Question

atomic mass of Al2(SO4)3 is 342 gm/mole. in which of the following option has 342 grams of soluble substances.
a. 1 molar,1 liter solution
b. 2 molar,1 liter solution
c. 3 molar,2 liter solution
d. 3 normal,1000 gram water..

Answer 1

As much as I understand the question.

Understanding :

Soluble substance = Al2(SO4)3

We have to check the mass of soluble substance is 342g or not.
I. e. Al2(SO4)3

That is, No. of moles of solute = 342 / 342 = 1 mole

So, we will check for options in which no. of moles of solute is 1 .

Formula :

1) Molarity = No. of moles of Solute / Volume of solution
--------(1)
-------------------------------------
1) Molarity = 1 M

Volume of Solution = 1 L

By using equation(1),

No. of moles of solute = 1M * 1L
= 1 mole

Yes, this option is correct.

2) Similarly,

No. of moles of solute = 2*1 = 2moles

implies this is incorrect.

3) Similarly,

No. of moles of solute = 3* 2= 6 moles

4) This is bit different option

Normality of solute = 3 N
No. of equivalents of Solute = Total cationic charge = 2*3 = 6

No. of moles of solute =
Normality / No. of equivalents = 3N/ 6 = 1/2

Here too No. of moles is 1/2 which is incorrect.

$$$$Danger:$$$$$$

I had written the solution according to my understanding , so there may be an error.

If you are not satisfied, Report this answer freely.

Answer 2

Molar mass of Al₂(SO₄)₃ = 342 g/mol
given mass of solute { Al₂(SO₄)₃} = 342g
so, mole of solute = given weight/molar mass = 342/342 = 1 mol
Assume Volume of solution = V L
and Molarity of Al₂(SO₄)₃ = M

We know, according formula of molarity , Molarity = mole of solute /volume of solution in Litre
So, M = 1/V , this is the expression help to find solution,.
Check by putting each option ,
You get only option (1) is satisfied .

Hence, option (1) is correct.

[4th option is little difference, but it's okay , normality = n × molarity . Here n - factors = 6 so, molarity = 1/2 and mass of water = 1000g so, volume of water = 1000mL = 1L ,it's not satisfied ]