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6 days ago

41. A 13.0% solution of K2CO3 by mass has a density of 1.09 g/cm3. Calculate the molality of the solution.

Chemistry

1 answer:

lions [985]6 days ago

5 0

Answer:

The molality of the solution is 1.08 m.

Explanation:

First, determine the mass of the solvent.

A 13% solution by mass indicates that 13 grams are found in every 100 grams of solution.

Thus, solution mass = solute mass + solvent mass

100 g = 13 g + solvent mass

Therefore, solvent mass = 100 g - 13 g → 87 g

Next, we calculate the moles of solute (mass / molar mass):

13 g / 138.2 g/mol = 0.094 moles

Finally, to find the molality, which is the moles of solute per 1 kg of solvent (mol/kg), we convert the solvent mass to kg:

87 g. 1 kg / 1000 g = 0.087 kg

Then, molality → 0.094 mol / 0.087 kg = 1.08 m

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66.667 mL of 3.000 M H2SO4 (aq) solution was neutralized by the stoichiometric amount of 4.000 M Al(OH)3 solution in a coffee cu

eduard [944]

Answer:

$\large \boxed{\Delta_{\textbf{r}}H =\text{-4600 J$\cdot$ mol}^{-1}}$

Explanation:

This scenario is unrealistic since Al(OH)₃ is not soluble in water.

The question consists of two parts:

A. Stoichiometry — where we determine volumes, masses, and moles for the products

B. Calorimetry — where we assess the enthalpy of the reaction.

A. Stoichiometry

1. Determine the volume of Al(OH)₃

(a) The balanced chemical equation:

2Al(OH)₃ + 3H₂SO₄ ⟶ Al₂(SO₄)₃ + 6H₂O

M/V: 66.667

c/mol·L⁻¹: 4.000 3.000

(b) Moles of H₂SO₄

$\rm \text{66.667 mL H$_{2}$}SO_{4} \times \dfrac{\text{3.000 mmol H$_{2}$SO}_{4}}{\text{1 mL H$_{2}$SO}_{4}} = \text{200.00 mmol H$_{2}$SO}_{4}$

(c) Moles of Al(OH)₃

The molar ratio stands at 2 mmol Al(OH)₃: 3 mmol H₂SO₄

$\text{Moles of Al(OH)}_{3} = \text{200.00 mmol of H$_{2}$SO}_{4} \times \dfrac{\text{2 mmol Al(OH)}_{3}}{\text{3 mmol H$_{2}$SO}_{4}}\\\\= \text{133.33 mmol Al(OH)}_{3}$

(d) Volume of Al(OH)₃

$\text{Moles of Al(OH)}_{3} = \text{200.00 mmol of H$_{2}$SO}_{4} \times \dfrac{\text{1 mL Al(OH)}_{3}}{\text{4 mmol H$_{2}$SO}_{4}} = \text{50.000 mL Al(OH)}_{3}$

B. Calorimetry

This reaction has two energy exchanges.

q₁ = heat from the reaction

q₂ = heat used to heat the calorimeter

q₁ + q₂ = 0

nΔH + mCΔT = 0

Data:

Moles of Al₂(SO₄)₃ = 0.066 667 mol

C = 1.10 J°C⁻¹g⁻¹

T_initial = 22.3 °C

T_final = 24.7 °C

Calculations

(a) Mass of solution

Assume solutions are as dense as water (though not realistic).

Mass of sulfuric acid solution = 66.667 g

Mass of aluminium hydroxide solution = 50.000

TOTAL = 116.667 g

(b) ΔT

ΔT = T_final - T_initial = 24.7 °C - 22.3 °C = 2.4°C

(c) ΔH

$\begin{array}{ccccl}n\Delta H & +& mC \Delta T& = &0\\\text{0.066 667 mol }\times \Delta H& + & \text{116.667 g} \times 1.10 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 2.4 \, ^{\circ}\text{C} & = & 0\\0.066667 \Delta H \text{ mol} & + & \text{310 J} & = & 0\\&&0.066667 \Delta H \text{ mol} & = & \text{-310 J} & & \\\end{array}\\$

$\begin{array}{ccccl}& &\Delta H & = & \dfrac{\text{-310 J}}{\text{0.066667 mol}}\\\\& &\Delta H & = & \textbf{-4600 kJ/mol}\\\end{array}\\\large \boxed{\mathbf{\Delta_{\textbf{r}}H} =\textbf{-4600 J$\cdot$ mol}^{\mathbf{-1}}}$

This result appears nonsensical, but it is derived from your given figures.

