If the vapor's volume were to be incorrectly recorded as 125ml, how will this error affect the calculated molar mass of the unkn
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The answer for the following problem has been mentioned below.
- <em><u>Therefore the mass of the water is 5.802 grams.</u></em>
Explanation:
Given:
volume of oxygen (V) = 4.50 L
Temperature (T) = 425 K
pressure of oxygen (P) = 2.50 atm
Gram molecular mass of oxygen (M) = 16.0 grams
To calculate:
mass of water (m)
We know;
According to the ideal gas equation;
P × V = n × R × T
As we know;
no of moles =
m represents the mass of oxygen (m)
M represents the Gram molecular mass (M)
According to above mentioned equation;
P × V = n × R × T
P represents the pressure of the oxygen
V represents the volume of the oxygen
n represents the no of moles of the oxygen
R represents the universal gas constant
where,
the value of R is 0.0821 L atm/K moles
Substituting the values in the above equation;
2.50 × 4.50 = × 0.0821 × 425
11.25 = × 34.8925
180 = m × 34.8925
m =
m = 5.158 grams
Therefore the mass of the of oxygen is 5.158 grams
Now;
As we know;
where;
represents the mass of the oxygen
represents the gram molecular mass of the oxygen
represents the mass of the water
represents the gram molecular mass of water
From the above given formula,
=
where;
Gram molecular weight of water = 18.0 u
= 5.802 grams
<em><u>Therefore the mass of the water is 5.802 grams.</u></em>
4.648 gm of solute is needed to make 37.5 mL of 0.750 M KI solution.
Solution:
We will start with the Molarity
Also we know 1000 ml = 1 L
Therefore 37.5 ml by 1000ml we obtained 0.0375L
Equation for solving mole of solute
Now, multiply 0.750M by 0.0375
Substitute the known values in the above equation we get
Also we know that Molar mass of KI is 166 g/mol
So divide the molar mass value to get the no of grams.
So 4.648 gm of Solute is required for make 37.5 mL of 0.750 M KI solution.