Difficulty: Easy

Average Score: 100%

Which of the following solutions below has a molarity of 1 M?

Rationale

53.6 g of NH4Cl in 1 L of water has a molarity of 1 M.

To determine the molarity of a solution, the formula used is \( \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \). Given that NH4Cl has a molar mass of 53.6 g/mol, dissolving 53.6 g in 1 L results in exactly 1 mole of solute, yielding a molarity of 1 M.

A (58.4 g of NaCl in 100 L of water. (Molar mass of NaCl = 58.4 g/mol)) YOUR ANSWER

To calculate the molarity, we first determine the number of moles of NaCl: \( \frac{58.4 \text{ g}}{58.4 \text{ g/mol}} = 1 \text{ mole} \). Dividing by the volume of 100 L gives \( \frac{1 \text{ mole}}{100 \text{ L}} = 0.01 \text{ M} \). This concentration is significantly lower than 1 M.

C (100 g of CaCO3 in 10 L of water. (Molar mass of CaCO3 = 100 g/mol)) YOUR ANSWER

Calculating moles of CaCO3 yields \( \frac{100 \text{ g}}{100 \text{ g/mol}} = 1 \text{ mole} \). Dividing this by 10 L results in \( \frac{1 \text{ mole}}{10 \text{ L}} = 0.1 \text{ M} \), which is also much less than 1 M.

D (180 g of C6H12O6 in 0.1 L of water. (Molar mass of C6H12O6 = 180 g/mol)) YOUR ANSWER

For C6H12O6, the number of moles is \( \frac{180 \text{ g}}{180 \text{ g/mol}} = 1 \text{ mole} \). However, dividing by 0.1 L gives \( \frac{1 \text{ mole}}{0.1 \text{ L}} = 10 \text{ M} \), which is significantly higher than 1 M.

Conclusion

Among the given options, only 53.6 g of NH4Cl in 1 L of water achieves a molarity of 1 M, confirming that it contains precisely one mole of solute per liter of solution. The other choices either yield concentrations far below or above 1 M, demonstrating the importance of accurately calculating molarity based on solute mass and solution volume.

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