Answer:
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
