How many mL of a 17.6 % solution of compound X (MW 142) would be required to prepare 450 mL of a 0.75 M solution?

Question

Quizplus · Accepted Answer

The equation to use to solve this problem is V₁C₁ = V₂C₂.However, the units for concentration do not match.The first step is to convert % to M: 17.6 % = 17.6 g/100 mL = 176 g/L.The definition of a 1 M solution is the number of gram molecular weights per liter of solution.
 Thus $ 1 \mathrm{M} $ compound $ \mathrm{X} $ equals $ 142 \mathrm{~g} / \mathrm{L} $.
$$
\frac{142 \mathrm{~g} / \mathrm{L}}{176 \mathrm{~g} / \mathrm{L}}=\frac{1 \mathrm{M}}{\mathrm{x}} \mathrm{x}=1.24 \mathrm{M}
$$
The $ 17.6 \% $ solution of compound $ \mathrm{X} $ is equal to a $ 1.24 \mathrm{M} $ solution. Now, solving for $ \mathrm{V}_{1}: \mathrm{V}_{\mathrm{l}}(1.24 \mathrm{M})=450 \mathrm{~mL}(0.75 \mathrm{M}), \mathrm{V}_{\mathrm{l}}=272 \mathrm{~mL} $ of the compound $ \mathrm{X} $ solution.

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