Question

A solution is made by dissolving 5.84 grams of NaCl in enough distilled water to give a final volume of 1.00 L. What is the molarity of the solution

Group of answer choices

0.0250 M

0.400 M

0.100 M

1.00 M

Answer 1

Step-by-step explanation:

1. A solution is made by dissolving 5.84g of NaCl is enough distilled water to a give a final volume of 1.00L. What is the molarity of the solution? a. 0.100 M b. 1.00 M c. 0.0250 M d. 0.400 M 2. A 0.9% NaCl (w/w) solution in water is a. is made by mixing 0.9 moles of NaCl in a 100 moles of water b. made and has the same final volume as 0.9% solution in ethyl alcohol c. a solution that boils at or above 100°C d. All the above (don't choose this one) 3. In an exergonic process, the system a. gains energy b. loses energy c. either gains or loses energy d. no energy change at all

Answer 2

`\boxed {\boxed {\sf 0.100 \ M }}`

Step-by-step explanation:

Molarity is a measure of concentration in moles per liter.

`molarity = (moles \ of \ solute)/(liters \ of \ solution)}`

The solution has 5.84 grams of sodium chloride or NaCl and a volume of 1.00 liters.

1. Moles of Solute

We are given the mass of solute in grams, so we must convert to moles. This requires the molar mass, or the mass of 1 mole of a substance. These values are found on the Periodic Table as the atomic masses, but the units are grams per mole, not atomic mass units.

We have the compound sodium chloride, so look up the molar masses of the individual elements: sodium and chlorine.

• Na: 22.9897693 g/mol
• Cl: 35.45 g/mol

The chemical formula (NaCl) contains no subscripts, so there is 1 mole of each element in 1 mole of the compound. Add the 2 molar masses to find the compound's molar mass.

• NaCl: 22.9897693 + 35.45 = 58.4397693 g/mol

There are 58.4397693 grams of sodium chloride in 1 mole. We will use dimensional analysis and create a ratio using this information.

`\frac {58.4397693 \ g\ \ NaCl} {1 \ mol \ NaCl}`

We are converting 5.84 grams to moles, so we multiply by that value.

`5.84 \ g \ NaCl *\frac {58.4397693 \ g\ NaCl} {1 \ mol \ NaCl}`

Flip the ratio. It remains equivalent and the units of grams of sodium chloride cancel.

`5.84 \ g \ NaCl *\frac {1 \ mol \ NaCl}{58.4397693 \ g\ NaCl}`

`5.84 *\frac {1 \ mol \ NaCl}{58.4397693 }`

`0.09993194823 \ mol \ NaCl`

2. Molarity

We can use the number of moles we just calculated to find the molarity. Remember there is 1 liter of solution.

`molarity= (moles \ of \ solute)/(liters \ of \ solution)`

`molarity= ( 0.09993194823 \ mol \ NaCl)/(1 \ L)`

`molarity= 0.09993194823 \ mol \ NaCl/L`

3. Units and Significant Figures

The original measurements of mass and volume have 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandths place. The 9 in the ten-thousandths place tells us to round the 9 to a 0, but then we must also the next 9 to a 0, and the 0 to a 1.

`molarity \approx 0.100 \ mol \ NaCl/L`

1 mole per liter is 1 molar or M. We can convert the units.

`molarity \approx 0.100 \ M \ NaCl`

The molarity of the solution is 0.100 M.