Chapter 5, Problem 5.32E

Interpretation Introduction

Interpretation:

Mass of glucose has to be calculated.

Concept introduction:

Molarity of the solution can be calculated by using following formula,

  $\begin{array}{l}\Pi \text{ = iRT }\times \text{ Molarity}\\ \\ \text{Molarity}\text{=}\frac{\Pi }{\text{ iRT}}\\ \\ \text{Molarity}\text{=}\frac{7.7atm}{\text{ 1}\times \left(\text{0}\text{.0821}\text{L}\cdot \text{atm}\cdot mo{l}^{-1}\cdot {\text{K}}^{\text{-1}}\right)\times \left(310\text{K}\right)}\\ \text{Molarity}\text{=}0.303mol\cdot L^{-1}\end{array}$