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3 years ago

What is the mass of 20 moles of fluorine?

Chemistry

1 answer:

Daniel [21]3 years ago

7 0

Answer:

380g

Explanation:

Given parameters:

Number of moles of Fluorine = 20moles

Unknown:

Mass of Fluorine = ?

Solution:

The number of moles is the unit of quantifying the amount of particles in a substance.

So;

Mass of a substance = number of moles x molar mass

Molar mass of F = 19g/mol

Now;

Mass of F = 20 x 19 = 380g

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A student dissolves 15.g of sucrose C12H22O11 in 300.mL of a solvent with a density of 1.13/gmL. The student notices that the vo

Kruka [31]

Answer: The molarity and molality of sucrose solution is 0.146 M and 0.129 m respectively

Explanation:

Calculating the molarity of solution:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of sucrose = 15 g

Molar mass of sucrose = 342.3 g/mol

Volume of solution = 300 L

Putting values in above equation, we get:

$\text{Molarity of sucrose solution}=\frac{15\times 1000}{342.3\times 300}\\\\\text{Molarity of sucrose solution}=0.146M$

Hence, the molarity of sucrose solution is 0.146 M

Calculating the molality of solution:

To calculate the mass of solvent, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solvent = 1.13 g/mL

Volume of solvent = 300 mL

Putting values in above equation, we get:

$1.13g/mL=\frac{\text{Mass of solvent}}{300mL}\\\\\text{Mass of solvent}=(1.13g/mL\times 300mL)=339g$

To calculate the molality of solution, we use the equation:

$\text{Molality of solution}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

$m_{solute}$ = Given mass of solute (sucrose) = 15 g

$M_{solute}$ = Molar mass of solute (sucrose) = 342.3 g/mol

$W_{solvent}$ = Mass of solvent = 339 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{15\times 1000}{342.3\times 339}\\\\\text{Molality of solution}=0.129m$

Hence, the molality of sucrose solution is 0.129 m

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