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mars1129 [50]
2 years ago
9

The density of a metal is 11.4 g/cm3. How much volume (in cm3) would a sample of 30.5 g have?

Chemistry
1 answer:
julia-pushkina [17]2 years ago
6 0

<u>The Concept:</u>

We are given the density of a sample of the metal = 11.4 grams / cm³

and we need to find the volume occupied by a sample of 30.5 grams

For this solution, we will use dimensional analysis

from the given information, we can also say that the density of the metal is:

1 cm³ / 11.4 grams

If we multiply this value by 30.5 grams, the 'grams' in the numerator and the denominator will cancel out and we will be left with the volume occupied by 30.5 grams of the metal

<u>Solving for the volume:</u>

\frac{ 1 cm^{3} }{11.4 grams}  X  30.5 grams  = (30.5 / 11.4) cm³

Volume of 30.5 grams of the sample = 2.68 cm³

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Kb →  1.56 °C / m

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This is all about boiling point elevation, the colligative property that shows that boiling point for a solution is higher than boiling point of pure solvent.

This is the formula: ΔT = Kb . m . i

where i is the Van't Hoff factor (ions dissolved in solution). As these are organic compounds, we assume they are non electrolytic,

m is molality (mol of solute / 1kg of solvent)

Kb is our unknown. The value for ebulloscopic constant, it is specific for each solvent.

ΔT = T° boiling from solution - T° boiling from solute

First of all, let's determine the moles of solute.

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Molality is mol of solute/ 1 kg of solvent

We must convert the mass from g to kg

195g . 1kg /1000 = 0.195 kg

Molality = 0.286 mol / 0.195 kg = 1.47 m

Let's replace the values in the formula

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