Calculate the number of moles of Li3PO4 in 2.2 L of a 0.60 M Li3PO4 solution.

1.32 mole

Explanation:

The following data were obtained from the question:

Volume of solution = 2.2L

Molarity of solution = 0.60M

Mole of Li3PO4 =..?

Molarity is simply defined as the mole of solute per unit litre of the solution. Mathematically, it is represented as:

Molarity = mole /Volume

With the above formula we can easily calculate the number of mole of Li3PO4 as shown below:

Molarity =mole /Volume

0.6 = mole of Li3PO4 /2.2

Cross multiply

Mole of Li3PO4 = 0.6 x 2.2

Mole of Li3PO4 = 1.32 mole

Therefore, 1.32 mole of Li3PO4 is contained in the solution.

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a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.