Which of the following pairs of elements belong to the same group

The density of a 3.37M MgCl2 (FW = 95.21) is 1.25 g/mL. Calulate the molality, mass/mass percent, and mass/volume percent. So fa

Dafna1 [17]

Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

Solution : Given,

Density of solution = 1.25 g/ml

Molar mass of $MgCl_2$ (solute) = 95.21 g/mole

3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.

The volume of solution = 1 L = 1000 ml

Mass of $MgCl_2$ (solute) = 3.37 g

First we have to calculate the mass of solute.

$\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2$

$\text{Mass of }MgCl_2=3.37mole\times 95.21g/mole=320.86g$

Now we have to calculate the mass of solution.

$\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g$

Mass of solvent = Mass of solution - Mass of solute = 1250 - 320.86 = 929.14 g

Now we have to calculate the molality of the solution.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg$

The molality of the solution is, 0.0381 mole/Kg.

Now we have to calculate the mass/mass percent.

$\text{Mass by mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{320.86}{1250}\times 100=25.67\%$

The mass/mass percent is, 25.67 %

Now we have to calculate the mass/volume percent.

$\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100=\frac{320.86}{1000}\times 100=32.086\%$

The mass/volume percent is, 32.086 %

Therefore, the molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

8 0

3 years ago

5.1169 mol of Ne is held at 0.9148 atm and 911 K. What is the volume of its container in liters?

sveticcg [70]

By applying the Boyle's equation and substituting our given data the volume of the container was found to be 418.14 Litres

<h3>Boyle's Law</h3>

Given Data

number of moles of Ne = 5.1169 mol

Pressure = 0.9148 atm

Temperature = 911 K

We know that the relationship between pressure and temperature is given as

PV = nRT

R = 0.08206

Making the volume subject of formula we have

V= nRT/P

Substituting our given data to find the volume we have

V = 5.1169*0.08206*911/0.9148

V = 382.522353554/0.9148

V = 418.14 L

Learn more about Boyle's law here:

brainly.com/question/469270

5 0

2 years ago

Which best describes the particles in a gas when the temperature rises from 23 °C to 46 °C?

mariarad [96]

I think the correct answer is b. Temperature is proportional to the average kinetic energy so when temperarure rises so will the average kinetic energy. I hope this helps. Let me know if anything is unclear.

5 0

3 years ago

Which of the following ions is least likely to form colored compounds?

andreyandreev [35.5K]

Answer:

i think d maybe correct me if im wrong

Explanation:

3 0

2 years ago

A chemist determines by measurements that 0.0550 moles of nitrogen gas participate in a chemical reaction. Calculate the mass of

shtirl [24]

The mass of nitrogen gas that participated in the chemical reaction is 1.54g

HOW TO CALCULATE MASS OF AN ELEMENT:

Mass of a substance can be calculated by multiplying the number of moles in mol of the substance by its molecular mass in g/mol. That is;

mass (M) = molar mass (MM) × number of moles (n)

According to this question, a chemist determines by measurements that 0.0550 moles of nitrogen gas (N2) participate in a chemical reaction.

The molecular mass of nitrogen gas (N2) = 14.01(2)

= 28.02g/mol

Hence, the mass of the nitrogen gas that participated in the chemical reaction is calculated as follows:

Mass (g) = 0.0550 mol × 28.02 g/mol

Mass = 1.5411

Therefore, the mass of nitrogen gas that participated in the chemical reaction is 1.54g

Learn more: brainly.com/question/18269198

6 0

3 years ago