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Scrat [10]
4 years ago
13

An aqueous solution contains the amino acid glycine (NH2CH2COOH). Assuming that the acid does not ionize in water, calculate the

molality of the solution if it freezes at −1.83°C. The freezing point depression constant for water is 1.86°C/m.
Chemistry
1 answer:
zmey [24]4 years ago
7 0

Answer:

Molality for this solution is 0.98 mol/kg

Explanation:

ΔT = Kf . m . i

Freezing point depression, here

ΔT =  Freezing point of pure solvent - Freezing point of solution

i = Van't Hoff factor.

This solute, the amino acid glycine does not ionize, so the i in this case is 1

Let's replace the data given

0° - (-1.83°C) = 1.86 °C/m . m . 1

1.83°C =  1.86 °C/m . m . 1

1.83°C / 1.86 m/°C = 0.98 m

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Assuming that an acetic acid solution is 12% by mass and that the density of the solution is 1.00 g/mL, what volume of 1 M NaOH
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Explanation:

Let us assume that total mass of the solution is 100 g. And, as it is given that acetic acid solution is 12% by mass which means that mass of acetic acid is 12 g and 88 g is the water.

Now, calculate the number of moles of acetic acid as its molar mass is 60 g/mol.

    No. of moles = \frac{mass}{\text{molar mass}}

                           = \frac{12 g}{60 g/mol}

                           = 0.2 mol

Molarity of acetic acid is calculated as follows.

              Density = \frac{mass}{volume}

                 1 g/ml = \frac{100 g}{volume}

                    volume = 100 ml

Hence, molarity = \frac{\text{no. of moles}}{volume}

                           = \frac{0.2 mol}{0.1 L}

                           = 2 mol/l

As reaction equation for the given reaction is as follows.

     NaOH + CH_{3}COOH \rightarrow CH_{3}COONa + H_{2}O

So,          moles of NaOH = moles of acetic acid

Let us suppose that moles of NaOH are "x".

          x \times 1 M = 10 mL \times 2 M     (as 1 L = 1000 ml)

                        x = 20 L

Thus, we can conclude that volume of NaOH required is 20 ml.                    

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