Chemistry, 27.07.2019 00:00 morkitus13

What is the mole fraction, x, of solute and the molality, m (or b), for an aqueous solution that is 15.0% naoh by mass?

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Answer from: lapointayyy6388

The mole fraction of the aqueous solution is 0.073

The molality of the aqueous solution is 4.41 m.

What is a mole fraction?

The ratio of one component's moles to the total number of moles representing all the components in a solution or other mixture.

Step 1: We are calculating the number of moles of NaOH and H_2O .

We assume the solution is 100 ml, now we have 15.0% NaOH and 85% of H_2O .

$\bold{Moles\; of\; NaOH = 15\;g\;NaOH\times \dfrac{1\;mol\;of NaOH}{39.997\;g\;NaOH} = 0.3750 moles }$

$\bold{Moles\; of\; H_2O = 85\;g\;NaOH\times \dfrac{1\;mol\;of\;H_2O}{ 18\;g\;NaOH} = 4.722 moles }$

where 39.997 g and 18 g are molecular mass of NaOH and H_2O respectively.

Step 2: Now calculate the mole fraction

It can be calculated by dividing the number of moles of NaOH by the total number of moles.

$\bold{X NaOH = \dfrac{Moles\;of \;NaOH}{Total\;moles}=\dfrac{0.3750}{0.3750+ 4.7222} = 0.073 }$

Step 3: Calculate the number of moles

we know that the solution has 15 grams of NaOH and 85 grams (0,085 kg) of water. We can calculate the moles of NaOH in the following way:

$\bold{moles\;NaOH= 15\;g\;NaOH\times \dfrac{1\;mol\;NaOH}{39.997\;g\;NaOH}= 0.3750\; moles }$

Step 4: Now calculate the molality

$\bold{m\;NaOH = \dfrac{moles\;NaOH}{kg\;of \;solvent} =\dfrac{0.3750}{0.085\;kg\;H_2O} =4.41\;m}$

Thus, the mole fraction of 0.073 and molality is 4.41 m.

Learn more about molality, here:

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Answer from: whitakers87

Hello!

a) The mole fraction of solute of a 15% NaOH aqueous solution can be calculated in the following way:

First, we have to assume that we have 100 grams of solution. This will simplify the calculations.

Now, we know that this solution has 15 grams of NaOH and 85 grams of water. We can calculate the number of moles of each one in the following way:

$molesNaOH= 15 g NaOH*\frac{1molNaOH}{39,997 g NaOH}=0,3750 moles NaOH \\ \\ moles H_2O=85 gH_2O* \frac{1 mol H_2O}{18 g H_2O}=4,7222 moles H_2O$

To finish, we calculate the mole fraction by dividing the moles of NaOH between the total moles:

$X_{NaOH}= \frac{moles NaOH}{total moles}= \frac{0,3750 moles NaOH}{0,3750 moles NaOH+4,7222 molesH_2O} =0,073$

So, the mole fraction of NaOH is 0,073

b) The molality (moles NaOH/ kg of solvent) of a 15% NaOH aqueous solution can be calculated in the following way:

First, we have to assume that we have 100 grams of solution. This will simplify the calculations.

Now, we know that this solution has 15 grams of NaOH and 85 grams (0,085 kg) of water. We can calculate the moles of NaOH in the following way:

$molesNaOH= 15 g NaOH*\frac{1molNaOH}{39,997 g NaOH}=0,3750 moles NaOH$

Now, we apply the definition of molality to calculate the molality of the solution:

$mNaOH= \frac{moles NaOH}{kg_{solvent}}= \frac{0,3750 moles NaOH}{0,085 kg H_2O}=4,41 m$

So, the molality of this solution is 4,41 m

Have a nice day!

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