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kakasveta [241]
4 years ago
7

How many carbon are contained 25.00 g of sugar (C12H22O11)

Chemistry
1 answer:
zaharov [31]4 years ago
7 0
342 gm sugar = 1 mol of sugar = 12 moles of C. Therefore 25 gm of sugar will form = 25x12/342 = 0.877 moles of C. Therefore number of carbon atoms = number of moles x 6.02x 10^23 = 0.877x 6.02x 10^23 = 5.28x10^23.
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A certain liquid has a normal freezing point of and a freezing point depression constant . Calculate the freezing point of a sol
katrin2010 [14]

The question is incomplete, the complete question is:

A certain liquid X has a normal freezing point of 0.80^oC and a freezing point depression constant K_f=7.82^oC.kg/mol . Calculate the freezing point of a solution made of 81.1 g of iron(III) chloride () dissolved in 850. g of X. Round your answer to significant digits.

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<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m

OR

\text{Freezing point of pure solvent}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}} ......(1)

where,

Freezing point of pure solvent = 0.80^oC

Freezing point of solution = ?^oC

i = Vant Hoff factor = 4 (for iron (III) chloride as 4 ions are produced in the reaction)

K_f = freezing point depression constant = 7.82^oC/m

m_{solute} = Given mass of solute (iron (III) chloride) = 81.1 g

M_{solute} = Molar mass of solute (iron (III) chloride) = 162.2 g/mol

w_{solvent} = Mass of solvent (X) = 850. g

Putting values in equation 1, we get:

0.8-(\text{Freezing point of solution})=4\times 7.82\times \frac{81.1\times 1000}{162.2\times 850}\\\\\text{Freezing point of solution}=[0.8-18.4]^oC\\\\\text{Freezing point of solution}=-17.6^oC

Hence, the freezing point of the solution is -17.6^oC

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