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<span>Molar mass (MM) of benzene C6H6
C = 6 * 12 = 72u
H = 6 * 1 = 6u
MM C6H6 = 72 + 6 = 78 g / mol
Benzene - Molar Mass = 78 g --------- 1 mol
Of A Mix has 468 g -------------- x
78x = 468
X = 468/78
X = 6 moles
Molar mass (MM) of Hydrochloric Acid HCl H = 1 * 1 = 1u
CI = 1 * 35 = 35u
MM HCl = 1 + 35 = 36 g / mol
Hydrochloric Acid - Molar Mass = 36 g ---------- 1 mol
Of A Mix has 72 g ------------ y
36y = 72
Y = 72/36
Y = 2 moles
Thus, a mixture has a total of 8 moles (6 mol + 2 mol).
Dividing One Mole Amount of Each Substance by the Number of Total Mole Amounts, Then we will obtain a Molar Fraction of Each:
Molar fraction make benzene = (6/8) simplify 2 = 3/4
Molar Fraction to make Hydrochloric Acid = (2/8) = simplify 2 = 1/4
Note:. The sum of the molar fractions of the always give goes 1, we have: 3/4 + 1/4 = 1
ANSWER:
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C = 6 * 12 = 72u
H = 6 * 1 = 6u
MM C6H6 = 72 + 6 = 78 g / mol
Benzene - Molar Mass = 78 g --------- 1 mol
Of A Mix has 468 g -------------- x
78x = 468
X = 468/78
X = 6 moles
Molar mass (MM) of Hydrochloric Acid HCl H = 1 * 1 = 1u
CI = 1 * 35 = 35u
MM HCl = 1 + 35 = 36 g / mol
Hydrochloric Acid - Molar Mass = 36 g ---------- 1 mol
Of A Mix has 72 g ------------ y
36y = 72
Y = 72/36
Y = 2 moles
Thus, a mixture has a total of 8 moles (6 mol + 2 mol).
Dividing One Mole Amount of Each Substance by the Number of Total Mole Amounts, Then we will obtain a Molar Fraction of Each:
Molar fraction make benzene = (6/8) simplify 2 = 3/4
Molar Fraction to make Hydrochloric Acid = (2/8) = simplify 2 = 1/4
Note:. The sum of the molar fractions of the always give goes 1, we have: 3/4 + 1/4 = 1
ANSWER:
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I wrote the answer on this paper and here is the calculations step by step
very sry the "V" must be replaced with 285 and the 285 must be replaced with "V" and the answer is 231.09 cm3. sorry for the inconvenience.
very sry the "V" must be replaced with 285 and the 285 must be replaced with "V" and the answer is 231.09 cm3. sorry for the inconvenience.