Question

73.0 mL of a 1.60 M solution is diluted to a total volume of 288 mL. A 144 mL portion of that solution is diluted by adding 129 mL of water. What is the final concentration? Assume the volumes are additive.

Determine the concentration of each of the individual ions in a 0.350 M K2SO4 solution.

A 0.110 L sample of an unknown HNO3 solution required 51.1 mL of 0.200 M Ba(OH)2 for complete neutralization. What was the concentration of the HNO3 solution?

Answer 1

1)

This question can be solved by relation,

M1V1 = M2V2

where M1, M2 = Molarities of concentrated and diluted solutions

V1, V2 = Volumes of concentrated and diluted solutions

Putting the respective values, we get

1.60 M x 73 mL= M2 x 288 mL

0.405 M = M2

New solution volume after 2nd time dilution = ( 144 + 129)mL

= 273 mL

Again using the same relation ie

M1V1 = M2V2

0.405 M x 144 mL = M2 x 273

0.213 M = M2

So, 0.213 M would be the final concentration of the resulting solution.