SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

CHEMISTRY, M2 EQ-Bank 11 MC

A 5.85 g sample of sodium chloride is dissolved in water to form a solution.

How many ions in total (sodium + chloride) are present in the solution?

  1. 1.5 \(\times\) 10\(^{22}\)
  2. 3.01 \(\times\) 10\(^{22}\)
  3. 6.02 \(\times\) 10\(^{22}\)
  4. 1.20 \(\times\) 10\(^{23}\)
Show Answers Only

\(D\)

Show Worked Solution

\(n\ce{(NaCl)} = \dfrac{m}{MM} = \dfrac{5.85}{58.44} = 0.100\ \text{mol}\)

\(N = n \times N_A = 0.100 \times 6.022 \times 10^{23} = 6.02 \times 10^{22}\)

  • Each sodium chloride compound will produce one of each ion (sodium and chloride) when dissolved in water, hence the total number of ions in the solution \(= 2 \times 6.02 \times 10^{22} = 1.20 \times 10^{23}\)

\(\Rightarrow D\)

CHEMISTRY, M2 EQ-Bank 9 MC

How many molecules are present in 124.0 L of oxygen gas at 298 K and 100 kPa?

  1. 3.0 \(\times\) 10\(^{24}\)
  2. 6.0 \(\times\) 10\(^{24}\)
  3. 3.0 \(\times\) 10\(^{23}\)
  4. 6.0 \(\times\) 10\(^{23}\)
Show Answers Only

\(A\)

Show Worked Solution

\(n = \dfrac{V}{V_m} = \dfrac{124.0}{24.79} = 5.00\ \text{mol}\)

Number of molecules present:

\(N = n \times N_A = 5.00 \times 6.022 \times 10^{23} = 3.0 \times 10^{24}\)

\(\Rightarrow A\)