What mass of CH3COOH is present in a 250 mL cup of 1.25 mol/L solution of vinegar?
1 answer:
If 1000 ml (1 L) of CH₃COOH contain 1.25 mol
let 250 ml of CH₃COOH contain x
⇒ x =
= 0.3125 mol
∴ moles of CH₃COOH in 250ml is 0.3125 mol
Now, Mass = mole × molar mass
= 0.3125 mol × [(12 × 2)+(16 × 2)+(1 × 4)] g/mol
= 18.75 g
∴ Mass of CH₃COOH present in a 250 mL cup of 1.25 mol/L solution of vinegar is <span>18.75 g</span>
let 250 ml of CH₃COOH contain x
⇒ x =
= 0.3125 mol
∴ moles of CH₃COOH in 250ml is 0.3125 mol
Now, Mass = mole × molar mass
= 0.3125 mol × [(12 × 2)+(16 × 2)+(1 × 4)] g/mol
= 18.75 g
∴ Mass of CH₃COOH present in a 250 mL cup of 1.25 mol/L solution of vinegar is <span>18.75 g</span>
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The value is
Explanation:
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The initial volume of the fluorocarbon gas is
The final volume of the fluorocarbon gas is
The initial temperature of the fluorocarbon gas is
The final temperature of the fluorocarbon gas is
The initial pressure is
The final pressure is
Generally the equation for adiabatically reversible expansion is mathematically represented as
Here R is the ideal gas constant with the value
So
=>
Generally adiabatic reversible expansion can also be mathematically expressed as
=>
=>
=>
So
=>