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$$ \begin{aligned} & \text { What is the molarity of pure water with a density of }\\\ &1.00 \mathrm{~g} / \mathrm{mL} ? \end{aligned} $$

Short Answer

Expert verified
The molarity of pure water is approximately 55.5 M.

Step by step solution

01

Understand Molarity

Molarity ( ext{M}) is defined as the number of moles of solute per liter of solution. To calculate molarity, we need to know the number of moles and the volume of the solution in liters.
02

Calculate the Volume of Water

Since the density of water is given as \(1.00 ext{ g/mL}\), 1 liter of water has a mass of 1000 grams because 1 liter = 1000 mL.
03

Determine Moles of Water

The molecular weight of water \( (H_2O) \) is approximately \(18.015 ext{ g/mol}\). To find moles of water in 1000 grams, we calculate: \[\text{Moles of water} = \frac{1000 ext{ grams}}{18.015 ext{ g/mol}}\approx 55.5 \text{ moles}\].
04

Calculate Molarity

Now, we can calculate the molarity by using the formula: \[\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} = \frac{55.5}{1} = 55.5 \text{ M}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Density of Water
Water is often considered the baseline for density measurements. The density of water is particularly significant because it is used as a standard reference. At room temperature, the density of water is typically around 1.00 grams per milliliter (g/mL). This means that for every milliliter of water, the mass is approximately 1 gram.
The high density of water allows it to be a versatile solvent in chemical calculations. For instance, knowing this density helps in determining how many grams exist in a given volume. If you take 1 liter of water, it will have a mass of about 1000 grams, because there are 1000 milliliters in a liter.
Understanding water's density is crucial when measuring or converting the mass and volume of water in scientific and everyday calculations.
Molecular Weight of Water
Water's molecular weight can be easily understood by breaking down its molecular composition. The chemical formula for water is \( H_2O \), which means each molecule of water consists of two hydrogen atoms and one oxygen atom. To calculate the molecular weight of water:
  • Hydrogen has an atomic weight of about 1.008 grams per mole.
  • Oxygen has an atomic weight of about 16.00 grams per mole.
To find the molecular weight of water, you combine the weight of two hydrogens and one oxygen: \[ (2 \times 1.008) + 16.00 = 18.015 \text{ g/mol} \]
This calculation results in the molecular weight of water being approximately 18.015 grams per mole. Knowing this value is essential for converting between mass and moles, particularly in chemistry, where precise measurements are necessary.
Moles of Solute
Understanding moles is foundational in chemistry. Moles express amounts of a chemical substance, allowing scientists to count particles by weighing them. When you calculate moles of solute, you're determining how many moles of a substance are in a given mass. To compute this, the formula is:\[ \text{Moles} = \frac{\text{Mass of substance (g)}}{\text{Molecular weight (g/mol)}} \] For water, if you have 1000 grams and the molecular weight is 18.015 g/mol, you calculate: \[ \text{Moles of water} = \frac{1000}{18.015} \approx 55.5 \text{ moles} \]This step allows us to further calculate molarity, which involves dividing these moles by the volume in liters of the solution. Such calculations are vital in various chemical processes, including reactions and concentrations adjustments.

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Most popular questions from this chapter

A solution contains \(6.00 \%\) (by mass) \(\mathrm{NaBr}\) (sodium bromide). The density of the solution is \(1.046 \mathrm{~g} / \mathrm{cm}^{3} .\) What is the molarity of \(\mathrm{NaBr}\) ?

Barium carbonate is the source of barium compounds. It is produced in an aqueous precipitation reaction from barium sulfide and sodium carbonate. (Barium sulfide is a soluble compound obtained by heating the mineral barite, which is barium sulfate, with carbon.) What are the molecular equation and net ionic equation for the precipitation reaction? A solution containing \(33.9 \mathrm{~g}\) of barium sulfide requires \(21.2 \mathrm{~g}\) of sodium carbonate to react completely with it, and \(15.6 \mathrm{~g}\) of sodium sulfide is produced in addition to whatever barium carbonate is obtained. How many grams of barium sulfide are required to produce 5.00 tons of barium carbonate? (One ton equals 2000 pounds.)

Decide whether a reaction occurs for each of the following. If it does not, write \(N R\) after the arrow. If it does, write the balanced molecular equation; then write the net ionic eauation. (a) \(\mathrm{Al}(\mathrm{OH})_{3}+\mathrm{HNO}_{3} \longrightarrow\) (b) \(\mathrm{NaBr}+\mathrm{HClO}_{4} \longrightarrow\) (c) \(\mathrm{CaCl}_{2}+\mathrm{NaNO}_{3} \longrightarrow\) (d) \(\mathrm{MgSO}_{4}+\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2} \longrightarrow\)

A 25.0-mL sample of sodium sulfate solution was analyzed by adding an excess of barium chloride solution to produce barium sulfate crystals, which were filtered from the solution. \(\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{BaCl}_{2}(a q) \longrightarrow 2 \mathrm{NaCl}(a q)+\mathrm{BaSO}_{4}(s)\) If \(5.719 \mathrm{~g}\) of barium sulfate was obtained, what was the molarity of the original \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) solution?

f \(38.2 \mathrm{~mL}\) of \(0.248 M\) aluminum sulfate solution is diluted with deionized water to a total volume of \(0.639 \mathrm{~L}\), how many grams of aluminum ion are present in the diluted solution?

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