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47.4 grams of calcium carbonate (CaCO3) is dissolved in 450 mL of water. What is the molarity of this aqueous solution?

Sagot :

Sebassandinidn

Answer:

[tex]M=1.05M[/tex]

Explanation:

Hello!

In this case, since the formula for the calculation of molarity is defined in terms of moles and volume in liters as shown below:

[tex]M=\frac{n}{V}[/tex]

Whereas the moles are computed by considering the molar mass of CaCO3 (100.09 g/mol):

[tex]n=47.4g*\frac{1mol}{100.09g}=0.4736mol[/tex]

Thus, we obtain:

[tex]M=\frac{0.4736mol}{0.450L}\\\\M=1.05M[/tex]

Best regards!

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