Question

3. Calculate the molarity of an NH3 aqueous solution made up of 30.0 g of NH3 and 70.0 g of water at 25 °C. The density of water at that temperature is 0.982 g mL−1. M(NH3) = 17.034 g mol−1 A. 17.3 mol L−1 B. 24.7 mol L−1 C. 5.78 × 10−2 mol L−1 D. 0.578 mol L−1 E. 1.73 mol L−1

Answer 1

Answer:  E. 1.73 mol/L

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in ml

moles of $NH_3$ (solute) = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{30.0g}{17g/mol}=1.76$

mass of water (solvent )= 70.0 g

mass of solution = moles of $NH_3$ (solute) + mass of water (solvent ) = (30.0 + 70.0) g= 100.0 g

density of solution = 0.982 g/ml

volume of solution = $\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100.0g}{0.982g/ml}=101.8ml$

Now put all the given values in the formula of molality, we get

$Molarity=\frac{1.76\times 1000}{101.8ml}=1.73M$

Therefore, the molarity of solution is 1.73 mol/L