A 850.0 mL bottle of Listerine is of a 24 % (v/v) ethyl alcohol. If the density of ethyl alcohol is 0.789 g/mL and the molar mass is 46.07 g/mol, calculate the molarity of ethyl alcohol in Listerine.

Explanation:

The given data is as follows.

       Volume of solution = 850.0 ml

    Volume percentage of ethyl alcohol = 24%

24%(v/v) solution of alcohol means that 24 ml solute of ethyl alcohol is present per 100 ml of solution.

Hence, volume of alcohol present in 850 ml solution is calculated as follows.

       =

       = 204 ml ethyl alcohol

Now, we will calculate the number of moles of ethyl alcohol as follows.

         No. of moles =

                               = 3.49 mol

Volume of solution 850 ml or 0.850 L (as 1 L = 1000 ml).

Concentration of ethyl alcohol will be calculated as follows.

             =

             = 4.10 mol/L

             = 4.10 M

Therefore, molarity of ethyl alcohol in Listerine is 4.10 M.

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied