Question

Calculate the new molarity if each of the following dilutions is made. Assume the volumes are additive (a) 49.4 mL of water is added to 28.2 mL of 0.108 M CaCl2 solution 4.0 M (b) 148 mL of water is added to 38.8 mL of 0.685 M NaOH solution 40 M (c) 490. mL of water is added to 102 mL of 32.3 M NaCl solution 4.0 M (d) 329 mL of water is added to 835 mL of 1.87 M KOH solution 4.0 M

Answer 1

a)

use dilution formula

M1*V1 = M2*V2

1---> is for stock solution

2---> is for diluted solution

Given:

M1 = 0.108 M

V1 = 28.2 mL

V2 = 28.2 mL + 49.4 mL = 77.6 mL

use:

M1*V1 = M2*V2

M2 = (M1*V1)/V2

M2 = (0.108*28.2)/77.6

M2 = 0.0392 M

Answer: 0.0392 M

b)

Given:

M1 = 0.685 M

V1 = 38.3 mL

V2 = 38.3 mL + 148 mL = 186.8 mL

use:

M1*V1 = M2*V2

M2 = (M1*V1)/V2

M2 = (0.685*38.3)/186.8

M2 = 0.1404 M

Answer: 0.140 M

c)

Given:

M1 = 32.3 M

V1 = 102 mL

V2 = 102 mL + 490 mL = 592 mL

use:

M1*V1 = M2*V2

M2 = (M1*V1)/V2

M2 = (32.3*102)/592

M2 = 5.5652 M

Answer: 5.57 M

d)

Given:

M1 = 1.87 M

V1 = 835 mL

V2 = 835 mL + 329 mL = 1164 mL

use:

M1*V1 = M2*V2

M2 = (M1*V1)/V2

M2 = (1.87*835)/1164

M2 = 1.3415 M

Answer: 1.34 M