What mass of KCl is contained in 3.00 L of solution that has a molarity of 1.50M

The solution contains 335.5 grams of KCl

Explanation:

Step 1: Data given

Volume of the KCl solution = 3.00 L

Molarity of the KCl solution = 1.50 M

Molar mass of KCl = 74.55 g/mol

Step 2: Calculate moles KCl

Moles KCl = molarity KCl solution * volume KCl solution

Moles KCl = 1.50 M * 3.00 L

Moles KCl = 4.50 moles

Step 3: Calculate mass of KCl

Mass KCl = moles KCl * molar mass KCl

Mass KCl = 4.50 moles * 74 .55 g/mol

Mass KCl = 335.5 grams

The solution contains 335.5 grams of KCl

a. hope i'm right

explanation:

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