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disa [49]
3 years ago
15

A mixture of nacl and sucrose (c12h22o11) of combined mass 10.2 g is dissolved in enough water to make up a 250 ml solution. the

osmotic pressure of the solution is 7.75 atm at 23°c. calculate the mass percent of nacl in the mixture.
Chemistry
1 answer:
Vilka [71]3 years ago
4 0
We use the osmotic pressure to determine the concentration of the solute in the solution. Then, we multiply the volume of the solution to determine the number of moles of solute particles. We need to establish to equations since we have two unknowns, the mass of of each solute. We do as follows:

osmotic pressure = CRT 
<span>C = 7.75 / 0.08205 (296.15) = 0.3189 mol / L</span>
<span>moles of particles = C*V = 0.3189*0.250 =0.0797 mol </span>
<span>0.0797 = moles of sucrose + 2*moles of salt </span>

<span>x + 2y = 0.0797 </span>
<span>and </span>
<span>x(MMsucrose) + y(MMNaCl) = 10.2</span>
<span>342x + 58.5y = 10.2
</span>
<span>solve for x and y
</span>
<span>x = 0.0252 mol sucrose</span>
<span>y = 0.0273 mol NaCl
 </span>
<span>mass Sucrose = 0.0252(342) = 8.6184 g </span>
<span>mass NaCl = 0.0273(58.5) = 1.5971 g </span>
<span>% NaCl = (1.5971 / 10.2)*100 = 15.66%</span>
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Answer:

172 g Al

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We know we will need a balanced equation with masses and molar masses, so let’s <em>gather all the information</em> in one place.  

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(a) Calculate the <em>moles of Al₂O₃ </em>

n = 325 g Al₂O₃ × 1 mol Al₂O₃ /39.10 g Al₂O₃

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(b) Calculate the <em>moles of Al </em>

The molar ratio is (4 mol Al/2 mol Al₂O₃)

n = 3.188 mol Al₂O₃ × (4 mol Al/2 mol Al₂O₃)

n = 6.375 mol Al

(c) Calculate the <em>mass of Al</em>

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Note: The answer can have only <em>three</em> significant figures because that is all you gave for the mass of Al₂O₃.

8 0
3 years ago
Suppose a current of is passed through an electroplating cell with an aqueous solution of agno3 in the cathode compartment for s
WARRIOR [948]

The given question is incomplete. The complete question is:

Suppose a current of 0.920 A is passed through an electroplating cell with an aqueous solution of agno3 in the cathode compartment for 47.0 seconds. Calculate the mass of pure silver deposited on a metal object made into the cathode of the cell.

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Explanation:

Q=I\times t

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t= time in seconds = 47.0 sec

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Thus the mass of pure silver deposited on a metal object made into the cathode of the cell is 0.0484 g

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3 years ago
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3 years ago
Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica
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<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

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To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = \frac{1}{2}\times 0.52=0.26mol of lead (II) iodide

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of lead (II) iodide = 0.26 moles

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Putting values in above equation, we get:

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Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0
3 years ago
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