Answer:
12.2329 atm
Explanation:
Applying
PV = nRT.................... Equation 1
Make P the subject of the equation
P = (n/V)RT................ Equation 2
Where P = Osmotic pressure, (n/V) = Molarity of urea, T = Temperature, R = molar gas constant.
From the question,
Given: (n/V) = 0.5 M, T = 25°C = (273+25) = 298 K, R = 0.0821 Latm/mol.K
Substitute these values into equation 2
P = (0.5)(298)(0.0821)
P = 12.2329 atm
Answer:
Your question is not complete, but use this answer as a guide for your solution.
Question: A chemistry student weighs out 0.112g of acetic acid (HCH₃CO₂) into a 250. mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1600 <em>M</em> NaOH solution. Calculate the volume of solution the student will need to add to reach the equivalence point. Be sure your answer has the correct number of significant digits.
Answer: Volume of NaOH is 11.6 mL
Explanation:
The reaction of acetic acid with NaOH is as follows:
CH3COOH + NaOH -----> CH3COONa + H2O
M1V1 = M2V2
Here M1 V1 are molarity and volume of acetic acid.
M2, V2 are molarity and volume of NaOH.
Number of moles of acetic acid:
0.112 g CH3COOH × (1 mol / 60.05 g) = 0.001865 mol
Molarity = moles of solute / Liters of solution
Molarity = 0.001865 mol / 0.250 L = 0.00746 M
Hence,
M1 = 0.00746 M
V1 = 250 mL
M2 = 0.160 M
V2 = ?
V2 = M1V1 / M2
V2 = 0.00746 M × 250 mL / 0.160 M
V2 = 11.6 mL
Hence the volume of NaOH is 11.6 mL
