Question

one litre solutions containing 20g of sucrose is isotonic with 450 ml solution containing 1.65g of boric acid. find the molar mass of boric acid​

Answer 1

Explanation:

Explanation:

The first thing that you need to do here is to use the density of the solution to determine the mass of

150 mL

of this hydrochloric acid solution.

The density of the solution is said to be equal to

1.1 g mL

−

1

, so right from the start, you know that every

1 mL

of solution will have a mass of

1.1 g

.

This means that your sample will have a mass of

150

mL

⋅

1.1 g

1

mL

=

165 g

Now, in order to find the solution's percent concentration by mass,

% m/m

, you need to figure out the mass of hydrochloric acid present in

100 g

of this solution.

Since you know that

165 g

of solution contain

30. g

of hydrochloric acid, you can say that

100 g

of this solution will contain

100

g solution

⋅

30. g HCl

165

g solution

=

18.18 g HCl

You can thus say that the solution has a percent concentration by mass equal to

% m/m = 18% HCl

−−−−−−−−−−−−−−−−−

Answer 2

Isotonic Solutions term refers for two or more than two solutions which have similar concentration or can say that have  similar osmotic pressure. 

Concentration of Sucrose = 20342.6×10001000= 0.05837 mol/L

This concentration must be equal for boric acid So, = 1.63Mw×10004501.63Mw×1000450

                                                Mw = 1.630.05837×1000450 = 62 g

So, the molecular weight of boric acid = 62 g