How many grams of NaCl are present in 350 mL of 0.020 M NaCL solution? (Remember that 1 mole NaCL = 58.44 grams g/mole.)?

As stated, 1 mole of NaCl weighs 58.44 g.

A 0.02 M solution contains 0.02 moles per liter.

Hence, 1 litre would contain:

58.44 x 0.02 = 1.1688 g

However, there are only 350 ml, which is 0.35 liters.

Therefore, there would be 0.35 x 1.1688 g in total, 0.409 g

For further help this blog post on moles and molarity may be useful.

The symbol for molarity, M, represents the units mol/L. When you perform equations involving molarity, use the units mol/L. The 0.02 M solution is said to be 0.02 molar, and is 0.02 mol/L.

Convert #"350 mL"# to liters.

#350cancel "mL"xx(1 "L")/(1000 cancel"mL")="0.35 L"#

In order to determine the moles of NaCl in the 0.35 L solution, multiply #"0.35 L"# by the molarity of the #"0.02 M NaCl"# solution. (Remember that the unit for molarity, M, is mol/L.)

#0.35 cancel"L"xx(0.02 "mol NaCl")/(1 cancel"L")="0.007 mol NaCl"#

In order to determine the mass of the NaCl in the 0.35 L NaCl solution, multiply the moles of #"NaCl"# by the molar mass of NaCl.

#0.007 cancel"mol NaCl"xx(58.44"g NaCl")/(1cancel"mol NaCl")="0.4 g NaCl"# (rounded to one significant figure, because of the 0.02 M solution)