Chemistry, 14.07.2019 09:00 john67728

Astudent dissolves 6.2g of aniline c6h5nh2 in 350.ml of a solvent with a density of 1.04/gml . the student notices that the volume of the solvent does not change when the aniline dissolves in it. calculate the molarity and molality of the student's solution

Answers: 1

Show answers

Answers

Answer from: jasmine77723

Answer : The correct answer for molarity = 0.19 $\frac{mol}{L}$ and Molality = 0.18 $\frac{mol}{Kg}$ .

Given : Mass of aniline = 6.2 g

Volume of solvent = 350 mL Density of solvent = 1.04 g/mL

1) Molarity : It is defined as number of moles of solute present in Litre of solution . It is expressed as :

$Molarity(M) = \frac{moles of solute(mole)}{volume of solution (L)}$ \

Molairty can be found in following steps :

Step 1 : To calculate mole :

Mole of solute can be calculate using mole formula :

$Mole of solute = \frac{given mass of solute (g)}{molar mass of solute\frac{g}{1 mol}}$

Molar mass of aniline (C₆H₅NH₂) = 93.13 $\frac{g}{1 mol}$

Mass of aniline = 6.2 g

Plugging values in mole formula :

$Mole = \frac{6.2 g}{93.13 \frac{g}{1 mol}}$

Mole = 0.0665 mol

Step 2 : To find volume of solution

Volume of solution = volume of solute + volume of solution

Since addition of aniline does not change final volume , so volume of solvent = volume of solution. Since volume is given in mL , so it need to be converted to L .

1 L = 1000mL

$Volume of solution = \frac{350 mL}{1000mL} * 1 L$

Volume of solution = 0.350 L

Step 3: Plug value of mole and volume in molarity formula :

$Molarity = \frac{0.0665 mol}{0.350 L }$

Molarity = 0.19 M or 0.19 $\frac{mol}{L}$

------------------------------------------------------------------------------------------

2) Molality : It is defined as mole of solute present in Kilogram of solvent . It can be expressed as :

$Molality (m) = \frac{mole of solute (mol)}{Kilogram of solvent (Kg)}$

Following are the steps to calculate molality :

Step 1: To find mole of Solute :

Mole of solute can be found out using mole formula . It is same as done for molarity .

Mole = 0.0665 mol

Step 2 : To find kilogram of solvent :

Mass of solvent can be calculated using density formula as :

$Density \frac{g}{mL} = \frac{mass (g) }{volume (mL)}$

Plugging value in density formula :

$1.04 \frac{g}{mL} = \frac{ mass }{350 mL}$

Multiplying both side by 350 mL

$1.04 \frac{g}{mL} * 350 mL = \frac{x}{350 mL } * 350 mL$

Mass of solvent = 364 g

Since mass is in g, it need to be converted to Kg . ( 1 Kg = 1000 g )

$Mass of solvent = \frac{364 g}{1000g} * 1 Kg$

Mass of solvent = 0.364 Kg

Step 3: Plug values of mole and Kg in molality formula :

$Molality = \frac{0.0665 mol}{0.364 Kg}$

Molality = 0.18 m or 0.18 $\frac{mol}{Kg}$

Answer from: Quest

No is the answer to the question

Answer from: Quest

The correct answer is c. orbital shape. "the angular momentum quantum number l describes the shape of an orbital. it corresponds to each orbital type, i.e. (0,1,2,−1)=(s,p,d,f,g,"

Another question on Chemistry

Chemistry, 21.06.2019 21:30

Which statements are true about electrolysis? check all that apply. electrolysis requires an acid be present. electrolysis is described by two half-reactions. electrolysis is not an industrial process. electrolysis results in commercially valuable products. electrolysis involves the transfer of electrons. reduction results in the loss of electrons. oxidation results in the loss of electrons.

Answers: 1

Answer

Chemistry, 22.06.2019 01:30

Agas is contained in a thick walled balloon when the pressure changes from 1.21 atm to 2.52 the volume changes from 3.75 l to 1.72 l and the temperature change from 293k to blank k

Answers: 3

Answer

Chemistry, 22.06.2019 12:50

Is heavenly controller dating someone?

Answers: 1

Answer

Chemistry, 22.06.2019 19:30

To calculate percent by mass, use the equation below: calculate the percent by mass of each element. %n = % %h = % %o = %

Answers: 3

Answer

You know the right answer?

Astudent dissolves 6.2g of aniline c6h5nh2 in 350.ml of a solvent with a density of 1.04/gml . the s...