Question

What volume of water should be added to 500 mL of a 1.0 mol/L $CuS{{O}_{4}}$ (aq) solution to dilute it to 0.5 mol/L?

Answer 1

Copper(II) sulphate, often known as copper sulphate, is an inorganic compound having the chemical formula $CuS{{O}_{4}}.{{\left( {{H}_{2}}O \right)}_{x}}$ , where x is a number between 0 and 5. The most frequent type becomes pentahydrate (x = 5). The most common salt, copper sulphate pentahydrate ( $CuS{{O}_{4}}.5{{H}_{2}}O$ ), is a brilliant blue colour. It dissolves in water exothermically to form the aquo complex ${{\left[ Cu{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}$ , which has an octahedral molecular geometry.

Complete answer:
The use of molar concentration in thermodynamics is frequently inconvenient since the volume of most solutions varies somewhat with temperature owing to thermal expansion. This difficulty is typically handled by utilising temperature adjustment factors or a temperature-independent concentration measure such as molality. The reciprocal amount denotes the dilution (volume) that may be found in Ostwald's dilution law.
In chemistry, the molarity (M) of a solution, which becomes the number of moles of solute per litre of solution, is commonly used to express its concentration. The molar concentration ( ${{c}_{i}}$ ) is determined by multiplying the number of moles of solute ( ${{n}_{i}}$ ) by the total volume (V) of the Solution
${{c}_{i}}=\dfrac{{{n}_{i}}}{V}$
The symbol for molarity, M, has the unit $\text{mol }{{\text{L}}^{-1}}$ .
Upon diluting the solutions, use equation $M_{1} V_{1} \times M_{2} V_{2}$ , where $M_{1}$ becomes initial molarity in $\text{mol }{{\text{L}}^{-1}},{{\mathbf{M}}_{2}}$ becomes final molarity, $V_{1}$ becomes initial volume in Litres, and $V_{2}$ becomes final volume in Litres.
Known tends to be ${{M}_{1}}=1.0~\text{mol }{{\text{L}}^{-1}}\text{ CuS}{{\text{O}}_{4}}$
${{V}_{1}}=500ml\times \dfrac{1~\text{L}}{1000ml}=0.5~\text{L}$
${{M}_{2}}=0.5~\text{mol }{{\text{L}}^{-1}}\text{CuS}{{\text{O}}_{4}}$
Let the Unknown $V_{2}$ .
To isolate $V_{2}$ , rearrange the equation. Solve the equation by substituting the known numbers.