## Chemistry 9th Edition

The amount of solid of NaOH we require to form 1.00 L of 1.00 M NaOH solution is: $$1.00 \text{ L} *\frac{1.00 \text{ mol NaOH}}{1.00 \text{ L}}*\frac{40.00 \text{ grams NaOH}}{1\text{ mol NaOH}}=40.00 \text{ grams NaOH}$$ We then add 1 liter of water to these 40.00 grams to create 1.00 L solution of 1.00 M NaOH. We apply the formula: $M_1V_1=M_2V_2$ to calculate the volume of 2.00 M NaOH solution we require: $2.00*V=1.00*1.00\Rightarrow V=0.50$ L. Therefore, we need to combine 0.50 L of 2.00 M NaOH solution with 0.50 L of water to create 1.00 L of 1.00 M solution. By the multiplicative and division rule concerning significant figures when we take the lowest number of significant figures of the factors, both the volume and mass should be determined with at least three significant figures