# Numerical: How will you make a 0.35 M glucose solution? Also, prepare a 0.12 M solution of glucose from the above stock.

Jul 9, 2015

We first work out the molar mass of glucose, ${C}_{6} {H}_{12} {O}_{6}$

#### Explanation:

$M \left({C}_{6} {H}_{12} {O}_{6}\right) = 6 \cdot 12 + 12 \cdot 1 + 6 \cdot 16 = 180 g / m o l$

So for one liter we need $0.35 \cdot 180 = 63 g$

If you want a given volume $V$ of the 0.12M solution, you use:

$\frac{0.12}{0.35} \cdot V = 0.343 \cdot V$ of the 0.35M solution, and fill up to $V$.

(the method is called reverse ratio)

Jul 9, 2015

Use glucose's molar mass and the target solution's molarity.

#### Explanation:

To make calculations easier, you can always assume your stock solution to have a volume of 1.0 L.

Since the molarity of the stock solution must be equal to 0.35 M, the sample must contain 0.35 moles of glucose.

1.0cancel("L") * "0.35 moles"/(1cancel("L")) = "0.35 moles"

Determine how many grams of glucose would contain this many moles by using the compound's molar mass

$0.35 \cancel{\text{moles") * "180.16 g"/(1cancel("mole")) = "63.06 g" = color(green)("63 g}}$

So, to prepare a 0.35-M, 1.0-L stock solution of glucose you must dissolve 63 g of glucose in enough water to make the total volume equal to one liter.

To prepare a 0.12 M solution from the stock solution, you need to determine exactly what volume of the stock solution would contain the same number of moles of glucose as the stock solution.

This is where the equation for dilution calculations comes in handy.

${C}_{1} \cdot {V}_{1} = {C}_{2} \cdot {V}_{2}$, where

${C}_{1}$, ${V}_{1}$ - the molarity and volume of the stock solution;
${C}_{2}$, ${V}_{2}$ - the molarity and volume of the target solution.

V_2 = C_1/C_2 * V_1 = (0.35cancel("M"))/(0.12cancel("M")) * "1.0 L" = color(green)("2.9 L")