# Question 37d0d

May 2, 2017

The molarity of the glucose solution is $\approx 0.6 \textcolor{w h i t e}{.} \text{mol/L}$.

Another way to say this is that it is a $\text{0.6 molar}$ glucose solution.

Using the symbol for molarity, $\text{M}$, you can write $\text{0.6 M}$.

#### Explanation:

Convert given mass of glucose to mol glucose.

Multiply the given mass by the reciprocal of the molar mass.

3color(red)cancel(color(black)("g C"_6"H"_12"O"_6))xx(1"mol C"_6"H"_12"O"_6)/(180color(red)cancel(color(black)("g C"_6"H"_12"O"_6)))="0.017 mol C"_6"H"_12"O"_6"

Determine the volume of the water in liters (L).

Use the density formula to determine the volume of the water.

The density of water is $1.00 \text{g/mL}$ to three significant figures.

$\text{density"="mass"/"volume}$

Solve for volume.

$\text{volume"="mass"/"density}$

$\text{volume"=(30color(red)cancel(color(black)("g H"_2"O")))/((1.00color(red)cancel(color(black)("g H"_2"O")))/(1"mL H"_2"O"))="30 mL H"_2"O}$

Convert $\text{30 mL H"_2"O}$ to liters.

$\text{1 L=1000 mL}$

30color(red)cancel(color(black)("mL H"_2"O"))xx(1"L H"_2"O")/(1000color(red)cancel(color(black)("mL H"_2"O")))="0.03 L H"_2"O"#

Determine molarity.

Molarity, $\text{M}$, is $\text{moles of solute"/"liters of solution}$, or $\text{mol/L}$.

Divide the mol glucose by the volume of the solution in liters.

${\text{M"=(0.017"mol C"_6"H"_12"O"_6)/(0.03"L")="0.6 mol/L C"_6"H"_12"O}}_{6}$ (rounded to one significant figure due to $\text{3 g}$ and $\text{30 mL}$)