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"#)
