Chapter 21, Problem 21.122CHP

Interpretation Introduction

Interpretation:

The amount of ${\text{Y(OCH}}_{\text{2}}{\text{CH3)}}_{\text{3}}$  and ${\text{Ba(OCH}}_{\text{2}}{\text{CH}}_{\text{3}}{\text{)}}_{\text{2}}$ that are needed to react with $\text{75}{\text{.4 g of Cu(OCH}}_{\text{2}}{\text{CH}}_{\text{3}}{\text{)}}_{\text{2}}$ has to be calculated.  Also the amount of ${\text{YBa}}_{\text{2}}{\text{Cu}}_{\text{3}}{\text{O}}_{\text{7}}$ obtained has to be determined.

Concept introduction:

The number of moles of solute in a solution calculated as

$\text{Mole}\text{=}\frac{\text{Given}\text{weight}}{\text{Molar}\text{mass}}$

Molarity can be defined as the moles of solute to the volume of the solution (in liters). The molarity of a solution can be given by the formula,

$\text{Molarity(M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{mol)}}{\text{Volume}\text{of}\text{solution(in}\text{L)}}$