posted by Dave .
A student dissolved 4.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution and then diluted it with water to give 275. mL of a final solution. How many grams of NO3- ion are there in the final solution?
Don't understand it. I keep getting 0.394 g as my final result.
1. (0.218639 M)(0.00400 L) = (x M)(0.275 L)
= 0.00318 M Co(NO3)2
2. 0.00318 M Co(NO3)2 x 2 mol NO3- / 1 mol Co(NO3)2 x 62 g NO3- / 1 mol NO3- = 0.394 g NO3-
The correct answer isn't listed. Your solution doesn't work because you have convert to M and your answer is moles/L of NO3. In addition, the 275 never enters into the problem The problem asks for grams.
Dissolve 4.00 g Co(NO3)2 in 100 mL. That gives us 4/100 = 0.04 g/mL of Co(NO3)2 or 0.08 g NO3^-/mL. The final solution was made by using 4.00 mL of that solution; therefore, the final solution contained 0.08 g/mL x 4.00 mL = 0.32 grams NO3^-.