To solve this problem we need to have an idea about molarity. The concentration can be expressed as:

Molarity = moles/ L

1dm^3 = 1L

According to the ration given, there are 15 g or N atoms per 100 grams of Fertilizer. So the percent mass would be 15% N....

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To solve this problem we need to have an idea about molarity. The concentration can be expressed as:

Molarity = moles/ L

1dm^3 = 1L

According to the ration given, there are 15 g or N atoms per 100 grams of Fertilizer. So the percent mass would be 15% N. Next, we have get the amount of nitrogen present in 14 grams of fertilizer.

14 x (15/100) = 2.1 grams N

There are 2.1 grams of nitrogen atoms in 14 grams of fertilizer. Now we convert 2.1 grams of N to moles of N.

2.1 grams N x ( 1mole N / 14 grams N) = 0.15 moles N

Finally, we can get the concentration in Molarity.

Molarity = 0.15moles/5 dm^3

**Molarity = 0.15 moles/dm^3 or moles/L**