How many grams of lithium nitrate,LiNO₃ (68.9 g/mol),are required to prepare 778.4mL of a 0.825 M LiNO₃ solution?

Question

Quizplus · Accepted Answer

First, we need to calculate the number of moles of LiNO3 in 778.4 mL of a 0.825 M solution:

0.825 moles/L x 0.7784 L = 0.642 moles

Next, we can use the molar mass of LiNO3 to convert moles to grams:

0.642 moles x 68.9 g/mol = 44.2 g

Therefore, we need 44.2 grams of LiNO3 to prepare 778.4 mL of a 0.825 M solution. The answer is B.

How Many Grams of Lithium Nitrate,LiNO3 (68