Answer:
54 days
Explanation:
We have to use the formula;
0.693/t1/2 =2.303/t log Ao/A
Where;
t1/2= half-life of phosphorus-32= 14.3 days
t= time taken for the activity to fall to 7.34% of its original value
Ao=initial activity of phosphorus-32
A= activity of phosphorus-32 after a time t
Note that;
A=0.0734Ao (the activity of the sample decreased to 7.34% of the activity of the original sample)
Substituting values;
0.693/14.3 = 2.303/t log Ao/0.0734Ao
0.693/14.3 = 2.303/t log 1/0.0734
0.693/14.3 = 2.6/t
0.048=2.6/t
t= 2.6/0.048
t= 54 days
The increase in the boiling point of a solvent is a colligative property.
That means that the increase in the boling point will be related to the number of particles (molecules or ions) present in the solution.
The higher the number of particles (molecules or ions) the higher the increase in the boiling point.
All the aqueous solutions presented are electrolytes, i.e. the solutes are ionic compounds.
Then, you have to compare the number of ions that you have in each solution.
A) 1.0 M KCl ---> 1.0 M K+ + 1.0 MCl- = 2 moles of particles / liter
B) 1.0 M CaCl2 --> 1.0M Ca(2+) + 1.0M * 2 Cl (-) = 3 moles of particle / liter
C) 2.0M KCl ---> 2.0 M K+ + 2.0 M Cl- = 4 moles of particle / liter
D) 2.0 M CaCl2 ----> 2.0 M Ca (2+) + 2.0M * 2 Cl (-) = 6 moles of particle / liter.
Then, the solution 2.0M CaCl2(aq) has the highest increase in the boiling point.
Answer: option D) 2.0 M Ca Cl2(aq)
