Chapter 6, Problem 6.105SP

Interpretation Introduction

Interpretation:

The molarity of glucose in normal blood has to be calculated.

Concept Introduction:

Moles:

$\text{No}\text{. of moles (n) = }\frac{\text{mass}}{\text{Molar mass}}$

Molarity:

$\begin{array}{l}\text{Molarity = }\frac{\text{Moles of solute}}{\text{Litres of solution}}\\ \\ \text{Volume of solution = }\frac{\text{Moles of solute}}{\text{Molarity}}\end{array}$