Chapter 13: Problem 55

How many milliliters of 1.0 \(\mathrm{M} \mathrm{AgNO}_{3}\) are needed to provide 168.88 \(\mathrm{g}\) of pure \(\mathrm{AgNO}_{3} ?\)

Short Answer

Expert verified

You would need 994 mL of 1.0 \(\mathrm{M} \mathrm{AgNO}_{3}\) to provide 168.88 g of pure \(\mathrm{AgNO}_{3}\).

Step by step solution

Determine moles of AgNO3

Find the molar mass of AgNO3 by adding the atomic masses of silver (Ag), nitrogen (N) and three atoms of oxygen (O). Then, apply the formula for moles, which is given mass divided by molar mass. So, the moles of AgNO3 would be \(\frac{168.88 \mathrm{g}}{169.87 \mathrm{g/mol}} = 0.994 \mathrm{mol}.\)

Calculate the volume needed

Use the formula for molarity \(M = \frac{n}{V} \) to find the volume (V). Rearrange the formula to find V and then substitute the values n = 0.994 mol and M = 1.0 M. So, the volume in liters would be \(V = \frac{0.994 \mathrm{mol}}{1.0 \mathrm{M}} = 0.994 \mathrm{L}\).

Convert volume to milliliters

Since 1 L equals 1000 mL, you need to multiply the calculated volume by 1000 to convert it to milliliters. Therefore, the required volume of 1.0 \(\mathrm{M} \mathrm{AgNO}_{3}\) in milliliters is \(0.994 \mathrm{L} \times 1000 = 994 \mathrm{mL}\).

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How many milliliters of 1.0 \(\mathrm{M} \mathrm{AgNO}_{3}\) are needed to provide 168.88 \(\mathrm{g}\) of pure \(\mathrm{AgNO}_{3} ?\)

Short Answer

Step by step solution

Determine moles of AgNO3

Calculate the volume needed

Convert volume to milliliters

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