6 0

4 days ago

What is the mass of 22.4 L of H2 at STP?

eduard [944]

A. 1.01 is the accurate result

Because

The formula used is Pv= nRT

P=1 atm

V= 22.4 L

N= x

R= 0.0821

T= 273 K (since it’s standard temperature)

Thus, (1)(22.4)=(x)(0.0821)(273)

X= 1.001

7 0

2 days ago

Read 2 more answers

he population of the Earth is roughly eight billion people. If all free electrons contained in this extension cord are evenly sp

eduard [944]

1) Drift velocity: $3.32\cdot 10^{-4}m/s$

2. $5.6\cdot 10^{13}$ electrons per individual

Explanation:

In a conducting material with an electric current, the drift velocity of electrons can be calculated using this equation:

$v_d=\frac{I}{neA}$

where

I stands for current

n represents the density of free electrons

$e=1.6\cdot 10^{-19}C$ indicates the charge of an electron

A signifies the wire's cross-sectional area

The wire's cross-sectional area can be determined as

$A=\pi r^2$

where r denotes the wire's radius. Thus, the equation transforms to

$v_d=\frac{I}{ne\pi r^2}$

In this scenario, we have:

I = 8.0 A as the current

$8.5\cdot 10^{28} m^{-3}$ indicates the free electron concentration

d = 1.5 mm is the diameter, making the radius

r = 1.5/2 = 0.75 mm = $0.75\cdot 10^{-3}m$

So, the resulting drift velocity is:

$v_d=\frac{8.0}{(8.5\cdot 10^{28})(1.6\cdot 10^{-19})\pi(0.75\cdot 10^{-3})^2}=3.32\cdot 10^{-4}m/s$

The entire length of the cord is

L = 3.00 m

And the cross-sectional area is

$A=\pi r^2=\pi (0.75\cdot 10^{-3})^2=1.77\cdot 10^{-6} m^2$

Consequently, the volume of the cord is

$V=AL$ (1)

The number of electrons per unit volume is $n$ , thus the total electrons in this cord would be

$N=nV=nAL=(8.5\cdot 10^{28})(1.77\cdot 10^{-6})(3.0)=4.5\cdot 10^{23}$

Overall, the Earth's population rounds to 8 billion individuals, equating to

$N'=8\cdot 10^9$

Hence, the number of electrons distributed to each person is:

$N_e = \frac{N}{N'}=\frac{4.5\cdot 10^{23}}{8\cdot 10^9}=5.6\cdot 10^{13}$

7 0

5 days ago

A chamber with a fixed volume is shown above. The temperature of the gas inside the chamber before heating is 25.2 C and it’s pr

KiRa [971]

Answer:

Explanation:

Given data:

Initial temperature T₁ = 25.2°C = 298.2K

Initial pressure P₁ = 0.6atm

Final temperature = 72.4°C = 345.4K

What we need to find:

Final pressure = ?

To determine this, we apply a modified version of the combined gas law with constant volume. This simplifies our calculations to:

$\frac{P_{1} }{T_{1} } = \frac{P_{2} }{T_{2} }$

Here, P and T signify pressure and temperatures, 1 refers to initial and 2 to final temperatures.

Now we can substitute the known variables:

$\frac{0.6}{298.2} = \frac{P_{2} }{345.4}$

P₂ = 0.7atm

3 0

14 days ago

n the table below, write the density of each object. Then predict whether the object will float or sink in each of the fluids. W

Anarel [852]

Answer:

0.5 g/mL----- will float

1.0 g/mL---- will float

2.0 g/mL----- will sink

Explanation:

Objects with a density less than or equal to that of water will float due to having a lower mass, while objects with a density exceeding that of water will sink because their mass is greater than that of water. Thus, objects with a density of 0.5 g/mL and 1.0 g/mL will float since they are less dense than water (1 g/mL), whereas an object with a density of 2.0 g/mL will sink.

3 0

11 days ago