A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in mol/L of the chemist's copper(II) fluoride solution. Round your answer to 3 significant digits.

<u>Answer:</u> The concentration of copper fluoride in the solution is $4.90\times 10^{-3}mol/L$

<u>Explanation:</u>

To calculate the molarity of solute, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Given mass of copper (II) fluoride = 0.0498 g

Molar mass of copper (II) fluoride = 101.54 g/mol

Volume of solution = 100.0 mL

Putting values in above equation, we get:

$\text{Molarity of copper (II) fluoride)=\frac{0.0498\times 1000}{101.54\times 100.0}\\\\\text{Molarity of copper (II) fluoride}=4.90\times 10^{-3}mol/L$

Hence, the concentration of copper fluoride in the solution is $4.90\times 10^{-3}mol/L$

4 0

3 years ago

Which element is most reactive sodium,magnesium, or aluminum and why?

ludmilkaskok [199]

Magnesium would be more reactive.

6 0

3 years ago

A 225. mL sample 3.5M solution of KBr is diluted to a final volume of 750. mL. What is the final concentration of this solution?

Allushta [10]

C= comcentration
V= volume
C1*V1=C2*V2
3.5M*225.mL = xM*750 mL
x=(3.5*225/750) =1.05M≈1.1M

7 0

4 years ago