Question

Normal saline solution is 0.009 g/mL NaCl. How many grams of sodium chloride are needed to prepare 3.5 L of a normal saline solution?

Answer 1

1) List the values we have.

Normal solution = 0.009 g/mL NaCl

Volume =3.5 L

Grams of NaCl = ?

2) Set the equation

`\text{Normal solution =}\frac{\text{grams of solute}}{\text{mililiters of solution}}`

3) Convert L into mL

`\text{mililiters of normal solution= }\frac{3.5\text{ L}}{\square}\cdot\frac{1000\text{ mL}}{1\text{ L}}=3500\text{ mL}`

4) Replace know values

`(0.009g)/(mL)=\text{ }\frac{\text{grams of NaCl}}{3500\text{ mL}}`
`\frac{0.009\text{ g}}{mL}\cdot3500\text{ mL=}\frac{\text{grams of NaCl}}{3500\text{ mL}}\cdot3500\text{ mL}`
`\frac{0.009\text{ g}}{mL}\cdot3500\text{ mL=grams of NaCl}`

Grams of NaCl = 0.009g/mL*3500 mL= 31.5 g of NaCl.