Chapter 4, Problem 4D.3E

Interpretation Introduction

Interpretation:

The molality and the molar concentration of sucrose in the given solution has to be calculated.

Concept introduction:

The number of moles of any substance can be determined using the equation

$\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mole (g/mol).

$Mass=Volume\times density$

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent.

$\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

Molarity (M): Molarity is number of moles of the solute present in the one litre of the solution.

$\text{Molarity (M) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1}\text{liter}\text{of}\text{solution}}$

The relation between molality and molarity is shown below,

$\frac{1}{m}=\frac{d}{M}-\frac{M_B}{1000}\boxed{\begin{array}{l}m\Rightarrow Molalityofthesolution\\ M\Rightarrow Molarityofthesolution\\ d\Rightarrow Densityofsolution\\ M_b\Rightarrow Molarmassofsolute\end{array}}$