# Question 4feb9

May 3, 2015

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 $\text{v/v}$ percent concentration is defined as volume of solute, in your case methanol, divided by volume of solution and multiplied by 100.

$\text{%v/v" = "volume of solute"/"volume of solution} \cdot 100$

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

$\text{20%" = V_"methanol"/"250.0 mL} \cdot 100$

${V}_{\text{methanol" = (20 * "250.0 mL")/100 = "50.0 mL}}$

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

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{\text{Tris-HCl" = 25 * 10^(-3)"M" * 250.0 * 10^(-3)"L" = "0.00625 moles}}$

${n}_{\text{glycine" = 192 * 10^(-3)"M" * 250.0 * 10^(-3)"L" = "0.048 moles}}$

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

0.00625cancel("moles Tris-HCl") * "157.60 g"/(1cancel("mole Tris-HCl")) = "0.985 g" $\text{Tris-HCl}$

and

0.048cancel("moles glycine") * "75.07 g"/(1cancel("mole glycine")) = "3.60 g"# $\text{glycine}$

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.