Question

A chemist prepares a solution of silver(I) nitrate AgNO3 by measuring out 62.3μmol of silver(I) nitrate into a 50.mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in /molL of the chemist's silver(I) nitrate solution. Be sure your answer has the correct number of significant digit

*please write the answer without any files disturbance*

Answer 1

0.0012 mol/L.

Step-by-step explanation:

From the question given above, the following data were obtained:

Number of mole of AgNO₃ = 62.3 μmol

Volume = 50 mL

Molarity of AgNO₃ =?

Next, we shall convert 62.3 μmol to mole. This can be obtained as follow as follow:

1 μmol = 10¯⁶ mole

Therefore,

62.3 μmol = 62.3 × 10¯⁶

62.3 μmol = 62.3×10¯⁶ mole

Next, we shall convert 50 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

50 mL = 50 mL × 1 L / 1000 mL

50 mL = 0.05 L

Finally, we shall determine the concentration of AgNO₃ in mol/L as follow:

Number of mole of AgNO₃ = 62.3×10¯⁶ mole

Volume = 0.05 L

Molarity of AgNO₃ =?

Molarity = mole /Volume

Molarity of AgNO₃ = 62.3×10¯⁶ / 0.05

Molarity of AgNO₃ = 0.0012 mol/L

Thus, the concentration of AgNO₃ is 0.0012 mol/L