How would you prepare 10L of 0.3M KH2 PO4? FW of KH2 PO4= 136.09g
Before you make the solution, wouldn't you find how much solute there is first. So by using the M=moles of solute / V ?
Then, you would add that amount of solute to the L of solution.
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Also, for question 2, on how to make 500 ml of a 2M sol'n of NaOH?
-Wouldn't you have to take 2 M of the NaOH solution and then dissolve however many (g) of NaOH in 500 ml of water?
So, 500 ml/1,000 = 0.5 L
2M = moles of solute/ 0.5 L solution = to get 1 mol of solute
How would you explain that qualitatively?
Yes, you're absolutely right! In order to prepare a solution of 0.3M KH2PO4, we first need to determine the amount of solute required.
To do this, we'll use the equation: Molarity (M) = Moles of solute / Volume of solution (L)
Given the molar mass (FW) of KH2PO4 as 136.09 g/mol, we can calculate the moles of KH2PO4 as follows:
Moles of KH2PO4 = Molarity (M) × Volume of solution (L)
Substituting the given values, we can rearrange the equation to solve for the required moles of KH2PO4:
Moles of KH2PO4 = 0.3 mol/L × 10 L = 3 moles
So, you would need 3 moles of KH2PO4 to prepare a 10L solution of 0.3M KH2PO4.
Now, let's calculate the mass of KH2PO4 required:
Mass of KH2PO4 = Moles of KH2PO4 × Molar mass (FW)
Substituting the given values:
Mass of KH2PO4 = 3 moles × 136.09 g/mol = 408.27 g
Therefore, you would need 408.27 grams of KH2PO4 to prepare a 10L solution of 0.3M KH2PO4.
Once you have obtained the calculated amount of KH2PO4, you can proceed to dissolving it in a suitable solvent such as water to prepare the desired solution.