A student dissolves 13.g of ammonia NH3 in 375.mL of a solvent with a density of 0.84/gmL. The student notices that the volume of the solvent does not change when the ammonia dissolves in it.Calculate the molarity and molality of the student's solution. Be sure each of your answer entries has the correct number of significant digits.molarity =

To calculate the molarity and molality of the solution, we first need to find the number of moles of ammonia and the volume of the solution in liters.

Step 1: Calculate the number of moles of ammonia (NH3) The molar mass of NH3 is approximately 17.03 g/mol. So, the number of moles of NH3 is given by the mass of NH3 divided by its molar mass.

Number of moles = mass / molar mass Number of moles = 13 g / 17.03 g/mol = 0.763 mol (rounded to three decimal places for significant figures)

Step 2: Calculate the volume of the solution in liters The volume of the solution is given in mL, so we need to convert it to liters. There are 1000 mL in 1 L, so:

Volume in L = 375 mL / 1000 = 0.375 L

Step 3: Calculate the molarity of the solution Molarity is defined as the number of moles of solute (in this case, NH3) divided by the volume of the solution (in liters).

Molarity = moles / volume Molarity = 0.763 mol / 0.375 L = 2.04 M (rounded to three significant figures)

Step 4: Calculate the molality of the solution Molality is defined as the number of moles of solute divided by the mass of the solvent (in kilograms). The mass of the solvent can be found by multiplying its volume by its density.

Mass of solvent = volume x density Mass of solvent = 375 mL x 0.84 g/mL = 315 g = 0.315 kg

Molality = moles / mass of solvent Molality = 0.763 mol / 0.315 kg = 2.42 m (rounded to three significant figures)

So, the molarity of the solution is 2.04 M and the molality is 2.42 m.