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Chapter 1: Problem 106

A 25.00-g sample of a solid is placed in a graduated cylinder, and then the cylinder is filled to the 50.0-mL mark with benzene. The mass of benzene and solid together is 58.80 g. Assuming that the solid is insoluble in benzene and that the density of benzene is \(0.880 \mathrm{g} / \mathrm{cm}^{3},\) calculate the density of the solid.

Short Answer

Expert verified
The density of the solid is \(2.16 \mathrm{g/cm^3}\).

Step by step solution

01

Find the mass of benzene in the mixture

We are given the mass of the mixture, which includes both the solid and the benzene. We are also given the mass of the solid. To find the mass of benzene, we need to subtract the mass of the solid from the total mass of the mixture: mass_benzene = mass_mixture - mass_solid Given: mass_mixture = 58.80 g mass_solid = 25.00 g mass_benzene = 58.80 g - 25.00 g = 33.80 g So, the mass of the benzene in the mixture is 33.80 g.
02

Find the volume of the benzene in the mixture

We are given the density of benzene and we have calculated the mass of benzene in the mixture. We can now use the density formula to find the volume of benzene: density = mass / volume We will rearrange the formula to solve for volume: volume = mass / density Given: density_benzene = 0.880 g/cm³ mass_benzene = 33.80 g volume_benzene = 33.80 g / 0.880 g/cm³ = 38.41 cm³ So, the volume of the benzene in the mixture is 38.41 cm³.
03

Find the volume of the solid

We know the total volume of the mixture and the volume of the benzene. We can now find the volume of the solid by subtracting the volume of benzene from the total volume: volume_solid = volume_mixture - volume_benzene Given: volume_mixture = 50.0 mL (1 mL = 1 cm³) volume_benzene = 38.41 cm³ volume_solid = 50.0 cm³ - 38.41 cm³ = 11.59 cm³ So, the volume of the solid is 11.59 cm³.
04

Calculate the density of the solid

Now that we have the mass and volume of the solid, we can calculate its density using the density formula: density_solid = mass_solid / volume_solid Given: mass_solid = 25.00 g volume_solid = 11.59 cm³ density_solid = 25.00 g / 11.59 cm³ = 2.16 g/cm³ So, the density of the solid is 2.16 g/cm³.

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Most popular questions from this chapter

Many times errors are expressed in terms of percentage. The percent error is the absolute value of the difference of the true value and the experimental value, divided by the true value, and multiplied by 100. Percent error $=\frac{ | \text { true value }-\text { experimental value } |}{\text { true value }} \times 100$ Calculate the percent error for the following measurements. a. The density of an aluminum block determined in an experiment was 2.64 \(\mathrm{g} / \mathrm{cm}^{3} .\) (True value 2.70 $\mathrm{g} / \mathrm{cm}^{3} . )$ b. The experimental determination of iron in iron ore was 16.48\(\% .\) (True value 16.12\(\% . )\) c. A balance measured the mass of a \(1.000-\mathrm{g}\) standard as 0.9981 \(\mathrm{g}\) .

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Precious metals and gems are measured in troy weights in the English system: $$\begin{aligned} 24 \text { grains } &=1 \text { pennyweight (exact) } \\ 20 \text { pennyweight } &=1 \text { troy ounce (exact) } \\ 12 \text { troy ounces } &=1 \text { troy pound (exact) } \\ 1 \operatorname{grain} &=0.0648 \mathrm{g} \\ 1 \text { carat } &=0.200 \mathrm{g} \end{aligned}$$ a. The most common English unit of mass is the pound avoirdupois. What is 1 troy pound in kilograms and in pounds? b. What is the mass of a troy ounce of gold in grams and in carats? c. The density of gold is 19.3 \(\mathrm{g} / \mathrm{cm}^{3} .\) What is the volume of a troy pound of gold?

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