# 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.)?

Aug 18, 2016

0.409 g

#### Explanation:

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.

Aug 22, 2016

The 350 mL solution will contain 0.4 g NaCl.

#### Explanation:

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 $\text{350 mL}$ to liters.

$350 \cancel{\text{mL"xx(1 "L")/(1000 cancel"mL")="0.35 L}}$

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

$0.35 \cancel{\text{L"xx(0.02 "mol NaCl")/(1 cancel"L")="0.007 mol NaCl}}$

In order to determine the mass of the NaCl in the 0.35 L NaCl solution, multiply the moles of $\text{NaCl}$ by the molar mass of NaCl.

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