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Juli2301 [7.4K]
2 years ago
10

a compound had a molecular weight of 112.124 atomic mass units and the empirical formula C3H4O what is the molecular formula of

the compound?
Chemistry
1 answer:
Snowcat [4.5K]2 years ago
8 0

Answer : The molecular of the compound is, C_6H_8O_2

Solution : Given,

Molecular weight of compound = 112.124 amu

The Empirical formula = C_3H_4O_1

First we have to calculate the empirical formula weight.

The empirical formula weight = 12(3) + 4(1) + 1(16) = 56 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}\\\\n=\frac{112.124}{56}=2

Molecular formula = (C_3H_4O_1)_n=(C_3H_4O_1)_2=C_6H_8O_2

Therefore, the molecular of the compound is, C_6H_8O_2

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Where Na+ = 0.25M

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The total concentration of fluoride ion in the solution is (2x + 0.25M), however, due to common ion effect i.e. 2x<0.25, 2x can be neglected. This means that concentration of fluoride ion will be 0.25M

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Ksp = {x}{0.25}^2

4.0 × 10^-11 = 0.25^2 × x

4.0 × 10^-11 = 0.0625x

x = 4.0 × 10^-11 ÷ 6.25 × 10^-2

x = 4/6.25 × 10^ (-11+2)

x = 0.64 × 10^-9

x = 6.4 × 10^-10

Therefore, the molar solubility of CaF2 in NaF solution is 6.4 × 10^-10M

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