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

a) the particles of the chair are strongly held together.

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole