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If 1000 ml (1 L) of CH₃COOH contain 1.25 mol
let 250 ml of CH₃COOH contain x
⇒ x =
= 0.3125 mol
∴ moles of CH₃COOH in 250ml is 0.3125 mol
Now, Mass = mole × molar mass
= 0.3125 mol × [(12 × 2)+(16 × 2)+(1 × 4)] g/mol
= 18.75 g
∴ Mass of CH₃COOH present in a 250 mL cup of 1.25 mol/L solution of vinegar is <span>18.75 g</span>
let 250 ml of CH₃COOH contain x
⇒ x =
= 0.3125 mol
∴ moles of CH₃COOH in 250ml is 0.3125 mol
Now, Mass = mole × molar mass
= 0.3125 mol × [(12 × 2)+(16 × 2)+(1 × 4)] g/mol
= 18.75 g
∴ Mass of CH₃COOH present in a 250 mL cup of 1.25 mol/L solution of vinegar is <span>18.75 g</span>
The question requires us to calculate the mass of NaNO3 necessary to prepare 506 mL of a 10.5 M solution of this salt.
To answer this question, we'll need to go through the following steps:
1) Calculate the molar mass of NaNO3, as it will be necessary during the question;
2) Calculate the number of moles necessary to prepare 506 mL of the 10.5 M solution;
3) Convert the number of moles calculated into the mass of NaNO3, using the molar mass of this salt.
Next, we'll follow the steps:
1) To calculate the molar mass of NaNO3, we need to consider the atomic masses of Na, N and O and the number of atoms of each of these elements. The atomic masses are 22.99 u, 14.01 u and 15.99 u for Na, N and O, respectively. With this information, we can calculate the molar mass:
molar mass of NaNO3 = (1 * 22.99) + (1 * 14.01) + (3 * 15.99) = 84.97 g/mol
2) Next, we use the molar concentration provided (10.5 M or 10.5 mol/L) and the volume required (506 mL = 0.506 L) to obtain the number of moles necessary to prepare this solution:
<em>1 L of solution --------------- 10.5 mol of NaNO3</em>
<em>0.506 L of solution -------- x</em>
Solving for x, we have:
Therefore, the number of moles of NaNO3 necessary to prepare the solution is 5.31 moles.
3) At last, we use the molar mass obtained in step 1 (84.97 g/mol) to calculate the mass that corresponds to 5.31 moles of NaNO3:
<em>1 mol of NaNO3 ---------------- 84.97 g</em>
<em>5.31 mol of NaNO3 ----------- y</em>
Solving for y, we have:
Therefore, the mass of NaNO3 necessary to prepare the solution given by the question is 451.2 g.