Question

0.585%nacl solution at 27°C has osmotic pressure?

Answer 1

0.585% NaCl solution at 27°C has osmotic pressure of 5 atm

Explanation:

$\pi=iCRT$

$\pi$ = osmotic pressure

i= vant hoff factor = 2  (for NaCl)

C= concentration in molarity

R= solution constant = 0.0821 Latm/molK

T= temperature in Kelvin =$27^0C=(27+273)K=300K$

Given : 0.585 g of NaCl is dissolved in 100 ml of solution

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

We are given:

Mass of solute (NaCl) = 0.585 g   (conversion factor: 1 g = 1000 mg)

Molar mass of NaCl = 58.5 g/mol

Volume of solution = 100 ml =0.1 L    (1L=1000ml)

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{0.585g}{58.5g/mol\times 0.1L}\\\\\text{Molarity of solution}=0.1M$

$\pi=2\times 0.1\times 0.0821\times 300$

$\pi=5atm$

Thus 0.585% NaCl solution at 27°C has osmotic pressure of 5 atm

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