Chapter U4.83, Problem 5E

Interpretation Introduction

Interpretation:

The solution with the highest molarity needs to be identified from 10 g NaNO₃ in 1.0 L water , 10.0 g Cd(NO₃)₂ in 1.0 L water and 10.0 g Pb(NO₃)₂ in 1.0 L

Concept Introduction:

Molarity is defined as the moles of solute in a 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 5E

The solution of 10 g NaNO₃ in 1.0 L water has the highest molarity of 1.2x10-1 mol/L and the highest density of 10.2 g/L

Explanation of Solution

a)

Given:

Mass of NaNO₃ = 10 g

Volume of water solvent = 1.0 L

Calculation:

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

  $\text{Molarity = }\frac{\text{10 g}}{\text{84}\text{.99 g/mol }\times \text{1}\text{.0 L}}=1.2\times {10}^{-1}\text{ mol/L}$

b)

Given:

Mass of Cd(NO₃)₂= 10 g

Volume of water solvent =1.0 L

Calculation:

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

  $\text{Molarity = }\frac{\text{10 g}}{\text{236}\text{.42 g/mol }\times \text{1}\text{.0 L}}=4.2\times {10}^{-2}\text{ mol/L}$

c)

Given:

Mass of Pb(NO₃)₂ = 10 g

Volume of water solvent = 1.0 L

Calculation:

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

  $\text{Molarity = }\frac{\text{10}\text{.0 g}}{\text{331}\text{.2 g/mol }\times \text{1}\text{.0 L}}=3.0\times {10}^{-2}\text{ mol/L}$

Therefore, the solution of 10 g NaNO₃ in 1.0 L water has the highest molarity of 1.2x10-1 mol/L

Molarity can be expressed in terms of solution density and solute molecular weight by combining equations (3) and (4) i.e.

  $\begin{array}{l}\text{M = }\frac{\text{D}}{\text{mol}\text{.wt }}\\ \text{D = M×mol}\text{.wt ----(5)}\end{array}$

a) NaNO₃ solution

  $\text{M}=1.2\times {10}^{-1}\text{ mol/L}$

Mol wt = 84.99 g/mol

Based on equation (5):

  $\begin{array}{l}\text{D = 1}\text{.2}\times {\text{10}}^{-1}\text{ mol/L}\times \text{84}\text{.99 g/mol}\\ \text{=10}\text{.2 g/L}\end{array}$

b) Cd(NO₃)₂ solution

  $\text{M}=4.2\times {10}^{-2}\text{ mol/L}$

Mol wt = 236.42 g/mol

Based on equation (5):

  $\begin{array}{l}\text{D = 4}\text{.2}\times {\text{10}}^{-2}\text{ mol/L}\times \text{236}\text{.42 g/mol}\\ \text{=9}\text{.92 g/L}\end{array}$

c) Pb(NO₃)₂ solution

  $\text{M}=3.0\times {10}^{-2}\text{ mol/L}$

Mol wt = 331.2 g/mol

Based on equation (5):

  $\begin{array}{l}\text{D = 3}\text{.0}\times {\text{10}}^{-2}\text{ mol/L}\times \text{331}\text{.2 g/mol}\\ \text{=9}\text{.94 g/L}\end{array}$

Thus, the solution of 10 g NaNO₃ in 1.0 L water has the highest density as well.

Conclusion

Thus, the solution of 10 g NaNO₃ in 1.0 L water has the highest molarity and the highest density.