Question

Determine the volume of a 0.035 M Li3PO4 solution that contains 0.0164 moles of Li3PO4.

A. 470 mL
B. 210 mL
C. 580 mL
D. 170 mL
E. 350 mL​

Answer 1

The volume of the Li3PO4 solution that contains 0.0164 moles of Li3PO4 is approximately 470 mL.

Step-by-step explanation:

To determine the volume of a solution, we can use the formula:

Volume = Moles / Concentration

In this case, we have 0.0164 moles of Li3PO4 and a concentration of 0.035 M. Plugging these values into the formula, we get:

Volume = 0.0164 / 0.035 = 0.468 mL

Therefore, the volume of the Li3PO4 solution is approximately 470 mL.

Answer 2

The correct option is:

A. 470 mL

Step-by-step explanation:

Molarity of the solution = 0.035M

`Molarity=(Number\:of\:moles)/(Volume\:of\:solution\:in\:L)\\\\ Number\:of\:moles\:of\:Li_(3)PO_(4)=0.0164\\\\Volume\:of\:solution=(number\:of\:moles)/(Molarity)=(0.0164)/(0.035)=0.46857L=468.57mL`

It is approximately equal to 470 mL.

Hence, 470 mL of 0.035 M
`Li_(3)PO_(4)` solution contains 0.0164 moles of
`Li_(3)PO_(4)`.