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3 years ago

If you take a 15.0-mL portion of the stock solution and dilute it to a total volume of 0.600 L , what will be the concentration

of the final solution?

Chemistry

1 answer:

alexira [117]3 years ago

6 0

$15 \div .6$

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3 years ago

PLEASE HELP ASAP!

choli [55]

Answer:

$\large \boxed{\text{D. 710 g}}$

Explanation:

1. Calculate the molar mass of Na₂SO₄

$\begin{array}{ccc}\textbf{Atoms} &\textbf{M}_{\textbf{r}} & \textbf{Mass/u}\\\text{2Na} & 23 & 46\\\text{1S} & 32 & 32\\\text{4O}&16 & 64\\&\text{TOTAL =} & \mathbf{142}\\\end{array}$

The molar mass of Na₂SO₄ is 142 g/mol.

2. Calculate the moles of Na₂SO₄

$\text{Moles of Na$_{2}$SO}_{4} = \text{2.5 L solution} \times \dfrac{\text{2.0 mol Na$_{2}$SO}_{4}}{\text{1 L solution}} = \text{5.0 mol Na$_{2}$SO}_{4}$

3. Calculate the mass of Na₂SO₄

$\text{Mass of Na$_{2}$SO}_{4} = \text{5.0 mol Na$_{2}$SO}_{4} \times \dfrac{\text{142 g Na$_{2}$SO}_{4}}{\text{1 mol Na$_{2}$SO}_{4}} = \text{710 g Na$_{2}$SO}_{4}\\\\\text{You need } \large \boxed{\textbf{710 g}} \text{ of Na$_{2}$SO}_{4}$

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3 years ago