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xxTIMURxx [149]
3 years ago
8

approximately how many grams of sodium chloride (NaCl) are required to prepare 500. mL of 3.00 M solution

Chemistry
1 answer:
matrenka [14]3 years ago
8 0
First convert mL to L:
(500. mL)/(1000L) = .500L

Then convert molarity to moles:
(3.00M)(.500L) = 1.5mol NaCl

Then convert moles of NaCl to grams of NaCl by multiplying it by its molar mass:
(1.5mol NaCl)(58.44g/mol NaCl) = 87.66g NaCl are needed
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Answer:

\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}

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\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}

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\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}

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