Question #37d0d

1 Answer
May 2, 2017

Answer:

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"#)