A student increases the temperature of a 556 cm3 balloon from 278 K to 308 K. Assuming constant pressure, what should the new vo
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The question is incomplete. Here is the complete question.
When 2.10 g of a certain molecular compound X are dissolved in 65.0 g of benzene (C₆H₆), the freezing point of the solution is measured to be 3.5°C. Calculate the molar mass of X. If you need any additional information on benzene, use only what you find in the ALEKS Data resource. Also, be sure your answer has a unit symbol, and is rounded to 2 significant digits.
Answer: MM = 47.30 g/mol.
Explanation: There is a relationship between <u>freezing</u> <u>point</u> <u>depression</u> and <u>molality</u>. With this last one, is possible to calculate <u>molar</u> <u>mass</u> or molar weight of a compound.
<u>Freezing</u> <u>Point</u> <u>Depression</u> occurs when a solute is added to a solvent: the freezing point of the solvent decreases when a non-volatile solute is incremented.
<u>Molality</u> or <u>molal</u> <u>concentration</u> is a quantity of solute dissolved in a certain mass, in kg, of solvent. Its symbol is m and it's defined as
Freezing point depression and molal are related as the following:
where
is freezing point depression of solution
is molal freezing point depression constant
m is molality
Now, to determine molar mass, first, find molality of the mixture:
For benzene, constant is 5.12°C/molal. Then
m = 0.683 molal
Second, knowing the relationship between molal and moles of solute, determine the last one:
mol(solute) = 0.683(0.065)
mol(solute) = 0.044 mol
The definition for <u>Molar</u> <u>mass</u> is the mass in grams of 1 mol of substance:
In the mixture, there are 0.044 moles of X, so its molecular mass is
MM = 47.30 g/mol
The molecular compound X has molecular mass of 47.30 g/mol.
There are three possible Isomers for Molecular Formula C₃H₄. Isomers are;
1) Propyne
2) Propa-1,2-diene
3) Cyclopropene
Structures of all three isomers are given below.
There are no geometrical isomers for given molecular formula because geometrical isomers are formed in compounds containing double bond with at least two bulky groups at both sp² hybridized carbons. In given isomers, propyne does not contain any double bond, Propa-1,2-diene is symmetrical in nature while cyclopropene is a 3-membered ring. And for cycloalkenes there must be other groups attached to ring other than Hydrogen atom, hence cyclopropene does not contain any other group, so it can not show geometrical isomerism.
1) Propyne
2) Propa-1,2-diene
3) Cyclopropene
Structures of all three isomers are given below.
There are no geometrical isomers for given molecular formula because geometrical isomers are formed in compounds containing double bond with at least two bulky groups at both sp² hybridized carbons. In given isomers, propyne does not contain any double bond, Propa-1,2-diene is symmetrical in nature while cyclopropene is a 3-membered ring. And for cycloalkenes there must be other groups attached to ring other than Hydrogen atom, hence cyclopropene does not contain any other group, so it can not show geometrical isomerism.