Question

Calculate the molality of a 1 M solution of sodium nitrate. The density of the solution is $1.25gc{m}^{-3}$.

• A. 0.66 mol $k{g}^{-1}$
• B. 0.86 mol $k{g}^{-1}$
• C. 0.46 mol $k{g}^{-1}$
• D. None of these

Answer 1

Answer: (B)

0.86 mol $k{g}^{-1}$

in this question 1M solution of $NaN{O}_3$,
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 $NaN{O}_3$ for 1000ml of solution.
Therefore, mass of solute is mass of 1 mole of $NaN{O}_3$ = 23+14+48 = 85 gm.
volume of solution = 1000 ml as per definition.
Now,Given that density is 1.25 g/cm3
Therefore, mass = $1.25\times 1000$ (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 = $\frac{1}{1165}\times 1000$ = $\frac{1000}{1165}$

$=0.85m$