Chapter 3, Problem 148QRT

Interpretation Introduction

Interpretation:

When mixing the $\text{60}\text{.0-mL}$ sample of $\text{2}\text{.00-M NaCl}$ and $\text{40}\text{.0-mL}$ sample of $\text{0}\text{.500-M NaCl}$ and make up to $\text{500}\text{mL}$ solution, the molarity of solution has to be calculated.

Concept introduction:

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

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

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

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