What is the boiling point elevation of a solution of nano3...

Transcript

00:01 So here the concept involved is boiling point elevation and we have been given nan no3 as the solute.

00:09 So nna and no 3 when it is dissolved in water it dissociates into ions completely that is na positive plus no3 negative that is nitrate ions.

00:29 This is sodium nitrate which is dissociated into sodium ion and nitrate ions.

00:34 So here the value for vent off factor that is i value will be equal to two the total number of dissociated ions we have.

00:43 This is 100 % dissociated as we have been given the complete dissociation and hence the value for i will be 2 here.

00:52 Now for boiling point elevation the mathematical representation is delta tb that is this is the elevation in boiling point.

01:05 Point we have is equal to i k b m so i is the vent of factor kb is the molal elevation constant which is specific for solvent to solvent and m is the molality we can further extend the formula delta t b is equal to i k b and instead of molality we can extend this formula molality is the method of determining the concentration and it is the number of moles of solute present in one kilogram of solvent so it can also be written as number of moles of solute that is n0 no n no 3 and n0 3 here divided by mass of the solvent that is water here in this case so this should be in kilogram or we can write delta t v is equal to i kb multiplied n mass of solvent here so instead of n instead of number of moles we can further write mass divided by molar mass that is mass of n nano3 divided by molar mass of nr3 so that is mass of nno3 is represented here by small w divided by molar mass of nr3 and now now the mass of water solvent is given to us in gram here.

02:49 So as i have written that it should be in kilogram...

Question

What is the boiling point elevation of a solution of NaNO3 (85.0 g/mol, complete dissociation) made by dissolving 10.0 g of NaNO3 into 250 g of water (Kb = 0.512°C/m)?

Question

What is the boiling point elevation of a solution of NaNO3 (85.0 g/mol, complete dissociation) made by dissolving 10.0 g of NaNO3 into 250 g of water (Kb = 0.512°C/m)?

Please give Ace some feedback