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Chapter 12: Problem 32

Battery acid has a specific gravity of \(1.285\) and is \(38.0\) percent sulfuric acid by weight. What mass of sulfuric acid is contained in a liter of battery acid?

Short Answer

Expert verified
488.3 grams of sulfuric acid in a liter.

Step by step solution

01

Understanding Specific Gravity

Specific gravity is the ratio of the density of a substance to the density of a reference substance, usually water. Water has a density of approximately 1,000 kg/m鲁, so the density of the battery acid can be calculated as: \[ \text{Density of Battery Acid} = 1.285 \times 1000 \text{ kg/m}^3 = 1285 \text{ kg/m}^3 \] This value can also be converted to g/L (which is the same as kg/m鲁), so the density of battery acid is **1285 g/L**.
02

Calculating Mass of Battery Acid in One Liter

Since density provides the mass per unit volume, the mass of one liter of battery acid can be directly taken from the density. Thus, the mass of battery acid in one liter is **1285 grams**.
03

Calculating Mass of Sulfuric Acid

Battery acid is 38.0% sulfuric acid by weight. Therefore, the mass of sulfuric acid in one liter of battery acid can be calculated using this percentage: \[ \text{Mass of Sulfuric Acid} = \frac{38.0}{100} \times 1285 \text{ g} = 488.3 \text{ g} \] Hence, there are 488.3 grams of sulfuric acid in a liter of battery acid.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Density Calculation
Density is a measure of how much mass exists within a particular volume of a substance. It is an intrinsic property that defines how tightly matter is packed together. When calculating density, one uses the formula:\[\text{Density} = \frac{\text{Mass}}{\text{Volume}}\]For our exercise, battery acid was said to have a specific gravity of 1.285. Specific gravity is the ratio of the density of a material to the density of water. The density of water is typically 1,000 kg/m鲁.
  • To find the density of battery acid, we multiply the specific gravity by the density of water.
  • This results in a density of 1,285 kg/m鲁 for battery acid.
  • Since 1 kg/m鲁 equals 1 g/L, the density is equivalently 1,285 g/L.
Understanding density is crucial, as it allows direct linkage not only between mass and volume, but also to explore subsequent calculations like mass percentage.
Mass Percentage
Mass percentage is a way of expressing the concentration of a component in a mixture. It tells us what fraction of the total mass is due to a particular component, and is often expressed as a percentage. The formula for mass percentage is:\[\text{Mass Percentage} = \left( \frac{\text{Mass of Component}}{\text{Total Mass of Mixture}} \right) \times 100\]In the battery acid example, sulfuric acid comprises 38.0% of the total mixture by weight.
  • We calculated this percentage directly from the given information about battery acid.
  • Mass percentage helps us determine how much sulfuric acid is present without analyzing each individual molecule.
Using the mass percentage, you find the mass of sulfuric acid in the solution. For this exercise, it was crucial as it tied together the density and the composition of the solution.
Chemical Concentration
Chemical concentration represents the abundance of a component in a mixture. It is frequently given in terms such as molarity, molality, and mass percentage. In this problem, we focused on the weight-specific concentration, informed by the mass percentage.
  • By knowing the density, we could determine the mass of the entire solution.
  • The mass percentage then allowed us to pinpoint the mass of sulfuric acid specifically.
  • This approach simplifies quantifying the amount of a substance present per unit volume.
These calculations are fundamental in chemistry, particularly in preparing solutions for experiments where exact concentrations are critical. Understanding concentrations aids in achieving the desired reactivity or preventing potential solution hazards.

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Most popular questions from this chapter

A 60 -kg motor sits on four cylindrical rubber blocks. Each cylinder has a height of \(3.0 \mathrm{~cm}\) and a cross-sectional area of 15 \(\mathrm{cm}^{2}\). The shear modulus for this rubber is \(2.0 \mathrm{MPa}\). \((a)\) If a sideways force of \(300 \mathrm{~N}\) is applied to the motor, how far will it move sideways? (b) With what frequency will the motor vibrate back and forth sideways if disturbed?

Find the density and specific gravity of ethyl alcohol if \(63.3 \mathrm{~g}\) occupies \(80.0 \mathrm{~mL}\).

A thin sheet of gold foil has an area of \(3.12 \mathrm{~cm}^{2}\) and a mass of \(6.50 \mathrm{mg}\). How thick is the sheet? The density of gold is 19300 \(\mathrm{kg} / \mathrm{m}^{3}\) One milligram is \(10^{-6} \mathrm{~kg}\), so the mass of the sheet is \(6.50 \times 10^{-6}\) kg. Its volume is $$ V=(\text { area }) \times(\text { thickness })=\left(3.12 \times 10^{-4} \mathrm{~m}^{2}\right)(d) $$ where \(d\) is the thickness of the sheet. We equate this expression for the volume to \(m / \rho\) to get $$ \left(3.12 \times 10^{-4} \mathrm{~m}^{2}\right)(d)=\frac{6.50 \times 10^{-6} \mathrm{~kg}}{19300 \mathrm{~kg} / \mathrm{m}^{3}} $$ from which \(d=1.08 \times 10^{-6} \mathrm{~m}=1.08 \mu \mathrm{m}\).

Two parallel oppositely directed forces, each \(4000 \mathrm{~N}\), are applied tangentially to the upper and lower faces of a cubical metal block \(25 \mathrm{~cm}\) on a side. Find the angle of shear and the displacement of the upper surface relative to the lower surface. The shear modulus for the metal is 80 GPa.

The compressibility of water is \(5.0 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{N}\). Find the decrease in volume of \(100 \mathrm{~mL}\) of water when subjected to a pressure of 15 MPa.
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