What amount (in moles) of NH4CI is needed to prepare 275 mL of a 0.35 M ammonium chloride solution?

Answer 1

To calculate the amount of NH4Cl needed to prepare 275 mL of a 0.35 M solution, we first need to convert the volume from mL to L:

275 mL = 0.275 L

Now, we can use the formula for calculating moles:

Moles = (Molarity) x (Volume in L)

Moles = 0.35 mol/L x 0.275 L = 0.09625 mol

Therefore, 0.09625 moles of NH4Cl is needed to prepare 275 mL of a 0.35 M ammonium chloride solution.