Question

10: The molarity of NaNO, solution is 1 M. The density of solution is 1.25 g/ml. Calculate mos
solution. (Molar Mass of NaNO, = 85 g)
in 85 m) is nresent in 1000 ml of solution.
I​

Answer 1

A = Atomic Number

N = No. of atom

P = Proton

A + N + A = 46

(A + A) + N = 46

2A + N = 46

N = 46 - 2A

According to the question

N = P + 2

N = A + 1

46 - 2A = A + 1

46 - 1 = A + 2A

45 = A + 2A

45 = 3A

A = 45/3

A = 15

Answer 2

We need to find the Molality of the solution. We know that , Molality is defined as ,

• The Number of moles of solute present in per kilogram of solvent is called Molality .
• The Number of moles of solute present in per litre of solution is called Molality .

By Question , its given that the

• Molarity of Sodium Nitrate ( $\sf NaNO_3$ ) is 1 M .
• Density of the solution is 1.25 g/ ml .

• Calculating the mass of the solution :-

$\sf \to Mass = V \times \rho \\\\\sf\to Mass = 1000 ml \times 1.25 g/ml \\\\\sf\to \boxed{\red{\sf Mass = 1250 \ grams }}$

• Now here the molarity was 1 M which signifies that one mole of Sodium Nitrate is present in One Litre of the solution.

$\sf\to Mass_{(solvent)}= Mass_{(sol.)} - Mass_{(solute)}\\\\\sf\to Mass_{(solvent)}= 1250 g - 85 g \\\\\sf\to\boxed{\red{\sf Mass_{(solvent)}= 1165 \ g }}$

$\rule{200}2$

• Calculating the Molality :-

$\sf \to m = \dfrac{n_{(solute)}}{Mass_{(solvent)} (in \ kg ) }\\\\\sf\to m = \dfrac{ 1}{\frac{1165}{1000}}\\\\\sf\to m = \dfrac{ (1000)(1)}{1165} \\\\\sf\to\boxed{\pink{\sf Molality = 0.85 \ m } }$