Question

Na2so4 is added in water to make 12% wlw
solution having density 1.2 g/mL. Correct statement
for resulting solution is/are
(a) Nearly 0.96 mole of Na2so4 dissolved per kg
of H20
(b) Normality of Naion is equals to normality of
sulphate ion
(c) Molarity of solution is less than molality
(1) (a) & (6)
(2) (b) only
(3) (c) only
(4) (a) only​

Answer 1

Answer:

(1) (a) & (b)

Explanation:

Given that :

$Na_2SO_4$ is 12 % w/w in solution.

It means that 12 g of the salt is present in 100 g of the solution.

Mass of water = 100 - 12 g = 88 g = 0.088 kg

Mass of solution = 100 g

Density = 1.2 g/mL

Volume of the solution = Mass / Density = 100 / 1.2 mL = 83.33 mL = 0.08333 L

Moles of $Na_2SO_4$ can be calculated as:

Molar mass of $Na_2SO_4$ =  142.04 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{12\ g}{142.04\ g/mol}$

Moles = 0.0844 moles

88 g of water contains 0.0844 moles of salt

1 g of water contains 0.0844/88 moles of salt

1000 g or 1 kg of water contains (0.0844/88)*1000 moles of salt

0.96 moles of the salt is dissolved in 1 kg of water.

Option A is correct.

As the cation and anion are produced from the salt, and same number of equivalents will be produced. So, option B is correct.

Molarity = moles/ Volume of solution = 0.0844 / 0.08333 = 1.01 M

Molality = Moles / Mass of solvent = 0.0844 / 0.088 = 0.9591 m

Option C is incorrect.

Answer 2

Answer:

(3) (c) only

Explanation:

for option A and B see the attached file

option C

molarity is greater than molality