Chapter 6, Problem 6.55QP

Interpretation Introduction

Interpretation:

Number of grams of solute present in $1.75L0.500M$ glucose solution has to be calculated.

Concept Introduction:

Molarity $\left(M\right)$ can be calculated using following equation,

    $M=\frac{molsolute\left(n\right)}{Lsolution\left(V_L\right)}$

And number of mol can be calculated using given mass of the solute and molecular mass of the solute as follows,

    $\text{number}\text{of}\text{mol,}\text{n}\text{=}\frac{\text{mass}\text{of}\text{solute}\left(\text{w}\right)}{\text{molecular}\text{mass}\left(\text{M}\right)}$

On combining the above two equations another expression for molarity is obtained as below.

    $Molarity=\frac{\text{mass}\text{of}\text{solute}}{molecularmass\times volume\left(L\right)}$