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2 years ago

How many grams of CaF2 are present in 1.25 L of a 0.15 M solution of CaF2? How do I find the grams I am confused on that part?

Chemistry

1 answer:

dybincka [34]2 years ago

7 0

Answer:

Mass = 14.64 g

Explanation:

Given data:

Volume of solution = 1.25 L

Molarity of Solution = 0.15 M

Mass of CaF₂ = ?

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

We will calculate the number of moles of CaF₂ and then determine the mass by using number of moles.

0.15 M = number of moles of solute / 1.25 L

number of moles of solute = 0.15 M × 1.25 L

number of moles of solute = 0.1875 mol/L × L

number of moles of solute = 0.1875 mol

Mass in gram:

Mass = number of moles × molar mass

Mass = 0.1875 mol ×78.07 g/mol

Mass = 14.64 g

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2 years ago

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3 years ago

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Explanation:

It is known that molality is the number of moles present in kg of solution.

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The given data is as follows.

Molar mass of ammonia = 17 g/mol

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Also, density = $\frac{1 g}{mL}$ = 1 kg/L

Therefore, molality will be calculated as follows.

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Concentration = 0.387 mg/L = $\frac{0.000412 g/L}{46 g/mol}$

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And, density = $\frac{1 g}{mL}$ = 1 kg/L

Hence, molality = $\frac{8.956 \times 10^{-6} mol/L}{1 kg/L}$

= $8.956 \times 10^{-6}$ mol/kg

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Hence, molality will be calculated as follows.

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Therefore, molality of given species is $5.89 \times 10^{-4} mol/kg$ for ammonia, $8.956 \times 10^{-6}$ mol/kg for nitrite, and 0.02181 mol/kg for nitrate ion.

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