A 2.950 × 10^{−2 }*M* solution of NaCl in water is at 20.0°C. The sample was created by dissolving a sample of NaCl in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 999.2 mL . The density of water at 20.0°C is 0.9982 g/mL.

**Part A. **Calculate the molality of the salt solution. Express your answer to four significant figures and include the appropriate units.

*m *NaCl =

**Part B. **Calculate the mole fraction of salt in this solution. Express the mole fraction to four significant figures.

*χ *NaCl =

**Part C. **Calculate the concentration of the salt solution in percent by mass. Express your answer to four significant figures and include the appropriate units.

percent by mass NaCl =

**Part D. **Calculate the concentration of the salt solution in parts per million. Express your answer as an integer to four significant figures and include the appropriate units.

parts per million NaCl =

Based on the information provided to prepare a 2.950x10^{-2} M NaCl solution in water at 20 °C, we´re asked to **calculate the concentration of the salt solution** **in 4 parts**:

Molality (m), mole fraction (X), percent by mass (%), and parts per million (ppm).

__For all cases, we need the amount of NaCl in the sample__.

We calculate the moles of NaCl from the definition of Molarity (M):

$\overline{){\mathbf{Molarity}}{\mathbf{\left(}}{\mathbf{M}}{\mathbf{\right)}}{\mathbf{=}}\frac{\mathbf{moles}}{\mathbf{Liters}}}$

ppm