Hello!

A molecule is a chemical combination of two or more atoms. which contains the largest number of molecules? Â

A.1.0g CH4 (MM=16 g/mol)

B.1.0g HNO3 (MM=63 g/mol)

C.1.0g H20 (MM=18g/mol)

D.1.0g N204 (MM=92 g/mol)

Solving:

A)

m1 (solute mass - CH4) = 1.0 g

MM (molar mass - CH4) = 16 g/mol

n (number of mols of CH4) = ? (in mol)

Let's find the number of mol, let's see:

Knowing that by Avogadro's Law for each mole of a substance we have 6.022 * 10Â²Â³ molecules, then:

1 mol -------------- 6.022*10Â²Â³ molecules

0.0625 mol ------------------ y molecules

multiply the means by the extremes

B)

m1 (solute mass - HNO3) = 1.0 g

MM (molar mass - HNO3) = 63 g/mol

n (number of mols of HNO3) = ? (in mol)

Let's find the number of mol, let's see:

Knowing that by Avogadro's Law for each mole of a substance we have 6.022 * 10Â²Â³ molecules, then:

1 mol -------------- 6.022*10Â²Â³ molecules

0.0158 mol ------------------ y molecules

multiply the means by the extremes

C)

m1 (solute mass - H2O) = 1.0 g

MM (molar mass - H2O) = 18 g/mol

n (number of mols of H2O) = ? (in mol)

Let's find the number of mol, let's see:

Knowing that by Avogadro's Law for each mole of a substance we have 6.022 * 10Â²Â³ molecules, then:

1 mol -------------- 6.022*10Â²Â³ molecules

0.0555 mol ------------------ y molecules

multiply the means by the extremes

D)

m1 (solute mass - N2O4) = 1.0 g

MM (molar mass - N2O4) = 92 g/mol

n (number of mols of N2O4) = ? (in mol)

Let's find the number of mol, let's see:

Knowing that by Avogadro's Law for each mole of a substance we have 6.022 * 10Â²Â³ molecules, then:

1 mol -------------- 6.022*10Â²Â³ molecules

0.0108 mol ------------------ y molecules

multiply the means by the extremes

Therefore:

(A) 3.76375*10Â²Â² > (C) 3.34221*10Â²Â² > (B) 9.51476*10Â²Â¹ > (D) 6.50376*10Â²Â¹

A.1.0g CH4 (MM=16 g/mol)

The "A" alternative contains the largest number of molecules (3.76375*10Â²Â² molecules).

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