Normal saline is 0.920% (w/v) nacl in water. how many grams of nacl are needed to prepare 15.0 l of normal saline? normal saline is 0.920% (w/v) nacl in water. how many grams of nacl are needed to prepare 15.0 l of normal saline? 807 g 53.8 g 1.38 g 16.3 g 138 g

To prepare a 15L 0.920 % NaCl solution we need 138 grams of NaCL

Explanation:

Step 1: Data given

Normal saline is 0.920% (w/v)

⇒ 0.92 g NaCl / 100 mL H2O

Step 2: Calculate grams of NaCl needed to prepare 15 L

We need 0.92 grams of NaCl for 100 mL

We need 9.2 grams of NaCL for 1L

For 15L we need 9.2 grams * 15 = 138 grams of NaCl

To prepare a 15L 0.920 % NaCl solution we need 138 grams of NaCL

in chemistry, a solution’s concentration is how much of a dissolvable substance, known as a solute, is mixed with another substance, called the solvent. the standard formula is c = m/v, where c is the concentration, m is the mass of the solute dissolved, and v is the total volume

im pretty sure its b