Question

What is the molality of ammonia in a solution containing 0.85 g of $N{H}_3$ in 100 $c{m}^3$ of a liquid of density $0.85gc{m}^{-3}$?

• A. 0.588 mol $k{g}^{-1}$
• B. 5.88 mol $k{g}^{-1}$
• C. 0.0588 mol $k{g}^{-1}$
• D. None of these

Answer 1

The correct option is A 0.588 mol $k{g}^{-1}$

$\begin{array}{c}\text{Given, mass of}N{H}_3\text{dissolved}=0.85gm\\ \text{So, moles of}{Nn}_3\text{dissolved}=\frac{0.85}{17}mol=0.05mol.\\ \text{Also, volume of solution}=100{cm}^3\\ \text{and, density of Solytion}=0.85gm/{cm}^3\\ \therefore \text{mass of solution}=(100\times 0.85)gm=85gm\\ \text{and, mass of solvent}=\frac{85-0.85kg=0.08415kg}{1000}\end{array}$

$\begin{array}{cc}\text{Hence, molality}=& \frac{\text{mol of solute}}{\text{mass of solvent inkg}}=\frac{0.05}{0.08415}m\\ & =0.59m\text{(or mol/kg)}\end{array}$

Option A is correct answer