A student prepares a solution of sodium chloride by dissolving 116.9 g of NaCl into enough water to make 1.00 L of solution. How
1 answer:
Answer:
<em>The accurate label for the solution is</em><em> </em><u><em>2.00 mol/L NaCl.</em></u>
Explanation:
Molrity = mass ÷ molar mass
Molar mass of NaCL = 58.44 g/mol
Weighed mass = 116.9g
⇒ Molarity of the solution = = 2.0M
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<em>Therefore, the accurate label for the solution is </em><u><em>2.00 mol/L NaCl.</em></u>
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Answer:
- <u>3.0M</u>
Explanation:
The dilution formula is:
Where M₁ and M₂ are the molar concentrations and V₁ and V₂ are the volumes of the concentrated and the diluted solutions.
Substitute:
- M₁ = <em>6.0M</em>
- M₂ = unknown
- V₁ = <em>251 mL</em>
- V₂ = <em>500. mL</em>
Clear M₂ and compute:
Answer:
690 g IrI₃
Explanation:
To convert from moles to grams, you have to use the molar mass of the compound. The molar mass of IrI₃ is 572.92 g/mol. You use this as the unit converter in this equation:
Round to the lowest number of significant figures which is 2 to get 690 g IrI₃.