Chemistry, 12.10.2019 06:30 YungToeSuck

How many grams of cuso4 would be needed to make 150 ml of 0.300 m cuso4 solution?

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$\boxed{{\text{7}}{\text{.182 g}}}$ of ${\text{CuS}}{{\text{O}}_{\text{4}}}$ would be needed to make 150 mL of 0.300 M ${\text{CuS}}{{\text{O}}_4}$ 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 ${\text{CuS}}{{\text{O}}_4}$ solution is as follows:

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

Rearrange equation (1) to calculate the moles of ${\text{CuS}}{{\text{O}}_4}$ .

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

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

${\text{1 mL}} = {10^{ - 3}}\;{\text{L}}$

So the volume of ${\text{CuS}}{{\text{O}}_4}$ solution is calculated as follows:

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

The molarity of ${\text{CuS}}{{\text{O}}_4}$ solution is 0.300 M.

The volume of ${\text{CuS}}{{\text{O}}_4}$ solution is 0.15 L.

Substitute these values in equation (2).

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

The formula to calculate the mass of ${\text{CuS}}{{\text{O}}_4}$ is as follows:

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

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

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

Therefore the mass of ${\mathbf{CuS}}{{\mathbf{O}}_{\mathbf{4}}}$ is 7.182 g.

Learn more:

1. Calculation of volume of gas:

2. Determine the moles of water produced:

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.

Answer from: Quest

The mass has escaped into a gas which you can not see so the law stays true.

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