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
<h3>What is molarity? </h3>
This is defined as the mole of solute per unit litre of solution. Mathematically, it can be expressed as:
Molarity = mole / Volume
<h3>1. How to determine the mass of RbCN required</h3>
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
<h3>2. How to determine the mass of RbCN required</h3>
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
<h3>3. What does a molar solution of CaCl₂ contains?</h3>
This is a solution that contains 1 mole of CaCl₂ in 1 L of the solution
<h3>4. How to determine the empirical formula</h3>
- 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
<h3>5. How to determine the empirical formula</h3>
- 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:
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Learn more about empirical formula:
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