Answer :
1. The molarity of 37.1 g of table sugar [tex](\(C_{12}H_{22}O_{11}\))[/tex] in 100 mL of solution is 1.08 M.
2. The molarity of 8.25 mL of 0.0521 M KI solution diluted to 12.0 mL with water is 0.0367 M.
The molarity of 37.1 g of table sugar [tex](\(C_{12}H_{22}O_{11}\))[/tex] in 100 mL of solution is calculated as follows:
[tex]\[\text{{Molarity}} = \frac{{\text{{moles of table sugar}}}}{{\text{{volume of solution}}}}\][/tex]
To determine the moles of table sugar, we divide the given mass by the molar mass of [tex]\(C_{12}H_{22}O_{11}\)[/tex], which is 342.3 g/mol:
[tex]\[\text{{moles of table sugar}} = \frac{{37.1 \, \text{{g}}}}{{342.3 \, \text{{g/mol}}}} = 0.1083 \, \text{{mol}}\][/tex]
The volume of the solution is given as 100 mL, which is equivalent to 0.1 L. Plugging the values into the molarity formula:
[tex]\[\text{{Molarity}} = \frac{{0.1083 \, \text{{mol}}}}{{0.1 \, \text{{L}}}} = 1.083 \, \text{{M}}\][/tex]
Rounding the answer to three significant figures gives us the final molarity:
[tex]\[\text{{Molarity}} = 1.08 \, \text{{M}}\][/tex]
2. The molarity of 8.25 mL of 0.0521 M KI solution diluted to 12.0 mL with water is calculated as follows:
[tex]\[\text{{Molarity}} = \frac{{\text{{moles of KI}}}}{{\text{{volume of solution}}}}\][/tex]
The moles of KI can be calculated by multiplying the molarity by the volume in liters:
[tex]\[\text{{moles of KI}} = 0.0521 \, \text{{M}} \times \left(\frac{{8.25 \, \text{{mL}}}}{{1000 \, \text{{mL/L}}}}\right) = 0.000430 \, \text{{M}}\][/tex]
The final volume of the solution is given as 12.0 mL, which is equivalent to 0.012 L. Plugging the values into the molarity formula:
[tex]\[\text{{Molarity}} = \frac{{0.000430 \, \text{{M}}}}{{0.012 \, \text{{L}}}} = 0.0358 \, \text{{M}}\][/tex]
Rounding the answer to three significant figures gives us the final molarity:
[tex]\[\text{{Molarity}} = 0.0367 \, \text{{M}}\][/tex]
To know more about molarity, refer:
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