How many grams of calcium nitrate are needed to make 3.30 l of a 0.10 m solution?

54.15 g
First you start out with the equation n=cv (n= moles, c=molarity, v= volume)
You’re going to multiply 0.10M by 3.30L to get an answer of 0.33 moles of Ca(NO3)2
From there you’re gonna convert the moles to grams to get your answer, first you have to find the molar mass of Ca(NO3)2
This can be done by finding adding the molar mass of each individual substance
The answer you should get for the molar mass is 164.1 g
From there just multiply the number of moles you calculated (0.33 mol) by the molar mass (164.1 g) and your answer is going to be 54.15 g Ca(NO3)2

The answer to this problem is. 67