An aqueous 0.300 M glucose solution is prepared with a total volume of 0.150 L. The molecular weight of
1 answer:
Answer:
8.11g
Explanation:
Given parameters:
Molarity of aqueous solution = 0.3M
total volume = 0.15L
Molecular weight of glucose = 180.6g/mol
Unknown:
Mass of glucose needed in the solution = ?
Solution:
To solve this problem, we need to understand molarity.
Molarity is the number of moles of solute in a given volume of solution. In this problem, the solute here is the glucose and the solvent is water.
Molarity =
A solution is made up of solute and solvent.
now, let us solve for the number of moles of the solute which is glucose;
Number of moles of glucose = molarity x volume of solution;
= 0.3 x 0.15
= 0.045mole
Now to find the mass of glucose;
mass of glucose = number of moles x molar mass
input the parameters;
Mass of glucose = 0.045 x 180.16 = 8.11g
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Given parameters:
Initial volume = 120ml
Initial temperature = 35°C
Initial pressure = 1.2bar
Final volume = 180ml
Final temperature = 35°C
Unknown:
Final pressure = ?
To solve this problem, we apply the combined gas law. The expression is given below;
Where P₁ is the initial pressure
P₂ is the final pressure
V₁ is the initial volume
V₂ is the final volume
T₁ is the initial temperature
T₂ is the final temperature
We need to convert the parameters to standard units
take the volume to dm³;
1000ml = 1dm³
120ml = dm³ = 0.12dm³ = initial volume
Final volume;
1000ml = 1dm³
180ml = dm³ = 0.18dm³
Now, the temperature;
K = 273 + °C
Initial temperature = 273 + 35 = 308k
Final temperature = 308k
We then input the parameters into the equation;
Solving for P₂;
P₂ = 0.8bar
The new pressure or final pressure in the vessel is 0.8bar