Question

What mass of Ni3(PO4)2 is needed to prepare 155.13 mL of a solution of 2.150 M Ni3(PO4)2 ? 122.08 g O A. 38.62 g OB. 49.73 g C. 119.23 g OD. 87.67 g O E.

Answer 1

The correct answer is:
`\[ \text{C. 119.23 g} \]`.

To find the mass of Ni₃(PO₄)₂ needed to prepare a solution, you can use the formula:

`\[ \text{Number of moles} = \text{Molarity} * \text{Volume (in liters)} \]`

Given that the molarity
`(\(M\))` is
`\(2.150 \, \text{mol/L}\)` and the volume
`(\(V\))` is
`\(155.13 \, \text{mL}\)`, convert the volume to liters:

`\[ V = 155.13 \, \text{mL} * \left(\frac{1 \, \text{L}}{1000 \, \text{mL}}\right) = 0.15513 \, \text{L} \]`

Now, calculate the number of moles:

`\[ \text{Number of moles} = 2.150 \, \text{mol/L} * 0.15513 \, \text{L} = 0.3330225 \, \text{mol} \]`

Next, use the molar mass of Ni₃(PO₄)₂ to find the mass:

`\[ \text{Mass} = \text{Number of moles} * \text{Molar mass} \]`

The molar mass of Ni₃(PO₄)₂ is calculated in a previous response as
`\(365.01 \, \text{g/mol}\)`.

`\[ \text{Mass} = 0.3330225 \, \text{mol} * 365.01 \, \text{g/mol} = 121.68 \, \text{g} \]`

The closest answer among the choices is
`\(119.23 \, \text{g}\)` (Option C). Therefore, the correct answer is:
`\[ \text{C. 119.23 g} \]`.