According to Raoult's law the relative lowering of vapour pressure of a solution made by dissolving non volatile solute is equal to the mole fraction of the non volatile solute dissolved.

the relative lowering of vapour pressure is the ratio of lowering of vapour pressure and vapour pressure of pure solvent

$\frac{p^{0-}p}{p^{0}}=x_{B}$

Where

xB = mole fraction of solute=?

$p^{0}=23.8torr$

p = 22.8 torr

$x_{B}=\frac{23.8-22.8}{23.8}=0.042$

mole fraction is ratio of moles of solute and total moles of solute and solvent

moles of solvent = mass / molar mass = 500 /18 = 27.78 moles

putting the values

$molefraction=\frac{molessolute}{molesolute+molessolvent}$

$0.042=\frac{molessolute}{27.78+molessolute}$

$1.167+0.042(molesolute)=molessolute$

$molessolute=1.218$

mass of glucose = moles X molar mass = 1.218 X 180 = 219.24 grams

3 0

3 years ago

Before we can use this equation for

AlexFokin [52]

Answer:

2C₂H₆ + [7]O₂ → [4]CO₂ + [6]H₂O

Explanation:

Chemical equation:

C₂H₆ + O₂ → CO₂ + H₂O

Balanced chemical equation:

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

Step 1:

2C₂H₆ + O₂ → CO₂ + H₂O

Left hand side Right hand side

C = 4 C = 1

H = 12 H = 2

O = 2 O = 3

Step 2:

2C₂H₆ + O₂ → 4CO₂ + H₂O

Left hand side Right hand side

C = 4 C = 4

H = 12 H = 2

O = 2 O = 9

Step 3:

2C₂H₆ + O₂ → 4CO₂ + 6H₂O

Left hand side Right hand side

C = 4 C = 4

H = 12 H = 12

O = 2 O = 14

Step 4:

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

Left hand side Right hand side

C = 4 C = 4

H = 12 H = 12

O = 14 O = 14

3 0

3 years ago

Uranium in the isotope below had 140 nutrons​

Uranium in the isotope below had 140 nutrons