Question

Question #37d0d

Answer 1

The molarity of the glucose solution is `~~0.6color(white)(.)"mol/L"`.

Another way to say this is that it is a `"0.6 molar"` glucose solution.

Using the symbol for molarity, `"M"`, you can write `"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"g/mL"` to three significant figures.

`"density"="mass"/"volume"`

Solve for volume.

`"volume"="mass"/"density"`

`"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 `"30 mL H"_2"O"` to liters.

`"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, `"M"`, is `"moles of solute"/"liters of solution"`, or `"mol/L"`.

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

`"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 `"3 g"` and `"30 mL"`)