How many grams of LiOH are required to prepare 100 mL of a 0.1 M solution?

• A. 0.40 g
• B. 0.07 g
• C. 1.00 g
• D. 0.17 g

Answer: (B)

Verified step by step guidance

1. Identify the given information: volume of solution (V) = 100 mL and molarity (M) = 0.1 M. Convert the volume from milliliters to liters because molarity is expressed in moles per liter. Use the conversion: \(V (L) = \frac{100 \text{ mL}}{1000}\).
2. Use the molarity definition formula to find the number of moles of LiOH needed: \(M = \frac{\text{moles of solute}}{\text{liters of solution}}\). Rearrange to find moles: \(\text{moles} = M \times V\).
3. Calculate the molar mass of LiOH by adding the atomic masses of lithium (Li), oxygen (O), and hydrogen (H). Use the periodic table values: Li ≈ 6.94 g/mol, O ≈ 16.00 g/mol, H ≈ 1.01 g/mol. Then, \(\text{molar mass of LiOH} = 6.94 + 16.00 + 1.01\).
4. Convert moles of LiOH to grams using the molar mass: \(\text{mass} = \text{moles} \times \text{molar mass}\).
5. The resulting mass from the calculation will give the grams of LiOH required to prepare 100 mL of a 0.1 M solution.