Respuesta :

Willchemistry
Hey there!

Molar mass CuSO4 = 159.609 g/mol

Note that the molarity is in the enunciation of the issue, then the problem asks the mass of the solute :

Volume in liters of solution:

150 mL / 1000 => 0.15 L

Calculation of quantity in Moles of the solute :

1 L ----------------- 0.300 M
0.15 L ------------  ( moles )

Moles CuSO4 = 0.15 * 0.300 / 1

Moles CuSO4 = 0.045 moles

Therefore :

1 mole CuSO4 --------------------- 159.609 g
0.045 moles CuSO4-------------- mass

mass = 0.045 * 159.609 / 1

mass = 7.182405 g of CuSO4 

[tex]\boxed{{\text{7}}{\text{.182 g}}}[/tex] of [tex]{\text{CuS}}{{\text{O}}_{\text{4}}}[/tex] would be needed to make 150 mL of 0.300 M [tex]{\text{CuS}}{{\text{O}}_4}[/tex] solution.

Further Explanation:

The proportion of substance in the mixture is called concentration. The most commonly used concentration terms are as follows:

1. Molarity (M)

2. Molality (m)

3. Mole fraction (X)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

Molarity is a concentration term that is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by M and its unit is mol/L.

The formula to calculate the molarity of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] solution is as follows:

[tex]{\text{Molarity of CuS}}{{\text{O}}_4}\;{\text{solution}}=\dfrac{{{\text{Moles}}\;{\text{of}}\;{\text{CuS}}{{\text{O}}_4}}}{{{\text{Volume }}\left( {\text{L}}\right){\text{ of}}\;{\text{CuS}}{{\text{O}}_4}\;{\text{solution}}}}[/tex]   ......(1)

Rearrange equation (1) to calculate the moles of [tex]{\text{CuS}}{{\text{O}}_4}[/tex].

[tex]{\text{Moles}}\;{\text{of}}\;{\text{CuS}}{{\text{O}}_4}= \left( {{\text{Molarity of HN}}{{\text{O}}_3}\;{\text{solution}}} \right)\left( {{\text{Volume of}}\;{\text{CuS}}{{\text{O}}_4}\;{\text{solution}}} \right)[/tex]   ......(2)

The volume of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] solution is to be converted into L. The conversion factor for this is,

[tex]{\text{1 mL}} = {10^{ - 3}}\;{\text{L}}[/tex]

So the volume of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] solution is calculated as follows:

[tex]\begin{aligned}{\text{Volume of CuS}}{{\text{O}}_4}\;{\text{solution}}&= \left( {{\text{150 mL}}}\right)\left({\frac{{{{10}^{ - 3}}\;{\text{L}}}}{{{\text{1 mL}}}}} \right)\\&= 0.15{\text{ L}}\\\end{aligned}[/tex]

The molarity of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] solution is 0.300 M.

The volume of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] solution is 0.15 L.

Substitute these values in equation (2).

[tex]\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{CuS}}{{\text{O}}_4}&=\left( {{\text{0}}{\text{.300 M}}} \right)\left( {0.15{\text{ L}}}\right)\\&= 0.04{\text{5 mol}}\\\end{aligned}[/tex]

The formula to calculate the mass of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] is as follows:

[tex]{\text{Mass of CuS}}{{\text{O}}_{\text{4}}}=\left( {{\text{Moles of CuS}}{{\text{O}}_{\text{4}}}} \right)\left( {{\text{Molar mass of CuS}}{{\text{O}}_{\text{4}}}} \right)[/tex]       ......(3)

Substitute 0.045 mol for the moles of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] and 159.609 g/mol for the molar mass of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] in equation (3).

[tex]\begin{aligned}{\text{Mass of CuS}}{{\text{O}}_{\text{4}}}&= \left({{\text{0}}{\text{.045 mol}}} \right)\left( {\frac{{{\text{159}}{\text{.609 g}}}}{{{\text{1 mol}}}}}\right)\\&= 7.1824{\text{05 g}}\\&\approx {\text{7}}{\text{.182 g}}\\\end{aligned}[/tex]

Therefore the mass of [tex]{\mathbf{CuS}}{{\mathbf{O}}_{\mathbf{4}}}[/tex]  is 7.182 g.

Learn more:

1. Calculation of volume of gas: https://brainly.com/question/3636135

2. Determine the moles of water produced: https://brainly.com/question/1405182

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: molarity, CuSO4, dilution, moles of CuSO4, volume, solution, 0.045 mol, 150 mL, 0.300 M, 7.182 g, 159.609 g/mol, 0.15 L.

Otras preguntas