Question

Calculate the volume in milliliters of a 5.11/mol silver nitrate solution that contains 175.g of silver nitrate AgNO3. Round your answer to 3 significant digits.

Answer 1

Answer:

$V=202mL$

Explanation:

Hello!

In this case, considering we have a 5.11-molar solution (5.11 M) that contains 175. g of silver nitrate, we first need to compute the moles of solute as its molar mass is 169.87 g/mol:

$n=175g*\frac{1mol}{169.87 g} =1.03mol$

Next, since the definition of molarity is moles over volume:

$M=\frac{n}{V}$

We solve for the volume and plug in the molarity and moles as shown below:

$V=\frac{n}{M} =\frac{1.03mol}{5.11mol/L}\\\\V =0.202L$

However, as it is needed in milliliters, we convert the L to mL:

$V=0.202L*\frac{1000mL}{1L}\\\\V=202mL$

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