Question

Based on the manufacturer’s claim of 115 mg Na per serving and a 28-g serving size, calculate the proper chip sample weight that will give you a solution whose Na content falls within the desired 0.5-5.0 ppm range of your calibration curve.

Answer 1

115 mg of Na is present per serving.

Amount of chip in 1 serving = 28 grams.

Thus, 28 grams of chip contain 115 mg of Na.

1mg of Na will be present in amount of chip = 28/115 = 0.2434 gm

Maximum limit = 5 ppm = 5mg/liter of solution i.e. 5mg of Na can be present, at maximum, per liter of solution

Since 1 mg of Na is present in 0.2434 gm of chip, 5 mg of Na will be present in chip = 0.2434 x 5 = 1.217 gm of chip

Thus, chip sample weight should be less than or equal to 1.217 gm per liter of solution.

Minimum limit = 0.5 ppm = 0.5mg/liter of solution i.e. 0.5mg of Na can be present, at minimum, per liter of solution

Since 1 mg of Na is present in 0.2434 gm of chip, 0.5 mg of Na will be present in chip = 0.2434 x 0.5 = 0.1217 gm of chip

Thus, chip sample weight should be more than or equal to 0.1217 gm per liter of solution.

Thus, the range of chip sample should be between 0.1217 gm to 1.217 gm