Question

A solution containing 0.102 g of an unknown compound dissolved in 100. mL of water has an osmotic pressure of 28.1 mmHg at 20.°C. What is the molar mass of the compound? [R = 0.08206 L • atm/K • mol, 1 atm = 760 mmHg]

Answer 1

Answer:

The molar mass is 680 $\frac{g}{moles}$

Explanation:

The molar mass is the mass of one mole of a substance, which can be an element or a compound. So the molar mass can be calculated as:

$molar mass=\frac{mass}{number of moles}$

Then you must know the number of moles n.

Osmosis is the passage of the solvent from an already diluted solution to another with a higher concentration, through a semipermeable membrane.

Osmotic pressure is the pressure that must be exerted on the solution to prevent the entry of the solvent. The higher the osmotic pressure, the greater the tendency of the solvent to enter the solution.

The osmotic pressure depends on the concentration in mol / L of the total number of dispersed particles of the solute (M) and the temperature in kelvin of the solution (T).

op = M R T

where R is the universal constant of ideal gases.

In this case, it is necessary to calculate the molarity M, which is:

$M=\frac{op}{R*T}$

where:

• op = osmotic pressure = 28.1 mmHg = 0.037atm (1 atm=760mmHg)

• R=0.08206 $\frac{L*atm}{mol*K}$

• T=20 °C= 293°K (0°C=273 °K)

Replacing:

$M=\frac{0.037 atm}{0.08206\frac{L*atm}{mol*K}*293 K }$

Solving:

M=0.0015 $\frac{mol}{L}$

This indicates that in 1 L of solution there are 0.0015 moles. Applying rule of three, it is possible to calculate how many moles are in 100 mL = 0.1 L (1 L = 1000 mL)

In 100 mL there are 0.00015 moles.

Finally, the molar mass can be calculated as:

$molar mass=\frac{0.102 g}{0.00015 moles}$

$molar mass=680 \frac{g}{moles}$

The molar mass is 680 $\frac{g}{moles}$

Answer 2

Answer:

MM = 680g / mol

Explanation:

Hello! To calculate the molar mass of the compound (g / mol), I first have to calculate the molarity.

Molarity can be calculated from the osmotic pressure equation.

op = M * R * T

op = osmotic pressure = 28.1mmHg * (1 atm / 760mmHg) = 0.037atm

M = molarity

R = gas constant

T = temperature (K) = 20 ° C + 273.15 = 293.15K

M (mol / L) = op / R * T

M = 0.037atm / ((0.082 (atm * L) / (K * mol)) * 293.15K) = 0.0015mol / L

As I have the volume = 100ml * (1L / 1000ml) = 0.1L

I can calculate the amount of moles

n = M * V = 0.0015 * 0.1 = 0.00015mol

n = m / MM

m = mass

MM = molar mass

MM = m / n = 0.102g / 0.00015mol

MM = 680g / mol