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:
(a). The given data is as follows.
Volume of one glass of water = ml = 220 ml
Volume of 4 glass of water = = 880 ml
We known that density of water is 1 g/ml. Therefore, calculate the mass of water as follows.
Mass of water =
= 880 gm
= 0.88 Kg (as 1 kg = 1000 g)
The relation between heat energy, mass and temperature change is as follows.
Q =
Putting the given values into the above formula as follows.
Q =
=
= 124.5 kJ
Hence, the body have to supply 124.5 kJ to raise the temperature of the water to 37 degree C.
(b). As we know that the heat of fusion of ice is 333 J/g.
So, energy required for g or 840 g is as follows.
= 279.72 kJ
Heat capacity of water= 4.184
Now, heat energy will be as follows.
Q =
= 130.03 kJ
Therefore, total heat required = (279.72 + 130.03) kJ
= 409.75 kJ
Hence, for the given situation your body should lose 409.75 kJ of heat.