What role does water play in a solution

1 answer:

eduard3 years ago

4 0

If there is more water than anything else, water is the solvent. If water is a minor component of the solution, the water is a solute. For example, "190 proof alcohol" is 95 % alcohol and 5 % water. In this solution, alcohol is the solvent and water is the solute.

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What is a solution?

beks73 [17]

Answer: Its B

Explanation: *Dabs*

6 0

3 years ago

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How many moles of copper(II)sulfate are in in 5.00 L of a 2.00 M solution?

Julli [10]

Answer:

There are 10 moles of copper(II)sulfate.

Explanation:

We know that molarity is a concentration measure that expresses the moles of solute per liter of solution: in this case, being 1.30 molar, there are 1.30 moles of solute in 1 liter of solution. We solve with the following simple rule of three:

1 L solution-----2 moles of copper(II)sulfate

5 L solution----x= (5 L solution x 2 moles of copper(II)sulfate)/1 L solution

x= 10 moles of copper(II)sulfate

3 0

4 years ago

A chemist prepares a solution of copper(II) fluoride by measuring out of copper(II) fluoride into a volumetric flask and filling

Simora [160]

The question is incomplete, here is the complete question.

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.

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

Explanation:

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:

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