Chapter 4.9, Problem 4.13PP

(a)

Interpretation Introduction

Interpretation:

The given $\text{moles}$ of water has to be converted to $\text{grams}\text{.}$

Concept Introduction:

Mass:

Mass of the compound is calculated by mole of the compound multiplied with molar mass of the compound.

  $\text{Mass}\text{=}\text{Molar}\text{mass}\text{×}\text{mole}$

(b)

Interpretation Introduction

Interpretation:

The given $\text{grams}$ of $\text{LiCl}$ has to be converted to $\text{moles}\text{.}$

Concept Introduction:

Moles:

Mole of the compound present in the solution is calculated by molarity, concentration of the compound multiplied with litre, volume of solution.

  $\text{No}\text{. of moles (n) = }\frac{\text{mass}}{\text{Molar mass}}$

(c)

Interpretation Introduction

Interpretation:

The given $\text{moles}$ of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ has to be converted to $\text{micrograms}\text{.}$

Concept Introduction:

Refer part (a)

(d)

Interpretation Introduction

Interpretation:

The given $\text{grams}$ of ${\text{MgCl}}_{\text{2}}$ has to be converted to $\text{moles}\text{.}$

Concept Introduction:

Refer part (b)