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oksian1 [2.3K]
3 years ago
10

what is the molality of a 10.5% by mass glucose (C6H12O6) the density of the solution is 1.03 g/ml. i got up to m = 0.058281527

mol/ kg... please help !!!!,
Chemistry
1 answer:
iVinArrow [24]3 years ago
7 0
Solution:
<span>1 L of solution and 1.03 g/mL = 1030 g/L </span>
<span>
1030 g/L time 10.5%
= 108.15 g of glucose/L</span>
And
<span>=108.15 g / 180.16 g/mol
= 0.6003 moles in 1L </span>
<span>=or 0.600 mol/kg</span>
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If the volume is doubled and the number of molecules is doubled, pressure is unchanged

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The question is incomplete, here is the complete question:

Iron (III) oxide and hydrogen react to form iron and water, like this:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

At a certain temperature, a chemist finds that a 8.9 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, Iron, and water at equilibrium has the following composition.

Compound             Amount

  Fe₂O₃                     3.95 g

     H₂                        4.77 g

     Fe                        4.38 g

    H₂O                      2.00 g

Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

<u>Answer:</u> The value of equilibrium constant for given equation is 1.0\times 10^{-4}

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

  • <u>For hydrogen gas:</u>

Given mass of hydrogen gas = 4.77 g

Molar mass of hydrogen gas = 2 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

\text{Molarity of hydrogen gas}=\frac{4.77}{2\times 8.9}\\\\\text{Molarity of hydrogen gas}=0.268M

  • <u>For water:</u>

Given mass of water = 2.00 g

Molar mass of water = 18 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

\text{Molarity of water}=\frac{2.00}{18\times 8.9}\\\\\text{Molarity of water}=0.0125M

For the given chemical equation:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

The expression of equilibrium constant for above equation follows:

K_{eq}=\frac{[H_2O]^3}{[H_2]^3}

Concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

K_{c}=\frac{(0.0125)^3}{(0.268)^3}\\\\K_{c}=1.0\times 10^{-4}

Hence, the value of equilibrium constant for given equation is 1.0\times 10^{-4}

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