Chapter 3.12, Problem 3.25PSP

Interpretation Introduction

Interpretation:

The molarity of the ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution when $\text{20}\text{.0}\text{mL}$ sample of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, was titrated to the end point with $\text{41}\text{.3 mL of 0}\text{.100 M NaOH}$ has to be calculated.

Concept introduction:

Molar mass of a substance is the mass in grams of one mole of a substance.

  $No.ofmoles=\frac{Mass\times 1mol}{Molarmass}$

Molarity can be calculated by using following formula

  $Molarity(M)=\frac{W\times 1000}{M\times V}$

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

  $\text{Molarity of the solution}\text{=}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litre}}$