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QuestionCalculate the molarity of a solution made by dissolving 0.026 g of CuF₂ in 250 mL of water. Round to 2 significant digits.

Studdy Solution

STEP 1

Assumptions1. The mass of copper(II) fluoride (Cu) used is0.026 g. The volume of the solution prepared is250 mL3. We need to find the concentration in mol/L4. The molar mass of Cu is101.55 g/mol (63.55 g/mol for Cu,19 g/mol for each F)

STEP 2

First, we need to convert the mass of Cu2 from grams to moles. We can do this using the molar mass of Cu2.
Moles of Cu2=Mass of Cu2Molar mass of Cu2Moles\, of\, Cu2 = \frac{Mass\, of\, Cu2}{Molar\, mass\, of\, Cu2}

STEP 3

Now, plug in the given values for the mass of Cu2 and the molar mass of Cu2 to calculate the moles of Cu2.
Moles of Cu2=0.026 g101.55 g/molMoles\, of\, Cu2 = \frac{0.026\, g}{101.55\, g/mol}

STEP 4

Calculate the moles of Cu2.
Moles of Cu2=0.026 g101.55 g/mol=0.00026 molMoles\, of\, Cu2 = \frac{0.026\, g}{101.55\, g/mol} =0.00026\, mol

STEP 5

Next, we need to convert the volume of the solution from mL to L. We can do this by dividing the volume in mL by1000.
Volume in L=Volume in mL1000Volume\, in\, L = \frac{Volume\, in\, mL}{1000}

STEP 6

Now, plug in the given value for the volume in mL to calculate the volume in L.
Volume in L=250 mL1000Volume\, in\, L = \frac{250\, mL}{1000}

STEP 7

Calculate the volume in L.
Volume in L=250 mL1000=0.25 LVolume\, in\, L = \frac{250\, mL}{1000} =0.25\, L

STEP 8

Now that we have the moles of Cu2 and the volume of the solution in L, we can calculate the concentration in mol/L. The concentration in mol/L is calculated by dividing the moles of solute by the volume of the solution in L.
Concentration in mol/L=Moles of soluteVolume in LConcentration\, in\, mol/L = \frac{Moles\, of\, solute}{Volume\, in\, L}

STEP 9

Plug in the values for the moles of Cu2 and the volume in L to calculate the concentration in mol/L.
Concentration in mol/L=.00026 mol.25 LConcentration\, in\, mol/L = \frac{.00026\, mol}{.25\, L}

STEP 10

Calculate the concentration in mol/L.
Concentration in mol/L=0.00026 mol0.25 L=0.00104 mol/LConcentration\, in\, mol/L = \frac{0.00026\, mol}{0.25\, L} =0.00104\, mol/LThe concentration of the chemist's copper(II) fluoride solution is0.00104 mol/L, rounded to two significant digits, it is0.0010 mol/L.

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