000333 g fluorescein (332.32 g/mol) is dissolved in 225 ml solution of ethanol. the density of ethanol is 0.785 g/ml. what is th

Question

garri49 [273]

3 years ago

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000333 g fluorescein (332.32 g/mol) is dissolved in 225 ml solution of ethanol. the density of ethanol is 0.785 g/ml. what is th

e concentration of the solution in molarity (m)? how many ppm? i already

Chemistry

2 answers:

andrey2020 [161] · Answer 1 · 2022-08-14T03:50:47+03:00

1) Molarity

M = n / V
n: number of moles of solute
V: volume of the solution in liters

n = mass / molar mass = 0.000333 g / 332.32 g / mol = 1*10 ^ - 6 moles

V = 225 ml * 1 liter / 1000 ml = 0.225 liter

M = 10^-6 mol / 0.225 liter = 0.00000444 M

2) ppm

ppm = parts per million

grams of solute: 0.000333 g

grams of solution = volume * density = 225 ml * 0.785 g / ml = 176.625 g

ppm = [0.00033 g / 176.625 g] * 1,000,000 = 1.868 ppm

Gnoma [55] · Answer 2 · 2022-08-14T05:32:11+03:00

The concentration of the solution in molarity and ppm is $\boxed{{\text{0}}{\text{.00000445 M}}}$ and $\boxed{{\text{1}}{\text{.88 ppm}}}$ respectively.

Further Explanation:

The proportion of substance in the mixture is called concentration. The most commonly used concentration terms are as follows:

1. Molarity (M)

2. Molality (m)

3. Mole fraction (X)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

Molarity is a concentration term that is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by M and its unit is mol/L.

The formula to calculate the molarity of the solution is as follows:

${\text{Molarity of solution}} = \frac{{{\text{Moles}}\;{\text{of}}\;{\text{solute}}}}{{{\text{Volume }}\left( {\text{L}} \right){\text{ of}}\;{\text{solution}}}}$ …… (1)

The formula to calculate the moles of fluorescein is as follows:

${\text{Moles of fluorescein}} = \frac{{{\text{Given mass of fluorescein}}}}{{{\text{Molar mass of fluorescein}}}}$ …… (2)

The given mass of fluorescein is 0.000333 g.

The molar mass of fluorescein is 332.32 g/mol.

Substitute these values in equation (2).

$\begin{aligned}{\text{Moles of fluorescein}}&=\left( {{\text{0}}{\text{.000333 g}}} \right)\left( {\frac{{{\text{1 mol}}}}{{{\text{332}}{\text{.32 g}}}}} \right)\\&=0.000001002\;{\text{mol}} \\ \end{aligned}$

Substitute 0.000001002 for the moles of solute and 225 mL for the volume of solution in equation (2).

$\begin{aligned}{\text{Molarity of solution}}&=\left( {{\text{0}}{\text{.000001002 mol}}} \right)\left( {\frac{1}{{{\text{225 mL}}}}} \right)\left( {\frac{{{\text{1 mL}}}}{{{\text{1}}{{\text{0}}^{ - 3}}{\text{ L}}}}} \right) \\&= {\text{0}}{\text{.0000044535 M}}\\&\approx{\text{0}}{\text{.00000445 M}} \\ \end{aligned}$

The molarity of the solution is 0.00000445 M.

The ${\text{ppm}}$ or parts per million is a concentration term equal to the mass of any substance divided by the mass of the solution, multiplied by ${10^6}$ .

The formula to calculate the concentration of fluorescein in ${\text{ppm}}$ is as follows:

${ppm}} = \left( {\frac{{{\text{mass}}\;{\text{of fluorescein}}}}{{{\text{mass}}\;{\text{of}}\;{\text{solution}}}}} \right){10^6}$ ...... (3)

The formula to calculate the density of the solution is as follows:

${\text{Density of solution}} = \frac{{{\text{Mass of solution}}}}{{{\text{Volume of solution}}}}$ …… (4)

Rearrange equation (6) to calculate the mass of the solution.

${\text{Mass of solution}} = \left( {{\text{Density of solution}}} \right)\left( {{\text{Volume of solution}}} \right)$ …… (5)

Substitute 225 mL for the volume of fluorescein and 0.785 g/mL for the volume of solution in equation (5).

$\begin{aligned}{\text{Mass of solution}}&=\left({\frac{{{\text{0}}{\text{.785 g}}}}{{{\text{1 mL}}}}} \right)\left( {{\text{225 mL}}} \right)\\&=176.6{\text{25 g}} \\ \end{aligned}$

Substitute 0.000333 g for the mass of fluorescein and 176.625 g for the mass of solution in equation (3).

$\begin{aligned}{\text{ppm}}&=\left( {\frac{{{\text{0}}{\text{.000333 g}}}}{{{\text{176}}{\text{.625 g}}}}} \right){10^6}\\&= 1.8853{\text{5 ppm}}\\&\approx {\text{1}}{\text{.88 ppm}} \\ \end{aligned}$