Question

A 275 mg sample of a nonelectrolyte compound was dissolved in water to produce 10.0 mL of a solution at 25°C. The osmotic pressure of this solution was measured and found to be 3.67 atm. What is the molar mass of this compound? O 183 g/mol 18.3 g/mol 101 g/mol 1560 g/mol 1830 g/mol Question 25 (1 point) Which statement below is true? O Nonpolar molecules can dissolve in polar solvents. Gases are soluble only in nonpolar solvents. lonic compounds cannot dissolve in any solvent other than water. Polar molecules cannot dissolve in nonpolar solvents. Solubility depends only slightly on the strength of solute-solute attractive forces.

Answer 1

1)

Answer

183g/mol

Explanation

Osmotic pressure($\prod$) is calculated by the following formula

$\prod$= iMRT

where,

i = Van^'t Hoff factor , for nonelectrolyte it is 1

R = gas constant , 0.082057( L atm /mol K)

M = molarity of the solute

T = Temperature , 298.15K

M = $\prod$/iRT

= 3.67atm / (1 × 0.082057(L atm/mol K) × 298.15K)

= 0.1500M

Molarity = number of moles of solute per liter of solution

0.1500 M = 0.1500moles of solute per liter of solution

Number of moles for 10.00ml = (0.1500mol/1000ml) ×10ml = 0.0015mol

Molar mass = mass / number of moles

molar mass of the compound = 0.275g/0.0015mol = 183g/mol

ii)

Polar molecules cannot dissolve in nonpolar solvents