Chapter U4.83, Problem 6E

Interpretation Introduction

Interpretation:

The mass of solute in each solution needs to be calculated. The solution with the highest density needs to be identified from 1.0 L of 0.10 M NaNO₃, 1.0 L of 0.10 M Cd(NO₃)₂ and 1.0 L of 0.10 M Pb(NO₃)₂. Whether it is possible to calculate the density of each solution or not needs to be explained.

Concept Introduction:

Molarity is defined as the moles of solute in liter volume of a solution

  $\begin{array}{l}\text{Molarity = }\frac{\text{moles of solute}}{\text{L of solution}}\text{ -----(1)}\\ \text{moles = }\frac{\text{mass of solute (m)}}{\text{molar mass of solute (mol}\text{. wt)}}\text{ -----(2)}\\ \text{i}\text{.e}\text{.}\\ \text{Molarity = }\frac{\text{mass of solute}}{\text{molar mass }\times \text{L of solvent}}\\ \text{i}\text{.e}\text{. M = }\frac{\text{m}}{\text{mol}\text{.wt × V}}\text{ ------(3)}\end{array}$

The density of a solution is the ratio of its mass and volume i.e.

  $\text{Desnity(D) = }\frac{\text{Mass (m)}}{\text{Volume (V)}}\text{ -----(4)}$

Answer to Problem 6E

a) Mass of solute in 1.0 L of 0.10 M NaNO₃ = 8.5 g

b) Mass of solute in 1.0 L of 0.10 M Cd(NO₃)₂ = 23.6 g

c) Mass of solute in 1.0 L of 0.10 M Pb(NO₃)₂=33.1 g

The solution of 1.0 L of 0.10 M Pb(NO₃)₂ will have the highest density.

It is possible to calculate the solution density from the solution mass and volume

Explanation of Solution

Volume (V) of NaNO₃ = 1.0 L

Molarity (M) of NaNO₃ = 0.1 M

Based on equation (1) and (2):

  $\text{Molarity = }\frac{\text{mass of solute}}{\text{molar mass }\times \text{L of solution}}$

Molar mass of NaNO₃ = 84.99 g/mol

  $\begin{array}{l}{\text{Mass NaNO}}_{\text{3}}\text{ = 0}\text{.1 mol/L}\times \text{84}\text{.99 g/mol }\times \text{1}\text{.0 L}\\ \text{= 8}\text{.5 g}\end{array}$

Also,

Volume of Cd(NO₃)₂= 1.0 L

Molarityof Cd(NO₃)₂= 0.1 M

Thus, based on equation (1) and (2):

  $\text{Molarity = }\frac{\text{mass of solute}}{\text{molar mass }\times \text{L of solution}}$

Molar mass of Cd(NO₃)₂ = 236.42 g/mol

  $\begin{array}{l}{\text{Mass Cd(NO}}_{\text{3}}{\text{)}}_2\text{ = 0}\text{.1 mol/L}\times \text{236}\text{.42 g/mol }\times \text{1}\text{.0 L}\\ \text{= 23}\text{.6 g}\end{array}$

Similarly,

Volume of Pb(NO₃)₂ = 1.0 L

Molarity of Pb(NO₃)₂ = 0.1 M

Based on equation (1) and (2):

  $\text{Molarity = }\frac{\text{mass of solute}}{\text{molar mass }\times \text{L of solution}}$

Molar mass of Pb(NO₃)₂ = 331.2 g/mol

  $\begin{array}{l}{\text{Mass Pb(NO}}_{\text{3}}{\text{)}}_2\text{ = 0}\text{.1 mol/L}\times \text{331}\text{.2 g/mol }\times \text{1}\text{.0 L}\\ \text{= 33}\text{.1 g}\end{array}$

Based on equation (4):

  $\text{Desnity(D) = }\frac{\text{Mass (m)}}{\text{Volume (V)}}$

Since all solutions have the same volume i.e. 1.0 L or 1000 ml, the density will be directly proportional to the mass of the solution.

The total mass of solution is the sum of the mass of solute and that of the solvent which is water for aqueous solutions

  $\text{Mass of solution = Mass of solute + Mass of solvent (water) ----(5)}$

Density of water = 1 g/ml

Therefore for a volume of 1000 ml, the corresponding mass of water would be 1000 g

Based on equation (5):

  $\begin{array}{l}{\text{Mass of NaNO}}_{\text{3}}\text{ solution = 8}\text{.5 + 1000 = 1008}\text{.5 g}\\ {\text{Density of NaNO}}_{\text{3}}=\frac{\text{1008}\text{.5 g}}{\text{1}\text{.0 L}}=\text{1008}\text{.5 g/L}\\ {\text{Mass of Cd(NO}}_{\text{3}}{\text{)}}_2\text{ solution = 23}\text{.6 + 1000 = 1023}\text{.6 g}\\ {\text{Density of Cd(NO}}_{\text{3}}{\text{)}}_2=\frac{\text{1023}\text{.6 g}}{\text{1}\text{.0 L}}=\text{1023}\text{.6 g/L}\\ {\text{Mass of Pb(NO}}_{\text{3}}{\text{)}}_2\text{ solution = 33}\text{.1 + 1000 = 1033}\text{.1 g}\\ {\text{Density of Pb(NO}}_{\text{3}}{\text{)}}_2=\frac{\text{1033}\text{.1 g}}{\text{1}\text{.0 L}}=\text{1033}\text{.1 g/L}\end{array}$

Since density is directly proportional to the mass of the solution for equal volume, here the solution of 1.0 L of 0.10 M Pb(NO₃)₂ will have the highest density. Thus, it is possible to calculate the density of the solutions.

Conclusion

The solution of 1.0 L of 0.10 M Pb(NO₃)₂has the highest mass of solute, 33.1 g. and the highest density. It is possible to calculate the solution densities using mass and volume.