CHEM6201 - Weekly Questions

Which of these is NOT the SI base unit for the given quantity?

• time – second (s)
• length – meter (m)
• temperature – kelvin (K)
• amount of substance – mole (mol)
• mass – gram (g)

Answer explanation

The SI base unit for mass is the kilogram (kg), not the gram. Therefore, "mass – gram (g)" is not an SI base unit.

Which of these prefixes represents the smallest factor of 10?

• kilo
• pico
• deci
• giga
• nano

Answer explanation

The prefix "pico" represents 10^-12, which is the smallest factor among the options

The value 3.5 × 10^-5 A is equal to

• 35 mA
• 3.5 µA
• 35 µA
• 0.35 mA
• 0.035 A

Answer explanation

3.5×10−5A=35μA3.5 \times 10^{-5} A = 35 \mu A3.5×10−5A=35μA. The correct conversion from amperes to microamperes leads to 35 µA.

The density of 70.0 wt% (aqueous solution) for ACS reagent grade HClO4 is 1.664 g/mL. Calculate the molarity.

• 0.0116 mol/L
• 23.7 mol/L
• 11.6 mol/L
• 1.16 mol/L
• 1.16 × 103 mol/L

Answer explanation

To calculate the molarity of the 70.0 wt% HClO4 solution, we first use the given density of 1.664 g/mL to determine that 1 liter (1000 mL) of the solution weighs 1664 grams. Since the solution is 70.0 wt% HClO4, 70% of this mass is HClO4, resulting in 1164.8 grams of HClO4 in 1 liter of solution. By dividing the mass of HClO4 by its molar mass (100.46 g/mol), we find that the solution contains approximately 11.6 moles of HClO4. Finally, since molarity is defined as moles of solute per liter of solution, the molarity of the solution is approximately 11.6 mol/L

The concentration of a 1 ppb dilute aqueous solution can be expressed as

• 1 mg/L
• 1 ng/mL
• 1 μg/mL
• 1 g/mL
• -

Answer explanation

1 ppb (part per billion) is equivalent to 1 ng/mL because 1 ppb corresponds to 1 part in 10^9, which in water (where 1 mL ≈ 1 g) equals 1 ng/mL

What is the concentration (in ppm) of Cu2+ in a 2.7 × 10^-4 mol/L aqueous CuSO4 solution?

• 34 ppm
• 4.2 ppm
• 8.4 ppm
• 17 ppm
• 43 ppm

Answer explanation

To find the concentration in ppm, multiply the molarity by the molar mass of Cu²⁺ (63.55 g/mol) and convert to mg/L (ppm). Thus, 2.7 × 10^-4 mol/L × 63.55 g/mol × 1000 mg/g ≈ 17.2 ppm

What volume of 0.0500 mol/L Ba(ClO4)2 is needed to obtain 250.0 mL of 0.0100 mol/L Ba(ClO4)2 solution?

• 50.0 mL
• 200. mL
• 150. mL
• 100. mL
• 125 mL

Answer explanation

Use the dilution equation C₁V₁ = C₂V₂, where C₁ = 0.0500 mol/L, C₂ = 0.0100 mol/L, and V₂ = 250.0 mL. Solving for V₁ gives V₁ = (C₂ × V₂) / C₁ = (0.0100 mol/L × 250.0 mL) / 0.0500 mol/L = 50.0 mL

Phenol (analytical standard) is available as a 5000. μg/mL solution in methanol. What aliquot do you have to take to make 25.0 mL of a 100. μg/mL solution?

• 500. μL
• 5.00 mL
• 125 mL
• 50.0 mL
• 0.100 mL

Answer explanation

Use the dilution formula C₁V₁ = C₂V₂, where C₁ = 5000 μg/mL, C₂ = 100 μg/mL, and V₂ = 25.0 mL. Solving for V₁ gives V₁ = (C₂ × V₂) / C₁ = (100 μg/mL × 25.0 mL) / 5000 μg/mL = 0.5 mL = 5.00 mL

Lead(II) chloride precipitates when a solution of Pb2+ and a solution of Cl− are mixed together. If you mix 100. mL of a 0.205 mol/L Pb2+ solution with 150. mL of a 0.245 mol/L Cl− solution, what is the limiting reagent?

• Pb2+
• Cl−
• PbCl2
• Both Pb2+ and Cl−
• The limiting reagent cannot be determined.

Answer explanation

Calculate the moles of each ion: Pb²⁺: 0.205 mol/L × 0.100 L = 0.0205 mol, Cl⁻: 0.245 mol/L × 0.150 L = 0.03675 mol. The balanced reaction is Pb²⁺ + 2Cl⁻ → PbCl₂, requiring twice as much Cl⁻ as Pb²⁺. Since 0.0205 mol Pb²⁺ would require 0.041 mol Cl⁻, but only 0.03675 mol Cl⁻ is available, Cl⁻ is the limiting reagent

Round 0.60043 to three significant figures.

• 0.6
• 0.60043
• 0.6
• 0.6004
• 0.6

Answer explanation

To round to three significant figures, look at the fourth digit. Since it's 4 (less than 5), the third digit remains the same, so the answer is 0.600