Question

At a certain temperature, the vapour pressure of water is 50 mm .The relative lowering of vapour pressure of a solution containing 36 g of glucose in 900 g of water isa) 0.1b) 0.5c) 0,004d) 3.75​

Answer 1

The relative lowering in vapor pressure of the solution is 0.004

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

$\frac{p^o-p_s}{p^o}=i\times x_2$

where,

$\frac{p^o-p_s}{p^o}$= relative lowering in vapor pressure

i = Van'T Hoff factor = 1 (for non electrolytes)

$x_2$ = mole fraction of solute

=$\frac{\text {moles of solute}}{\text {total moles}}$

Given : 36g of glucose is present in 900 g of water

moles of solute (glucose) = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{36g}{180g/mol}=0.2moles$

moles of solvent (water) = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{900g}{18g/mol}=50moles$

Total moles = moles of solute (glucose)  + moles of solvent (water) = 0.2 + 50 = 50.2

$x_2$ = mole fraction of solute =$\frac{0.2}{50.2}=0.004$

$\frac{p_0-p_s}{p_0}=1\times 0.004$

$\frac{p_0-p_s}{p_0}=0.004$

Thus the relative lowering in vapor pressure of the solution is 0.004

Learn More about lowering in vapor pressure

https://brainly.in/question/10016774

https://brainly.com/question/12941286