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Basile [38]
3 years ago
5

A solution of a sugar with the chemical formula (c12h22o11) is prepared by dissolving 8.45 g in 250.0 ml of water at 25

Chemistry
2 answers:
lapo4ka [179]3 years ago
6 0
A. <em>Concentration in terms of molarity</em>

First, calculate for the molar mass of the given solution, C12H22O11.
    m = 12(12) + 22(1) + 11(16) = 232
Calculate for the number of moles.
   
    n = 8.45 g / 232 g/mol  = 0.036422 moles

Divide the determined number of moles by volume of solution in L.
    V(solution) = (250 mL)(1 L/1000 mL) = 0.25 L

The concentration in molarity is determined by dividing the number of moles by the volume of solution in liters.
          M = 0.03622 moles / 0.25 L = 0.146 moles/L

b. <em>Concentration in terms of molality</em>
Calculate the mass of the solvent in kilogram given the volume and its density.
              m(solvent) = (1.21 g/cm3)(250 mL)(1 kg/1000 g) = 0.3025 kg
Divide the calculated number of moles by the mass of the solvent in kilograms.

         molality = 0.036422 moles/0.3025 kg = 0.12 molal

c. <em>Concentration in terms of weight percent.</em>
The total weight of the solution is equal to 302.5 g.
    
           weight percent of solute = (8.45 g / 302.5 g)(100%) = 2.79%

d. <em>Weight percent in ppm.</em>
           weight percent of solute x (10000 ppm/1%) 
       = (2.79%)(10000 ppm/1%) = 27933.88 ppm
tigry1 [53]3 years ago
6 0

Answer : The molarity, molality, weight percent and ppm are, 0.0987 mole/L, 0.0839 mole/Kg, 2.79% and 27933.8843 respectively.

Solution : Given,

Mass of solute (C_{12}H_{22}O_{11}) = 8.45 g

Molar mass of solute (C_{12}H_{22}O_{11}) = 342.3 g/mole

Volume of solvent (water) = volume of solution = 250 ml

Density of solution = 1.21g/cm^3=1.21g/ml

<u>Calculation for molarity </u>

Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution in ml}}

Molarity=\frac{8.45g\times 1000}{342.3g/mole\times 250ml}=0.0987mole/L

Now we have to calculate the mass of solution.

\text{Mass of solution}=Density\times Volume=1.21g/ml\times 250ml=302.5g

Now we have to calculate the mass of solvent.

\text{Mass of solvent}=\text{Mass of solution}-\text{Mass of solute}=302.5-8.45=294.05g

<u>Calculation for molality </u>

Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent in g}}

Molality=\frac{8.45g}\times 1000}{342.3g/mole\times 249.05g}=0.0839mole/Kg

<u>Calculation for weight percent</u>

weight\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{8.45g}{302.5g}\times 100=2.79\%

<u>Calculation for ppm</u>

ppm=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6=\frac{8.45g}{302.5g}\times 10^6=27933.8843

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If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a
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Answer:

The concentration of the Potassium Dichromate solution is 0.611 M

Explanation:

First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:

KBr (aq) → K+ (aq) + Br- (aq)

K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)

According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.

Having said that, we calculate the moles of potassium ions coming from the KBr solution:

0.19 M KBr: this means that we have 0.19 moles of KBr in 1000 mL solution. So:

1000 mL solution ----- 0.19 moles of KBr

253 mL solution ----- x = 0.04807 moles of KBr

As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.

Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):

1000 mL solution ----- 0.846 moles K+

694 mL solution ----- x = 0.587124 moles K+

This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:

0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7

If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:

2 K+ moles ----- 1 K2Cr2O7 mole

0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles

Now this quantity of potassium dichromate moles came from the respective  solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:

441 mL ----- 0.269527 K2Cr2O7 moles

1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M

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