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>
Answer:
The answer is
<h2>2.00 %</h2>
Explanation:
The percentage error of a certain measurement can be found by using the formula
From the question
actual measurement = 46.37 g
error = 47.25 - 46.37 = 0.88
The percentage error of the measurement is
We have the final answer as
<h3>2.00 %</h3>
Hope this helps you
Answer:
2.6%
Explanation:
As, 1 ounce (oz) = 0.0625 pounds (lb)
Therefore, weight of baby at discharge = 7 lb,1 oz = 7+0.0625 lb = 7.0625 lb
Since, 1 oz = 0.0625 lb
⇒ 4 oz = 4×0.0625 = 0.25 lb
Therefore, weight of baby at birth = 7 lb,4 oz = 7+0.25 lb = 7.25 lb
The <u>amount of weight lost</u> is equal to the difference of weight of the baby at birth and discharge.
Therefore, <u>weight lost</u> = 7.25 lb - 7.0625 lb = <u>0.1875 lb</u>
Now, the <u>percentage of weight lost</u> by the baby is given by the amount of weight lost divided by the weight of the baby at birth.
Therefore, <u>the percentage of weight los</u>t = weight lost ÷ weight at birth = 0.1875 lb ÷ 7.25 lb × 100 = <u>2.6% </u>
Answer:
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