A teacher has 12MHCL and needs to make 2.5MHCL for a lab. There are 15 groups and each group need 75ml of the acid. What does the teacher need to do?

4.2 mL

Explanation:

To make the problem more interesting, let's assume that you don't know the formula for dilution calculations.

The idea with diluting a solution is that the number of moles of solute will remain constant after the initial solution is diluted. The only thing that changes in such cases is the volume of the solution.

This means that if you know how many moles of solute you have in the target solution, you also know how many moles of solute were present in the stock solution sample.

Use the molarity and volume of the target solution to determine how many moles of hydrochloric acid,

HCl

, you need in that solution

c=nV⇒n=c⋅VnHCl=0.10 M⋅500⋅10−3L=

0.050 moles HCl

Now the question is - what volume of stock solution would contain this many moles of hydrochloric acid?

c=nV⇒V= Nvm stock =0.050 moles 12 moles L= 0.0041667 L

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part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years