What volume of a 5.0 M stock solution of NH3 is needed to

make 3.0 L of 1.40 M NH3?_ _mL

840mL

Explanation:

Given parameters:

Concentration of stock = 5M

New volume = 3L

New concentration = 1.40M

Unknown:

Volume of stock needed = ?

Solution:

A concentrated standard solution which is the stock may become diluted to a lower concentration before using.

One important thing is that the number of moles of the substance in the solution before and after is the same.

Therefore;

C₁ V₁ = C₂ V₂

where C and V are concentration and volume

1 and 2 are initial and final states.

C₁ = concentration of stock

V₁ = volume of stock

C₂ = final concentration

V₂ = final volume

5 x V₁ = 1.4 x 3

V₁ = 0.84L

This gives 840mL