a mixture of NaCl and sucrose that has a combined mass of 10.20g is dissolved in enough water to make up to a 250mL solution. osmotic presdure of solution is 7.32atm at 23℃. calculate the mass percent of NaCl
To calculate the mass percent of NaCl in the solution, we need to know the mass of NaCl present in the solution.
First, we need to calculate the number of moles of NaCl present in the solution using the ideal gas law. The ideal gas law can be rearranged to solve for the number of moles:
PV = nRT
Where:
P = osmotic pressure (in atm)
V = volume of solution (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
Given:
osmotic pressure (P) = 7.32 atm
volume of solution (V) = 250 mL = 0.250 L
temperature (T) = 23℃ = 23 + 273 = 296 K
Using these values, we can calculate the number of moles of NaCl:
n(NaCl) = (P * V) / (R * T)
n(NaCl) = (7.32 * 0.250) / (0.0821 * 296)
n(NaCl) = 0.0606 moles
Next, we need to calculate the mass of NaCl present in the solution.
The molar mass of NaCl is approximately 58.44 g/mol. We can use this value to convert the number of moles to grams:
mass(NaCl) = n(NaCl) * molar mass(NaCl)
mass(NaCl) = 0.0606 * 58.44
mass(NaCl) = 3.56 g
Now that we have the mass of NaCl in the solution, we can calculate the mass percent:
mass percent(NaCl) = (mass(NaCl) / mass of solution) * 100
Given:
mass of solution = 10.20 g
mass percent(NaCl) = (3.56 / 10.20) * 100
mass percent(NaCl) = 34.9%
Therefore, the mass percent of NaCl in the solution is approximately 34.9%.
I believe this is a two equation to be solved simultaneously problem. I suppose you want mass % NaCl IN THE SAMPLE.
Let x = g NaCl
and y = g sucrose
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eqn 1 is x + y = 10.20
pi = iMRT.
Substitute pi, i, R, and T and solve for M. Remember i for NaCl is 2 and sucrose is 1. Then M x 0.250 = z mols
Use mols to set up eqn 2.
(x/mm NaCl) + (y/mm sucose) = z mols. where mm is molar mass.
Solve for x and y, then convert x to weight percent.
%NaCl = (mass NaCl/10.20)*100 = ?