how many grams of solid NaOH are required to prepare 200 ml of a .05 M solution?
mass=MolmassNaOH*Molarity*Volumeinliters
Memorize that relationship.
thanks
To calculate the grams of solid NaOH required to prepare a given volume and concentration of solution, you need to use the equation:
\( \text{moles of NaOH} = \text{volume (in liters)} \times \text{concentration (in moles per liter)} \)
Let's break down the calculation step-by-step:
1. Convert the given volume from milliliters (ml) to liters (L):
\( 200 \text{ ml} = 200 \times 10^{-3} \text{ L} = 0.2 \text{ L} \)
2. Plug the volume and concentration values into the equation to find the moles of NaOH:
\( \text{moles of NaOH} = 0.2 \text{ L} \times 0.05 \text{ M} = 0.01 \text{ moles} \)
3. Calculate the molar mass of NaOH:
Sodium (Na) has a molar mass of approximately 22.99 g/mol, Oxygen (O) has a molar mass of approximately 16.00 g/mol, and Hydrogen (H) has a molar mass of approximately 1.01 g/mol. Add these values together:
\( \text{NaOH molar mass} = 22.99 \text{ g/mol} + 16.00 \text{ g/mol} + 1.01 \text{ g/mol} = 39.99 \text{ g/mol} \)
4. Calculate the grams of NaOH required using the moles of NaOH and the molar mass of NaOH:
\( \text{grams of NaOH} = \text{moles of NaOH} \times \text{NaOH molar mass} \)
Substituting the values:
\( \text{grams of NaOH} = 0.01 \text{ moles} \times 39.99 \text{ g/mol} = 0.3999 \text{ g} \)
Thus, you would require approximately 0.3999 grams of solid NaOH to prepare a 200 ml solution with a concentration of 0.05 M.