Question #194498

A 0.674\,\text M

0.674M

0, point, 674, start text, M, end text

cobalt(II) chloride (\text{CoCl}_2

CoCl2

start text, C, o, C, l, end text, start subscript, 2, end subscript

) solution is prepared with a total volume of 0.0750\,\text{L}

0.0750L

0, point, 0750, start text, L, end text

. The molecular weight of \text{CoCl}_2

CoCl2

start text, C, o, C, l, end text, start subscript, 2, end subscript

is 128.9\,\dfrac{\text{g}}{\text{mol}}

128.9mol

g

128, point, 9, start fraction, start text, g, end text, divided by, start text, m, o, l, end text, end fraction

.

**What mass of \text{CoCl}_2**

CoCl2

start text, C, o, C, l, end text, start subscript, 2, end subscript

** (in grams) is needed for the solution?**

*Express the answer using 3 significant figures.*

Expert's answer

A 0.674 M cobalt(II) chloride (CoCl_{2}) solution is prepared with a total volume of 0.0750 L. The molecular weight of CoCl_{2} is 128.9 g/mol. What mass of CoCl_{2} (in grams) is needed for the solution?

Express the answer using 3 significant figures.

**Solution:**

Molarity of CoCl_{2} solution = Moles of CoCl_{2} / Solution volume

Moles of CoCl_{2} = Mass of CoCl_{2} / Molecular weight of CoCl_{2}

Molarity of CoCl_{2} solution = Mass of CoCl_{2} / (Molecular weight of CoCl_{2} × Solution volume)

Therefore,

Mass of CoCl_{2 }= Molarity of CoCl_{2} solution × Molecular weight of CoCl_{2} × Solution volume

Mass of CoCl_{2 }= (0.674 mol L^{-1}) × (128.9 g mol^{-1}) × (0.0750 L) = 6.5159 g = 6.52 g

**Mass of CoCl**_{2 }**= 6.52 g**

**Answer: 6.52 grams of CoCl**_{2}** is needed for the solution.**

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