Question

How many grams of NaCl (MM = 58.44g/mol) are in 250mL of a 0.75 molar solution?

Answer 1

`\boxed {\boxed {\sf 11 \ grams \ NaCl}}`

Step-by-step explanation:

We are asked to find how many grams of sodium chloride are in a solution.

1. Moles of Solute

Molarity is a measure of concentration in moles per liter.

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

We know the molarity is 0.75 molar. 1 molar is the same as 1 mole per liter, so the solution contains 0.75 moles of sodium chloride per liter.

There are 250 milliliters of solution but molarity uses liters for volume. We must convert milliliters to liters. Remember that 1 liter contains 1000 milliliters. Set up a ratio and use dimensional analysis to convert.

`250 \ mL * \frac{1 \ L} {1000\ mL} = ( 250)/(1000) \ L = 0.250 \ L`

Now we know the molarity and the liters of solution, but the moles of solute are unknown.

• molarity = 0.75 mol NaCl/L
• moles of solute =x
• liters of solution = 0.25 L

Substitute the values into the formula.

`0.75 \ mol \ NaCl/L = (x)/(0.250 \ L)`

We are solving for the moles of solute, so we must isolate the variable x. It is being divided by 0.250 liters. The inverse of division is multiplication, so multiply both sides of the equation by 0.250 L.

`0.250 \ L *0.75 \ mol \ NaCl/L = (x)/(0.250 \ L) * 0.250 \ L`

`0.250 \ L *0.75 \ mol \ NaCl/L = x`

The units of liters cancel.

`0.250 * 0.75 \ mol \ NaCl`

`\bold {0.1875 \ mol \ NaCl}`

2. Grams of Solute

Now that we have calculated the moles of solute, we must convert this to grams. We use the molar mass or the mass of 1 mole of a substance. Sodium chloride's molar mass is given and it is 58.44 grams per mole. This means there are 58.44 grams of sodium chloride in 1 mole of sodium chloride.

Set up a ratio so we can convert using dimensional analysis.

`\frac {58.44 \ g \ NaCl}{1 \ mol \ NaCl}`

Multiply by the number of moles we calculated.

`0.1875 \ mol \ NaCl *\frac {58.44 \ g \ NaCl}{1 \ mol \ NaCl}`

The units of moles of sodium chloride cancel.

`0.1875 *\frac {58.44 \ g \ NaCl}{1}`

`\bold {10.9575 \ g \ NaCl}`

3. Round using Significant Figures

The original measurements of molarity and volume have 2 significant figures, so our answer must have the same. For the number we calculated, that is the ones place. The 9 in the tenths place tells us to round the 0 up to a 1.

`11 \ g \ NaCl`

There are approximately 11 grams of sodium chloride in 250 mL of a 0.75 molar solution.