

$\displaystyle \begin{aligned}n &= c(\text{Before})\cdot V(\text{Before}) \\&= 0.050\;\text{L} \times 0.200\;\text{mol}\cdot\text{L}^{-1}\\ &= 0.0100\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{0.0100\;\text{mol}}{0.100\;\text{L}}\\ &= 0.100\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 0.100\;\text{M}\end{aligned}$ .

Hydrochloric acid HCl:

$\begin{aligned}n &= c(\text{Before})\cdot V(\text{Before})\\ &= 0.050\;\text{L} \times 1.00\times 10^{-3}\;\text{mol}\cdot\text{L}^{-1}\\ &= 5.00\times 10^{-5}\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{5.00\times 10^{-5}\;\text{mol}}{0.100\;\text{L}}\\ &= 5.00\times 10^{-4}\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 5.00\times 10^{-4}\;\text{M}\end{aligned}$ .

HCl is a strong acid. It will completely dissociate in water to produce H⁺. The H⁺ concentration in the solution before acetic acid dissociates shall also be $5.00\times 10^{-4}\;\text{M}$ .

The Ka value of acetic acid is considerably small. Acetic acid is a weak acid and will dissociate only partially when dissolved. Construct a RICE table to predict the portion of acetic acid that will dissociate. Let the change in acetic acid concentration be $-x\;\text{M}$ . $x > 0$ .

$\begin{array}{c|ccccc}\textbf{R}&\text{CH}_3\text{COOH}\;(aq) &\rightleftharpoons &\text{CH}_3\text{COO}^{-}\;(aq) &+& \text{H}^{+}\;(aq)\\\textbf{I}&0.100\;\text{M} & & & & 5.00\times 10^{-4}\;\text{M}\\\textbf{C}&-x\;\text{M} & & +x\;\text{M} & & +x\;\text{M} \\ \textbf{E}&0.100\;\text{M}-x\;\text{M} & & x\;\text{M} & & 5.00\times 10^{-4}\;\text{M} + x\;\text{M}\end{array}$ .

$\displaystyle K_a = \frac{[\text{CH}_3\text{COO}^{-}\;(aq)]\cdot[\text{H}^{+}\;(aq)]}{[\text{CH}_3\text{COOH}\;(aq)]} = \frac{x\cdot(x + 5.00\times 10^{-4})}{0.100 - x}$ .