# Question #4feb9

##### 1 Answer

You have to prepare a solution that has a given volume of **250.0 mL** and is **20%** **v/v** methanol, so the first thing you need to do is determine how much water and how much methanol this final solution must contain.

A

You know what the *volume of the solution* must be, so use the above equation to determine how much methanol you must use

This means that your final solution must contain **50.0 mL** of methanol and *enough* water to make the total volume equal to **250.0 mL**.

Since you know what the molarities of the Tris-HCl and glycine must be, you can use the volume of the solution to see how many moles of each you'd need

Now use the two compounds' molar masses to determine how many grams of each you'd need

and

And that's it. Start by adding **0.985 g** of tris-HCl and **3.60 g** of glycine to, say **100 mL** of water.

Add **50 mL** of methanol, and then adjust the volume of the solution to **250.0 mL** by adding extra water.

**SIDE NOTE** *As practice, you could do the calculations for a final volume of 1.00 L. To get your target solution, you'd take a 250-mL sample of that 1.00-L solution.*

*Hint: since the volume would be 4 times bigger (1.00 L vs. 250.0 mL), you'd need 4 times more of each ingredient.*