Answer:
6.85 g
Explanation:
Step 1: Given data
Step 2: Calculate the moles of sucrose (solute) required
Molarity is equal to the moles of solute divided by the liters of solution.
M = moles of solute / liters of solution
moles of solute = M × liters of solution
moles of solute = 0.200 mol/L × 0.10000 L = 0.0200 mol
Step 3: Calculate the mass corresponding to 0.0200 moles of sucrose
The molar mass of sucrose is 342.3 g/mol.
0.0200 mol × 342.3 g/mol = 6.85 g
The amount of sucrose (in grams) must be weighed out for the given reaction is 6.846g.
Mass of any substance will be calculated by using the moles as:
n = W/M, where
W = required mass
M = molar mass
Given that, molarity of sucrose = 0.2 M
Volume of solution = 100mL = 0.1 L
Relation between moles and molarity is represented as:
M = n/V
On putting values on the above equation we get,
n = (0.2)(0.1) = 0.02 moles
We know that molar mass of sucrose (C₁₂H₂₂O₁₁) = 342.3 g/mole
Now we calculate the required mass by putting values on the first equation as:
W = (0.02)(342.3) = 6.846g
Hence, the required mass of sucrose is 6.846g.
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https://brainly.com/question/24639749
