You make two dilutions. You take 83.52 mL of the stock solution and dilute to 500.0 mL to make Solution #1. You then add 52.27 mL of Solution #1 to 46.77 mL of water to make Solution #2. What is the concentration of the Solution #1 in terms of molarity?

The answer to the question is;

The concentration of the Solution #1 in terms of molarity is

0.16704X moles/litre.

Explanation:

Let the concentration of the stock solution be X moles/liter

Therefore, 83.52 ml of the stock solution contains

83.52×(X/1000) moles

Dilution of 83.52 ml of X to 500 ml gives solution 1 with a concentration of

500 ml of solution 1 contains 83.52×(X/1000) moles

Therefore 1000 ml or 1 litre contains 2×83.52×(X/1000) moles = 0.16704X moles/litre

The molarity of solution 1 is 0.16704X moles/litre.

icc, international association for cereal chemistry

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2