A student takes a stock solution that is 50 mM (solute formula weight is 120.5 g/mol) and prepares a series of solutions. The first solution is made by diluting 1 mL of the stock in water to a final volume of 15 mL (sample 1). They then take a 2 mL aliquot of sample 1 and dilute in water to a final volume of 25 mL (sample 2). Finally, they take a 1.5 mL aliquot of sample 2 and dilute to a final volume of 250 mL. Calculate the final molar concentration for the analyte in molarity, molality, ppm, and ppb. State any assumptions.

1.6x10⁻⁶M

1.6x10⁻⁶m

0.1928ppm

192.8ppb

Explanation:

The first dilution of the solution is from 1mL to 15mL. The second from 2mL to 25mL and the third from 1.5mL to 250mL. That means molarity is:

50mM =

0.050M * (1mL / 15mL) * (2mL / 25mL) * (1.5mL / 250mL) = 1.6x10⁻⁶M

Molality is defined as the ratio between moles and kg. Assuming the density of the solution is 1kg/L, molality is 1.6x10⁻⁶m

ppm is the ratio between milligrams and Liters:

1.6x10⁻⁶mol / L * (120.5g /mol) * (1000mg / g) = 0.1928mg/L = 0.1928ppm

And ppb = 1000*ppm;

0.1928ppm*1000 = 192.8ppb

the temperature of the gas, in k, not celsius.