A student prepares a 100.0 mL solution using 44.7 grams of potassium nitrite. They then take 11.9 mL of this solution and dilute it to a final volume of 200.0 mL. How many grams of potassium nitrite are in a 19.7 mL sample of this final diluted solution?

0.52 g of KNO₃ are contained in 19.7 mL of diluted solution.

Explanation:

We can work on this problem in Molarity cause it is more easy.

Molarity (mol/L) → moles of solute in 1L of solution.

100 mL of solution = 0.1 L

We determine moles of solute: 44.7 g . 1mol /101.1 g = 0.442 mol of KNO₃

Our main solution is 0.442 mol /0.1L = 4.42 M

We dilute: 4.42 M . (11.9mL / 200mL) = 0.263 M

That's concentration for the diluted solution.

M can be also read as mmol/mmL, so let's find out the mmoles

0.263 M . 19.7mL = 5.18 mmol

We convert the mmol to mg → 5.18 mmol . 101.1 mg / mmol = 523.7 mg

Let's convert mg to g → 523.7 mg . 1 g / 1000 mg = 0.52 g

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.