What mass of Li3PO4 is needed to prepare 0.500 L of a 0.400 M solution of Li3PO4? The molar mass of Li3PO4 is 115.8 g∙mol–1.单项选择题
A
5.80 g
B
11.6 g
C
17.4 g
D
23.2 g
E
None of these

Question

What mass of Li3PO4 is needed to prepare 0.500 L of a 0.400 M solution of Li3PO4?  The molar mass of Li3PO4 is 115.8 g∙mol–1.

BlackTom AI · Accepted Answer

We begin by identifying what the problem asks: the mass of Li3PO4 needed to make 0.500 L of a 0.400 M solution.
First, compute the amount of substance (moles) required using M = n/V, or more directly n = M × V. Here, n = 0.400 mol/L × 0.500 L = 0.200 mol.
Next, convert moles to mass using the molar mass: mass = moles × molar mass. The given molar mass of Li3PO4 is 115.8 g/mol, so mass = 0.200 mol × 115.8 g/mol = 23.16 g.
Rounding to the appropriate significant figures (three sig figs in the concentrations and volume), we obtain 23.2 g.
Now evaluate the answer options one by one:
- 5.80 g: This value would correspond to far fewer moles, specifically 5.80 g ÷ 115.8 g/mol ≈ 0.0500 mol, which would yield a solution of only 0.100 M if the volume were 1.0 L, or a proportionally smaller concentration for 0.500 L. It does not satisfy the required 0.500 L at 0.400 M.
- 11.6 g: Converting back, 11.6 g ÷ 115.8 g/mol ≈ 0.100 mol. If dissolved in 0.500 L, the resulting molarity would be 0.100 mol ÷ 0.500 L = 0.200 M, which is too low by a factor of 2.
- 17.4 g: This mass corresponds to 17.4 g ÷ 115.8 g/mol ≈ 0.150 mol. In 0.500 L, the molarity would be 0.150 mol ÷ 0.500 L = 0.300 M, still below the target 0.400 M.
- 23.2 g: This mass gives 0.200 mol (since 23.2 g ÷ 115.8 g/mol ≈ 0.200 mol). When dissolved in 0.500 L, the concentration is 0.200 mol ÷ 0.500 L = 0.400 M, which matches the target specification exactly.
- None of these: Based on the calculation above, at least one option aligns with the required mass, so this choice is not appropriate.
In summary, the reasoning hinges on converting between mass and moles using the molar mass, then applying the definition of molarity to relate moles, volume, and concentration. The only option that yields a 0.400 M solution in 0.500 L is the mass that corresponds to 0.200 mol of Li3PO4, which is 23.2 g.

What mass of Li3PO4 is needed to prepare 0.500 L of a 0.400 M solution of Li3PO4? The molar mass of Li3PO4 is 115.8 g∙mol–1.单项选择题

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