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3 years ago

The concentration of a solution can be expressed in

Chemistry

2 answers:

puteri [66]3 years ago

3 0

The concentration of a solution can be expressed in (4) <span>moles per liter~</span>

nordsb [41]3 years ago

3 0

The concentration of a solution can be expressed in <u>(4) moles per liter</u>

<h3>Further explanation </h3>

Stoichiometry in Chemistry learn about chemicals mainly emphasizes quantitative, such as the calculation of volume, mass, number, which is related to numbers, molecules, elements, etc.

In the reaction there are also manifestations of reagent substances namely gas (g), liquid (liquid / l), solid (solid / s) and solution (aqueous / aq).

The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight / volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.

Kelvin

Temperature is used as a guide to the level of heat energy from objects

Temperature cannot be measured by the surface of the hand because it is not accurate but can be measured by a thermometer. Temperature units include Celsius, Fahrenheit, Reamur or Kelvin

milliliters

Volume is a derivative quantity derived from the length

The unit of volume can be expressed in liters or milliliters or cubic meters

joules per kilogram

Energy can be interpreted as the ability to do work

Energy units can be stated in joules or calories (J or Cal)

moles per liter

Mole

The mole itself is the number of particles contained in a substance amounting to 6.02.10²³

Mole is the result of dividing the amount of substance mass and its molar mass

Molarity is a way to express the concentration of the solution

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

$\large{\boxed {\bold {M ~ = ~ \frac {n} {V}}}$

Where

M = Molarity

n = Number of moles of solute

V = Volume of solution

So to find the number of moles can be expressed as

<h3>Learn more </h3>

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Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

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