Some of the reactants or the products are in the gaseous phase.
According to Raoult's law the relative lowering of vapour pressure of a solution made by dissolving non volatile solute is equal to the mole fraction of the non volatile solute dissolved.
the relative lowering of vapour pressure is the ratio of lowering of vapour pressure and vapour pressure of pure solvent
Where
xB = mole fraction of solute=?
p = 22.8 torr
mole fraction is ratio of moles of solute and total moles of solute and solvent
moles of solvent = mass / molar mass = 500 /18 = 27.78 moles
putting the values
mass of glucose = moles X molar mass = 1.218 X 180 = 219.24 grams
Answer:
b)4.46 L/hr
Explanation:
To solve this question we need to convert the mL to liters (Using the conversion of 1000mL = 1L) and convert the time from seconds to hours (3600s = 1hr)
<em>mL to L:</em>
1.24mL/s * (1L / 1000mL) = 0.00124L/s
<em>seconds to hours:</em>
0.00124L/s * (3600s / 1hr) = 4.46L/hr
Right answer is:
<h3>b)4.46 L/hr
</h3>
Here you go! Feel free to ask questions!
2.2311 moles of gas are there in a 50. 0 l container at 22. 0 °c and 825 torrs.
<h3>What is an ideal gas?</h3>
An Ideal gas is a hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas laws exactly.
Assuming the gas is ideal, we can solve this problem by using the following equation:
PV = nRT
Where:
P = 825 torr ⇒ 825 / 760 = 1.08 atm
V = 50 L
n = ?
R = 0.082 atm·L·mol⁻¹·K⁻¹
T = 22 °C ⇒ 22 + 273.16 = 295.16 K
We input the data:
1.08 atm x 50 L = n x 0.082 atm·L·mol⁻¹·K⁻¹ x 295.16 K
And solve for n:
24.20312
n = 2.2311 mol
Hence, 2.2311 moles of gas are there in a 50. 0 L container at 22. 0 °c and 825 torrs.
Learn more about ideal gas here:
brainly.com/question/23580857
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