Calculate the molality for each of the following solutions. Then, calculate the freezing-point depression ΔTF = iKFcm produced by each of the salts. (Assume the density of water is 1.00 g/mL and KF = 1.86°C/m.) (a) 3.2 M NaCl (Assume the density of the solution is 1.00 g/mL.) Cm = WebAssign will check your answer for the correct number of significant figures. 3.9 Correct: Your answer is correct. m ΔTF = WebAssign will check your answer for the correct number of significant figures.14.5 Incorrect: Your answer is incorrect. Check the number of significant figures.°C (b) 24 g of KCl in 1.5 L of water Cm = WebAssign will check your answer for the correct number of significant figures. 0.21 Correct: Your answer is correct. m ΔTF = WebAssign will check your answer for the correct number of significant figures. 0.78 Incorrect: Your answer is incorrect. °C

a) Cm= 3.9 m  ; ΔTf= 14.51 ºC

b) Cm= 0.21 m ; ΔTf= 0.79ºC

Explanation:

In order to solve the problems, we have to remember that the molality (m) of a solution is equal to moles of solute in 1 kg of solvent.

m= mol solute/kg solvent

a) In this case we have molarity, which is moles of solute in 1 liter of solution. We have to know how many kg of solvent (water) we have in 1 L of solution.

3.2 M NaCl= 3.2 mol NaCl/ 1 L solution

1 L solution= 1000 ml solution x 1.00 g/ml= 1000 g

A solution is composed by solute (NaCl) + solvent, so:

1000 g solution = g NaCl + g solvent

g NaCl= 3.2 mol NaCl x 58.44 g/mol= 187 g NaCl

g solvent= 1000 g - 187 g NaCl= 813 g= 0.813 kg

Cm= 3.2 g NaCl/0.813 kg solvent= 3.9 m

NaCl is an electrolyte and it dissociates in water in two ions: Na⁺ anc Cl⁻, si the van't Hoff factor (i) is 2.

ΔTf= i x KF x Cm= 2 x 1.86ºC/m x 3.9 m= 14.51ºC

b) In this case we have 24 g of solute in 1.5 L of solvent. We have to convert the liters of solvent to kg, and to convert the mass of solute to mol by using the molecular weight of KCl (74.55 g/mol):

24 g KCl x 1 mol KCl/74.55 g= 0.32 mol

1.5 L solvent= 1500 g solvent x 1.00 g/ml= 1500 g = 1.5 kg

Cm= 0.32 g KCl/1.5 kg solvent= 0.21 m

KCl is an electrolyte and when it dissolves in water, it dissociates in 2 ions: K⁺ and Cl⁻. For this, van't Hoff factor (i) is equal to 2.

ΔTf= i x KF x Cm= 2 x 1.86ºC x 0.21 m= 0.79ºC

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