Answer: The molar mass of the gas is 9.878 g/mol.
Explanation:
According to Graham's law, the rate of diffusion is inversely proportional to square root of molar mass of gas.
where,
M = molar mass of gas
As given gas diffuses 1/7 times faster than hydrogen gas. So, its molar mass is calculated as follows.
where,
= molar mass of hydrogen gas
= molar mass of another given gas
= rate of diffusion of hydrogen
= rate of diffusion of another given gas = 
Substitute the values into above formula as follows.
Thus, we can conclude that the molar mass of the gas is 9.878 g/mol.
The concentration in weight percent when 6 g of sugar is mixed with 9 g of water is 40%.
There are several ways to denote the concentration of a solution like
- Molarity
- Molality
- Mass percent
- Mole Fraction
The formula for calculating mass percent is as follows
Mass per cent = (Mass of solute/Mass of solute + Mass of solvent) x 100%
In the given situation sugar is the solute and water is the solvent.
Putting the given values in the above formula
Mass per cent = (6/6+9) x 100%= 6/15 x 100% = 40%
Hence, the concentration in weight percent when 6 g of sugar is mixed with 9 g of water is 40%.
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Answer:
[C] = 0.4248M
Explanation:
A + B ⇄ 2C
C(i) 1.68M 1.68M 0.00
ΔC -x -x +2x
C(eq) 1.68-x 1.68-x 2x
Keq = [C]²/[A][B] = (2x)²/(1.68 - x)²= 8.98 x 10⁻²
Take SqrRt of both sides => √(2x)²/(1.68 - x)² = √8.98 x 10⁻²
=> 2x/1.68 - x = 0.2895
=> 2x = 0.2895(1.68 - x)
=> 2x = 0.4863 - 0.2895x
=> 2x + 0.2895x = 0.4863
=> 2.2895x = 0.4863
=> x = 0.4863/2.2895 = 0.2124
[C] = 2x = 2(0.2124)M = 0.4248M in 'C'
