Question

what molality of a resultant solution prepared by mixing 50 ml each of 2.5 molal and 3.5 molal H2 s o4​

Answer 1

Explanation:

 in this question 1M solution of NaNO3,

By definition if Molarity we know that it is the no. of moles present in per litre of solution. = 1M molar solution has 1 mole of NaNO3 for 1000ml of solution.

Therefore, mass of solute is mass of 1 mole of NaNO3 = 23+14+48 = 85 gm.

                  volume of solution = 1000 ml as per definition.

Now,Given that density is 1.25 g/cm3

Therefore, mass = 1.25×1000 (m = density X volume)

=> mass = 1250 gm (solution mass)

Mass of solvent = mass of solution - mass of solute

                          = 1250 - 85= 1165gm.

Molarity is the no. of mole of solute per kg of solvent

=> m = 11651×1000 = 11651000 

=0.85m