ADA 2009 Sample GC #41

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Exactly 492 g of Kerosene are required to fill a certain container. If the density of the Kerosene was 0.820 g *mL-1, and at the same temperature the density of mercury was 13.5 g * mL-1, then what mass of mercury, in grams, would be necessary to fill the container?

the answer is :
(492)(13.5)/(0.820)

help??🙂

 

[rockclock]

First, with kerosene:
D = m / V
0.820 g/mL = 492 g / V
V = 492 g / (0.820 g/mL) = ... mL (now you have the volume of the container)

Then, with mercury:
D = m / V
13.5 g/mL = m / [492 g / (0.820 g/mL)]
m = (13.5 g/mL) * 492 g / (0.820 g/mL) = ... g (the mass of mercury)

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Or:
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D1 = m1 / V1
V1 = m1 / D1

D2 = m2 / V2
V2 = m2 / D2

V1 = V2 since it's the same container, so:

m1 / D1 = m2 / D2
m2 = m1 / D1 * D2

 

[Jab1113]

Exactly 492 g of Kerosene are required to fill a certain container. If the density of the Kerosene was 0.820 g *mL-1, and at the same temperature the density of mercury was 13.5 g * mL-1, then what mass of mercury, in grams, would be necessary to fill the container?

the answer is :
(492)(13.5)/(0.820)

help??🙂

Given:

Kerosene

m = 492 g
d = 0.820

Mercury

m = ?
d = 13.5

Next, we need to know the density formula which is d = (m/v)

Deriving the equation a bit, we can find v for Kerosene which:

v = (m/d) = (492/.820)

Take that and plug into the density formula for Mercury:

d = (m/v)
13.5 = (m/(492/.820))
m = 13.5(492)/.820

Hope that made sense, good luck. 👍

 

[bittersweet008x]

ooh okay gotcha thanks! .. so simple once explained! 👍