When you need to produce a variety of diluted solutions of a solute, you can dilute a series of stock solutions. A stock solution has a significantly higher concentration of the given solute (typically 101 to 104 times higher than those of the diluted solutions). The high concentration allows many diluted solutions to be prepared using minimal amounts of the stock solution. What volume of a 6.57 M stock solution do you need to prepare 500. mL of a 0.0490 M solution of HCl

3.73 mL

Explanation:

To solve this problem we can use the equation C₁V₁=C₂V₂, where:

We input the data:

And solve for V₁:

So 3.73 mL of a 6.57 M stock solution would be required to prepare 500 mL of a 0.0490 M solution.

1. 1 mole of anything contains 6.022x10^23 pieces  

d  

2. a  

3. c, number  

5. 1.24x10^24atoms / 6.022x10^23atoms/mole = 2moles h2 atoms  

6. fe2br3 mass = 2moles x mass fe + 3 moles x mass br  

molar mass = grams / number of moles = g/mole  

to find moles = g / molar mass  

to find mass = g = molar mass / moles