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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Explanation:
Assuming that the final (equilibrium) temperature of the system is above the melting point of ice, such that all ice in the container melts in this process thus
and
Let the final temperature of the system be . Thus
(converted to kilojoules)
-
The fact that energy within this system (assuming proper insulation) conserves allows for the construction of an equation about variable .
Confirm the uniformity of units, equate the two expressions and solve for :
which goes against the initial assumption. Implying that the final temperature does <em>not</em> go above the melting point of water- i.e.,
. However, there's no way for the temperature of the system to go below
; doing so would require the removal of heat from the system which isn't possible under the given circumstance; the ice-water mixture experiences an addition of heat as the hot block of lead was added to the system.
The temperature of the system therefore remains at ; the only macroscopic change in this process is expected to be observed as a slight variation in the ratio between the mass of liquid water and that of the ice in this system.