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Sedaia [141]
3 years ago
6

When a 10 mL graduated cylinder is filled to the 10 mL mark, the mass of the water was measured to be 9.925 g. If the density of

water is taken to be 0.9975 g/mL, what is the percent error for the 10 mL of water,
Chemistry
1 answer:
Vlada [557]3 years ago
5 0
Measured volume = 10 ml 
mass = 9.925 g 
density = 0.9975 g/ml
density = \frac{mass}{volume} 
so actual volume = \frac{mass}{density} = \frac{9.925}{0.9975} = 9.95 mL
Percentage Error = \frac{measured volume - Actual volume}{actual volume} x 100 
= \frac{10 - 9.95}{9.95} x 100 = 0.5 % error
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Water has a boiling point of 100.0°C and a Kb of 0.512°C/m. What is the boiling
Reil [10]

Answer:

The boiling  point of a 8.5 m solution of Mg3(PO4)2 in water is<u> 394.91 K.</u>

Explanation:

The formula for molal boiling Point elevation is :

\Delta T_{b} = iK_{b}m

\Delta T_{b} = elevation in boiling Point

K_{b} = Boiling point constant( ebullioscopic constant)

m = molality of the solution

<em>i =</em> Van't Hoff Factor

Van't Hoff Factor = It takes into accounts,The abnormal values of Temperature change due to association and dissociation .

In solution Mg3(PO4)2 dissociates as follow :

Mg_{3}(PO_{4})_{2}\rightarrow 3Mg^{2+} + 2 PO_{4}^{3-}

Total ions after dissociation in solution :

= 3 ions of Mg + 2 ions of phosphate

Total ions = 5

<em>i =</em> Van't Hoff Factor = 5

m = 8.5 m

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Insert the values and calculate temperature change:

\Delta T_{b} = iK_{b}m

\Delta T_{b} = 5\times 0.512\times 8.5

\Delta T_{b} = 21.76 K

Boiling point of pure water = 100°C = 273.15 +100 = 373.15 K

\Delta T_{b} = T_{b} - T_{b}_{pure}

T_{b}_{pure} = 373.15 K[/tex]

21.76 = T - 373.15

T = 373.15 + 21.76

T =394.91 K

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2 years ago
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Also this should be listed for history
3 0
3 years ago
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