Question

What would be osmotic pressure of a solution containing 5.85g NaCl and 3.42g sugar in 500ml at 27°C?

Answer 1

Answer:

Explanation:

Total pressure p =p1 + p2

Where p1  consider it for nacl

P1= 5.85×1000×0.082×300×2÷500×58 .5

We get9.84

Now for sugar

P2= 3.42×1000×0.082×1×300÷500×342

We get 0.492

Summing up p1+p2 we get 10.34atm

Answer 2

Answer: The osmotic pressure of a solution containing 5.85 g NaCl and 3.42g sugar in 500 ml at 27°C is 10.8 atm

Explanation:

Osmotic pressure is a colligative property which depends on the amount of solute added.

$\pi=iCRT$

$\pi$= osmotic pressure

i= vant hoff factor = 2 (for NaCl)

C= concentration in Molarity

R= solution constant = 0.0821 Latm/Kmol

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

a) for 3.42 g sugar in 500 ml

$moles of solute =\frac{\text {given mass}}{\text {molar mass}}=\frac{3.42g}{342g/mol}=0.01moles$

b) for 5.85 g NaCl in 500 ml

$moles of solute =\frac{\text {given mass}}{\text {molar mass}}=\frac{5.85g}{58.5g/mol}=0.1moles$

Total moles = 0.01 +0.1= 0.11

Now put all the given values in the formula:

$Molarity=\frac{0.11moles\times 1000}{500ml}=0.22mole/L$

$\pi=iCRT$

$\pi=2\times 0.22\times 0.0821\times 300=10.8atm$

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