Calculate the mass percent by volume of 330.1 g of glucose (C₆H₁₂O₆, MM = 180.2 g/mol) in 325 mL of solution.
1 answer:
Answer: The mass percent by volume is 101.6%
Explanation:
The solution concentration expressed in percent by volume means that the amount of solute present in 100 parts volume of solution.
It is represented in formula as :
mass percent by volume =
Given : mass of glucose = 330.1 g
volume of solution = 325 ml
Thus mass percent by volume =
Thus the mass percent by volume is 101.6%
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Answer:
0.2g
Explanation:
Given parameters:
Number of moles of H₂ = 0.4mol
Number of moles of O₂ = 0.15mol
Unknown:
Mass of reactant that would remain = ?
Solution:
To solve this problem, we need to know the limiting reactant which is the one in short supply in the given reaction.
The expression of the reaction is :
2H₂ + O₂ → 2H₂O
2 mole of H₂ will combine with 1 mole of O₂
But given; 0.4 mole of H₂ we will require = 0.2mole of O₂
The given number of oxygen gas is 0.15mole and it is the limiting reactant.
Hydrogen gas is in excess;
1 mole of oxygen gas will combine with 2 mole of hydrogen gas
0.15 mole of oxygen gas will require 0.15 x 2 = 0.3mole of hydrogen gas
Now, the excess mole of hydrogen gas = 0.4 mole - 0.3 mole = 0.1mole
Mass of hydrogen gas = number of mole x molar mass
Molar mass of hydrogen gas = 2(1) = 2g/mol
Mass of hydrogen gas = 0.1 x 2 = 0.2g