Question

how many moles of nacl are in 120 g nacl? assume 1.0 mol nacl = 58 g
\\( \frac { 1.0 \mathrm { mol } } { 58 \mathrm { g } } = \frac { \square \mathrm { mol } } { 120 \mathrm { g } } \\)
6.02×10²³mol
2.1mol
0.50mol

Explanation:

Step1: Set up the proportion

Let \( x = \) moles of NaCl. The proportion is \(\frac{1.0\ \text{mol}}{58\ \text{g}} = \frac{x}{120\ \text{g}}\)

Step2: Solve for x

Rearrange to \( x = \frac{1.0\ \text{mol} \times 120\ \text{g}}{58\ \text{g}} \approx 2.1\ \text{mol} \)

Answer:

2.1 mol