In the laboratory you dissolve 19.1 g of ammonium fluoride in a volumetric flask and add water to a total volume of 375 . mL. What is the molarity of the solution

1.376 M

Explanation:

The following data were obtained from the question:

Mass of ammonium fluoride (NH₄F) = 19.1 g

Volume of solution = 375 mL

Molarity of ammonium fluoride (NH₄F) =?

Next, we shall convert 375 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

375 mL = 375 mL × 1 L / 1000 mL

375 mL = 0.375 L

Next, we shall determine the number of mole in 19.1 g of ammonium fluoride (NH₄F). This can be obtained as follow:

Mass of NH₄F = 19.1 g

Molar mass of NH₄F = 14 + (4×1) + 19

= 14 + 4 + 19

Molar mass of NH₄F = 37 g/mol

Mole of NH₄F =?

Mole = mass /Molar mass

Mole of NH₄F = 19.1 / 37

Mole of NH₄F = 0.516 mole

Finally, we shall determine the molarity of the solution. This can be obtained as follow:

Volume of solution = 0.375 L

Mole of NH₄F = 0.516 mole

Molarity of NH₄F =?

Molarity = mole /Volume

Molarity of NH₄F = 0.516 / 0.375

Molarity of NH₄F = 1.376 M

Therefore, the molarity of the ammonium fluoride (NH₄F) is 1.376 M

answer: at room temperature (roughly 20 degrees c), you can dissolve 203.9 grams (or 200 grams) of sugar in 100 ml of water. as water temperature increases, the amount of sugar that will dissolve increases, up to about 500 grams of sugar in 100 ml water at 100 degrees c.

explanation: