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2 years ago

1. How many mg of lithium phosphide are in 18.0 mL of a 1.25 M solution?

Chemistry

1 answer:

Temka [501]2 years ago

6 0

Answer:

1160mg

Explanation:

Molarity = number of moles ÷ volume

According to the information in the question, molarity = 1.25 M, volume = 18.0 mL = 18/1000 = 0.018L

M = n/V

n = M × V

n = 1.25 × 0.018

n = 0.0225moles.

Using mole = mass/molar mass, to find the mass of lithium phosphide (Li3P)

Molar mass of Li3P = 6.9(3) + 31 = 51.7g/mol

mole = mass/molar mass

0.0225 = mass/51.7

mass = 1.16grams.

In milligrams (mg), mass of Li3P = 1.16 × 1000 = 1160mg

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