When of a certain molecular compound X are dissolved in of benzene , the freezing point of the solution is measured to be . 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 the correct number of significant digits.

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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.

MM = 47.30 g/mol.

Explanation: There is a relationship between freezing point depression and molality. With this last one, is possible to calculate molar mass or molar weight of a compound.

Freezing Point Depression occurs when a solute is added to a solvent: the freezing point of the solvent decreases when a non-volatile solute is incremented.

Molality or molal concentration 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 Molar mass 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.

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when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml