Following the procedure, a student recorded the initial volume of naoh in his buret as 2.82 ml. he then titrated a 15.00 ml sample of 0.1027 m hcl until a persistent pale pink appeared. he recorded the final volume of naoh in his buret as 17 ml. what is the concentration (m) of his naoh solution?

The concentration (M) of the students NaOH solution is 0.1086M

HOW TO CALCULATE MOLARITY:

CaVa = CbVb

Where;

0.1027 × 15 = Cb × 14.18

1.5405 = 14.18Cb

Cb = 0.1086M

Therefore, the concentration (M) of the students NaOH solution is 0.1086M.

Learn more at: link

The concentration of this NaOH solution was 0.104 M

Explanation:

Step 1: Data given

Initial volume of NaOH = 2.82 mL = 0.00282 L

The sample is titrated with 15.00 mL of 0.1027 M HCl

Final volume NaOH = 17 mL = 0.017 L

Step 2:  Calculate change in volumes

17 mL - 2.82 mL = 14.18 mL = 0.01418 L

Step 3: Calculate the concentration of the NaOH

C1*V1 = C2*V2

⇒ with C1 = the initial concentration = TO BE DETERMINED

⇒ with V1 = the initial volume =  0.0148 L

⇒ with C2 = the final concentration = 0.1027

⇒ with V2 = the final volume = 0.015 L

C1 = (0.1027*0.015) /0.0148

C1 = 0.104 M

The concentration of this NaOH solution was 0.104 M

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.