Astudent performed a serial dilution on a stock solution of 1.33 m naoh. a 2.0 ml aliquot of the stock naoh (ms) was added to 18 ml of water to make the first dilution (m1). next, 2.0 ml of the m1 solution was added to 18 ml of water to make the second solution (m2). the same steps were repeated for a total of 5 times. what is the final concentration of naoh (m5)?

Explanation:

Data:

           c₀ = 1.33 mol·L⁻¹

Dilutions = 2 mL stock + 18 mL water

            n = five dilutions

Calculations:

The general formula is for calculating a single dilution ratio (DR) is

For your dilutions,

(Note: This is the same as a dilution factor of 10:1)

The general formula for the concentration cₙ after n identical serial dilutions is

So, after five dilutions

answer; ///i do believe that the correct answer is (b); ///particle composed of two or more atoms bonded together;

nacl or sodium chloride is an ionic compound. this is because an ionic compound pertains to a compound that consists of positively and negatively charged ions. in this case, na (sodium) has a tendency to lose one electron when it reacts with with other atoms, so the resulting sodium ion contains 11 protons and 10 electrons, thus making it positively charged (na+), when chlorine reacts, it usually gains one electron to form an ion called chloride, thus making it negatively charged ( since the ions are oppositely charged, they attract.