Question

recall that 7.41 g of ca(oh)₂ dissociated in this experiment. calculate the number of moles dissociated. \\(\frac{7.41\text{ g }\ce{ca(oh)2}}{1}\times\frac{1\text{ mol }\ce{ca(oh)2}}{74.096\text{ g }\ce{ca(oh)2}}\\) \\(?\text{ mol }\ce{ca(oh)2}\\)

Explanation:

Step1: Multiply the given mass by the molar mass conversion factor

We have the mass of \( \text{Ca(OH)}_2 \) as \( 7.41 \, \text{g} \) and the conversion factor \( \frac{1 \, \text{mol}}{74.096 \, \text{g}} \). So we calculate \( 7.41 \times \frac{1}{74.096} \).
\( 7.41 \div 74.096 \approx 0.100 \) (rounded to three significant figures)

Answer:

\( 0.100 \) (or approximately \( 0.10 \) depending on significant figure requirements, but the calculation gives approximately \( 0.100 \) when considering the given values)