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Chem Questions

1) Determine the mass of RbCN required to prepare a 370 mL solution which is 0.12 M RbCN (MW = 112 g/mol).


2) Determine the mass of RbCN required to prepare a 246 mL solution which is 0.12 M RbCN (MW = 112 g/mol).

3) What does a 1-molar solution of CaCl_2 contain?

4) Calculate the empirical formula for a chemical compound that gives the following analysis: 12.6 percent lithium, 29.2 percent sulfur, and 58.2 percent oxygen.

5) Calculate the empirical formula for a chemical compound that gives the following analysis: 65.2 g scandium and 34.8 g oxygen.

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1. The mass of RbCN required to prepare the solution is 4.97 g

2. The mass of RbCN required to prepare the solution is 3.31 g

3. 1 molar solution of CaCl₂ contains 1 mole of CaCl₂ in 1 L of the solution

4. The empirical formula of the compound is Li₂SO₄

5. The empirical formula of the compound is Sc₂O₃

What is molarity?

This is defined as the mole of solute per unit litre of solution. Mathematically, it can be expressed as:

Molarity = mole / Volume

1. How to determine the mass of RbCN required

We'll begin by calculating the mole of RbCN in the solution

  • Volume = 370 mL = 370 / 1000 = 0.37 L
  • Molarity = 0.12 M
  • Mole = ?

Mole = molarity × volume

Mole = 0.12 × 0.37

Mole = 0.0444 mole

Finally, we shall determine the mass of RbCN required as illustrated below

  • Mole = 0.0444 mole
  • Molar mass = 112 g/mol
  • Mass of RbCN =?

Mass = mole × molar mass

Mass of RbCN = 0.0444 × 112

Mass of RbCN = 4.97 g

2. How to determine the mass of RbCN required

We'll begin by calculating the mole of RbCN in the solution

  • Volume = 246 mL = 246 / 1000 = 0.246 L
  • Molarity = 0.12 M
  • Mole = ?

Mole = molarity × volume

Mole = 0.12 × 0.246

Mole = 0.02952 mole

Finally, we shall determine the mass of RbCN required as illustrated below

  • Mole = 0.02952 mole
  • Molar mass = 112 g/mol
  • Mass of RbCN =?

Mass = mole × molar mass

Mass of RbCN = 0.02952 × 112

Mass of RbCN = 3.31 g

3. What does a molar solution of CaCl₂ contains?

This is a solution that contains 1 mole of CaCl₂ in 1 L of the solution

4. How to determine the empirical formula

  • Li = 12.6%
  • S = 29.2%
  • O = 58.2%
  • Empirical formula =?

Divide by their molar mass

Li = 12.6 / 7 = 1.8

S = 29.2 / 32 = 0.9125

O = 58.2 / 16 = 3.6375

Divide by the smallest

Li = 1.8 / 0.9125 = 2

S = 0.9125 / 0.9125 = 1

O = 3.6375 / 0.9125 = 4

Thus, the empirical formula of the compound is Li₂SO₄

5. How to determine the empirical formula

  • SC = 65.2 g
  • O = 34.8 g
  • Empirical formula =?

Divide by their molar mass

SC = 65.2 / 45 = 1.449

O = 34.8 / 16 = 2.175

Divide by the smallest

SC = 1.449 / 1.449 = 1

O = 2.175 / 1.449 = 3/2

Multiply by 2 to express in whole number

SC = 1 × 2 = 2

O = 3/2 × 2 = 3

Thus, the empirical formula of the compound is Sc₂O₃

Learn more about molarity:

https://brainly.com/question/15370276

Learn more about empirical formula:

https://brainly.com/question/24297883

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