Question

when 3.20 g of a nonelectrolyte solute is dissolved in water to make 765 ml of solution at 23 °c, the solution exerts an osmotic pressure of 809 torr. what is the molar concentration of the solution? concentration: blank m how many moles of solute are in the solution? moles of solute: blank mol what is the molar mass of the solute? molar mass: blank g/mol

Explanation:

Step1: Convert units

First, convert osmotic - pressure from torr to atm and temperature from Celsius to Kelvin.
$809\ torr\times\frac{1\ atm}{760\ torr}=1.0645\ atm$
$T=(23 + 273.15)\ K=296.15\ K$

Step2: Use the osmotic - pressure formula to find molar concentration

The osmotic - pressure formula is $\Pi = MRT$, where $\Pi$ is osmotic pressure, $M$ is molar concentration, $R$ is the ideal gas constant ($R = 0.0821\ L\cdot atm/(mol\cdot K)$), and $T$ is temperature.
We can solve for $M$: $M=\frac{\Pi}{RT}$
$M=\frac{1.0645\ atm}{0.0821\ L\cdot atm/(mol\cdot K)\times296.15\ K}=0.0437\ mol/L$

Step3: Calculate moles of solute

The volume of the solution $V = 765\ mL=0.765\ L$.
The number of moles $n$ of solute is given by $n = M\times V$.
$n=0.0437\ mol/L\times0.765\ L = 0.0334\ mol$

Step4: Calculate molar mass

The mass of the solute $m = 3.20\ g$.
The molar mass $MM$ is given by $MM=\frac{m}{n}$.
$MM=\frac{3.20\ g}{0.0334\ mol}=95.8\ g/mol$

Answer:

concentration: $0.0437\ M$
moles of solute: $0.0334\ mol$
molar mass: $95.8\ g/mol$